We’ll start with a simple problem involving two sets, from 1996:
How Many are in the Group?
Mr. Jackson polled the class to see how many students had been to one or both of the nearby state parks. He found that everyone in Katie's group had been to at least one of the parks. When he asked how many had been to Punta de las Cuevas, four children raised their hands. When he asked how many had been to Salt Bay, three raised their hands. When he asked how many had been to both state parks, two raised their hands. How many people were in Katie's group?
This is a very straightforward example, and two of us answered it. First, Doctor Dennis:
Questions like this are often very easy using Venn Diagrams. Imagine you have a big rectangle. In that big rectangle are two circles which are partially overlapping. Now, say that all the students at the table are represented by the stuff within the rectangle. Call one circle people who have gone to Punta de las Cuevas and the other circle people who have gone to Salt Bay.
Now, since two have been to both parks there are two students in the area that is the intersection of the circles. The Salt Bay circle should have 3 people in it, but there are already two people in the part of the circle that intersects with the other circle, so there is just one person in the part of the Salt Bay circle that is not within the other circle. Similarly, there will be 2 people in the Punta de las Cuevas circle that are not in the part that overlaps the Salt Bay circle. so there are 2 + 2 + 1 = 5 kids total.
This is typical of many more complicated problems, in that the key idea is subtraction. We know the whole of a set, and one part, so we subtract to find the other part.
Simultaneously, Doctor Keith answered, without using a Venn diagram, showing that this is just a tool for organizing information, not an essential part of the work:
This is an example of everyone's favorite type of problem, the word problem. The real trick to word problems is trying to turn the words into equations. Let's summarize what we have in a table for easy reference: Number of students Place who have been there Punta de las Cuevas 4 Salt Bay 3 Both 2 Neither 0 We have several ways of looking at this but probably the easiest is a logical argument. The two students who have been to both parks would have raised their hands for both Punta de las Cuevas and Salt Bay. Thus two of the students in the Punta de las Cuevas group are also in the Salt Bay group. So let's find out how many have been to only one. For Punta de las Cuevas, four students have been but two also went to Salt Bay, so we have to remove them (remember we want students who have only been to Punta de las Cuevas). 4  2 = 2 no. of students who have only been to Punta de las Cuevas Thus only two students have only been to Punta de las Cuevas. Similarly, for Salt Bay we have three students who have gone, but two also went to Punta de las Cuevas so we must remove those two from the three who have been. 3  2 = 1 no. of students who have only been to Salt Bay Thus only one student has gone to only Salt Bay. To get the final answer we note that the sum of the students who have only been to Punta de las Cuevas, plus those who have only been to Salt Bay, plus those who have been to both must be the total in the group. So: 2 + 1 + 2 = 5 Thus there are 5 students in the group.
You may notice that the work here is identical; it just took more words, in the absence of pictures!
He went on to repeat the work by giving names to the students, showing the benefit of using math and ignoring such details! Math can often be best learned by starting with more concrete images (an actual group of individual children), and gradually reducing that to mere numbers as one understands relationships better.
Let’s look at one more example with two sets, this one using a table rather than a Venn diagram. This is from 2009:
Alternate Solution Path to Venn Diagram Problem
There are 30 students in a math class. Twelve belong to the computer club, and eight belong to the photography club. Three belong to both clubs. How many belong to neither club? I know that the correct answer (according to the book) is 13. I can't come up with that answer. I have either 10 or 7 as my answers, not knowing if the "both club" members are included already in the listed number of members. This question is listed in a section introducing proportions. I can't figure out a way to come up with the correct answer or use a proportion to solve it. Sometimes when I know the correct answer I can work backwards to solve it, but I can't do that with this problem. It is driving me CRAZY!
Doctor Ian responded:
This sort of problem is often solved using a Venn diagram. You can read about them at these locations: Wolfram Mathworld: Venn Diagram http://mathworld.wolfram.com/VennDiagram.html Venn Diagram: Fast Food Restaurants http://mathforum.org/library/drmath/view/60772.html But here's another way to organize the data you have and solve the problem. Suppose we make a table like this: photography ~photography +++ computer    +++ ~computer    +++ (I'm using "~" as an abbreviation for "not".) Can we agree that everyone in the class must be in EXACTLY one of these boxes? We're told that there are 30 students. So the numbers in the boxes have to add up to 30. We're told that 3 students belong to both clubs: photography ~photography +++ computer  3   +++ ~computer    +++ Now, if 12 students belong to the computer club, and we've already accounted for 3 of them, there must be 9 who are in the computer club, but NOT the photography club, right? photography ~photography +++ computer  3  9  < These have to +++ add up to 12 ~computer    +++ Can you use similar reasoning to fill in the lower left box, given that 8 students belong to the photography club? And once you know that, you know how many belong to neither club. Does this make sense?
After this start, Autumn was able to finish; she wrote back:
Thank you so much. This makes perfect sense and now I feel silly. I was so stuck on the numbers that I couldn't think beyond to get over the hump. Great visual and clues! I appreciate your help.
I’d like to add one more thing that I find helpful with this kind of table (which is also useful in probability problems). I include a column and a row that hold totals, so we can put all the information in the table. Here is my table, with only the given information:
photography ~photography total +++ computer  3   12 +++ ~computer    +++
total 8 30
After filling in the rest, by subtraction, we have
photography ~photography total +++ computer  3  9  12 +++ ~computer  5  13  18 +++
total 8 22 30
And there’s our answer: 13. (I’m still curious why this problem appeared in a section about proportions; possibly they had used tables of proportional values, and this was given as a contrast.)
Here is a question from 2017 that serves as a good introduction to the more complicated 3set case:
Some Drank All; Did All Drink Some?
At a meeting, 35 teachers were offered Fanta, Coke, and Pepsi. Fifteen drank Fanta, 6 drank both Fanta and Coke, 18 drank Coke, 8 drank both Coke and Pepsi, 20 drank Pepsi, and 2 drank all three beverages. How many of the teachers drank (i) Coke only? (ii) Fanta and Pepsi but not Coke? I tried a little, but didn't get the answers. I have just forgotten how to solve these kinds of problems.
I started by making a Venn diagram, in which each circle represents all the people who drink a given kind of soda:
I usually start these problems by drawing a Venn diagram, and labeling what I can. We need three sets, Fanta (F), Coke (C), and Pepsi (P):
I also like to make a table so I can see all the information clearly: Fanta: 15 Fanta and Coke: 6 Coke: 18 Coke and Pepsi: 8 Pepsi: 20 all three: 2 When it says 15 teachers drank Fanta, that means Fanta AND possibly other refreshments. So that refers to the entire circle for Fanta, not just the region of the diagram that represents those who drank ONLY Fanta. Similarly, "Fanta and Coke" doesn't mean they didn't also drink Pepsi; it includes both the region for ONLY Fanta and Coke AND the "all three" region in the middle. While the middle is all that we can fill in immediately, I'll also indicate each set's total next to its label:
(I didn’t mention it at the time, because I was making ASCIIgraphic “circles” for which this is awkward, but when I do these problems on paper, especially some that are more complicated than this one, I like to label as much of the given data as I can, so I use arrows to represent the intersections of two circles, which will help in the next step.)
But now we can take each of the facts about TWO sets, and use subtraction to fill in other regions. For example, since 6 drank both Fanta and Coke, but 2 of those ALSO drank Pepsi, there must be (6  2 =) 4 who drank ONLY Fanta and Coke:
See if you can finish, by filling in the other two "double" regions, and then again using subtraction to fill in the three "only one" regions. Then you can use the diagram to answer the questions. In trickier problems of this sort, more than just subtraction may be needed; but this problem is the straightforward kind, where you are given exactly the facts needed to make the diagram.
In preparing these pictures, I have realized that I missed an important feature of this problem: it doesn’t say how many drink Fanta and Pepsi, so we can’t continue quite as I said — this is one of the “trickier problems”, but I didn’t see that. What we can do is to label the missing intersection with a variable, and fill in everything else we can.
Now we realize that we can only proceed if we assume that everyone drank something. (The headline writer apparently noticed this!) If so, then the sum of the seven regions in the diagram is 35, and we have this equation to solve:
\((9x)+4+6+x+2+6+(12x) = 35\)
Solving it, we find that \(x=4\). Looking back at the questions, the first is already answered: the number who drank Coke only is the blue region, 6. The answer to the second question (how many drank Fanta and Pepsi but not Coke, is our \(x\): 4.
The typical basic problem would have included the number who drank Fanta and Pepsi, from which we would have more directly filled in everything, including how many drank nothing.
For an example of the basic kind (but with an algebraic method suggested), see
Venn Diagram  Fast Food Restaurants
Here is an example like mine above where a variable is needed:
Venn Diagram  Choose One of Three Options
A key idea in the wording of these problems is that when they mention those who choose, say, Coke and Pepsi, they include those who choose all three. Sometimes, like the last link I gave, the wording is done very carefully. Other times, it can be unclear, as in this 2002 question:
Solving Questions
I got this challenging question in my Math course: In a poll of 37 students, 16 felt confident solving quantitative comparison questions, 20 felt confident solving multiple choice questions, and 18 felt confident solving gridded response questions. Of all the students, 4 were confident solving gridded response and quantitative comparison questions, 5 were confident solving multiple choice and quantitative comparison questions, and 6 were confident with gridded response and multiple choice questions. Of those students, only 1 felt confident with all three types of questions. What number of students felt confident solving only multiple choice questions and no others? I tried to solve this riddle, and I got 10. My teacher told me that the answer is actually 8. I just can't figure this out. Would you please tell me which one is the right answer, and explain to me how to solve such questions?
I started by pointing out the potential problem, and the conventional interpretation:
I am always uneasy with these puzzles, because they are never worded clearly enough for me. Which numbers are meant to be inclusive? The first three numbers are clearly inclusive (that is, 16 like AT LEAST QC, but possibly others), since otherwise the answer would be given directly. Then the next three numbers seem likely to be inclusive as well, so that 4 like both GR and QC, AND possibly also MC. That implication is further supported by the wording "of those students [the 4, 5, and 6?], only 1 ...." So my assumption, probably like yours, is that the 1 is included in each of those numbers, so that there are 3, 4, and 5 who like ONLY two types of questions. That makes it easy to fill in a Venn diagram, and we quickly see that the circle for MC, whose total size is 20, has 1 shared with both QC and GR, 4 with QC alone, and 5 with GR, leaving 20(1+4+5) = 10 for MC alone. So I agree with your answer...
... except that if we then add up the total, we find that 40 students like something, out of 37 students in all! So our assumptions do not fit all the facts. Looking for something to reinterpret, I consider the possibility that the single student who liked everything might NOT be included in the 4, 5, and 6. Taking it that way, the number for MC alone is 20(1+5+6) = 8, agreeing with your teacher's answer.
(And this time, the sum is 37 — meaning that every student felt competent in something.)
I think the problem is very poorly stated. You might like to try rewording it to make everything clear. That's a good exercise in communication, which mathematicians, pollsters, and puzzlemakers should pay attention to as well! (Except when a puzzle is deliberately vague to mislead you, which this one should not be.)
For a similar issue, see
Venn Diagram: Two Possibilities
Next time, we’ll look at a more complicated case.
]]>Sometimes we give only minimal help on a question, to let a student puzzle over the problem and learn to figure it out herself. That may be enough, or we may go back and forth for some time, guiding her thinking until she finds the answer. In the latter case, do we put the whole discussion in our archive, so the answer is given away along with all the thinking? Or do we archive only hints? Recently a student wrote to ask about the answer to a puzzle we didn’t work to the end, and the discussion gives some interesting insights into the process — so here come the spoilers.
First, here is the question from our archive, asked by 12yearold Subby in 2015:
All in a Row: Divisible Except for One Consecutive Pair In a class, 25 students were lined up. The teacher wrote a number on the board. The first student said the number was divisible by 1. Student number 2 said it was divisible by 2. Student 3 said it was divisible by 3. This went on until the 25th student. The teacher said everyone was correct except two boys, named Ben and Danny, who were wrong. They were standing one after the other in the line. What are their positions in the line? Not sure if I have to find a common factor between 1 and 25, then divide the found number by each number between 2 and 25 to see who has divided correctly. I think I am not approaching this right.
Doctor Greenie answered this, restating the problem and giving some hints:
This is a terrific problem  quite challenging, I would think, for a 12yearold. According to the problem description, there are two consecutive numbers from 1 to 25 which do not evenly divide the teacher's number; and they are the only two numbers that do not divide the teacher's number. The key to solving the problem is to look at divisibility rules for the numbers 1 to 25. For a couple of examples: 25 = 5^2 since the only prime factor of 25 is 5, we can't relate divisibility by 25 to divisibility by any smaller numbers 24 = (2^3)*3 = 8*3 the number is divisible by 24 if it is divisible by both 8 and 3 23 is prime we can't relate divisibility by 23 to divisibility by any numbers smaller than 23 22 = 2*11 the number is divisible by 22 if it is divisible by both 2 and 11 Continue this list down to 1; there will only be one place where it is not possible to relate the divisibility rules for two consecutive numbers to divisibility rules for smaller numbers. Those two consecutive numbers are Ben's and Danny's places in the line.
The hint was (intentionally, I assume) not very detailed; a year later, another student wrote to ask about it, focusing too much on the rules for recognizing when a number is divisible by another, such as that “a number is divisible by 3 if the sum of its digits is divisible by 3”. That type of rule was not what Doctor Greenie emphasized; but rather, others, like “a number is divisible by 6 if it is divisible by both 3 and 2”, which is a particular case of a very general rule. I responded with some more specific hints:
I think you are misinterpreting what Doctor Greenie said. I imagine he deliberately kept his hint rather cryptic to avoid spoiling it for others. (For that matter, I'm not sure I follow his hint completely, myself!) What I would say is that, rather than get caught up in the details of the divisibility rules, you want to consider just the relationships that some of them imply  the rules like yours for 6. This rule implies two things: 1. If 6 does NOT evenly divide the number, then either 2 or 3 must also NOT be a divisor (since if both did divide it, 6 would also); since 2 or 3 and 6 are not consecutive, 6 can't be one of the two nondivisors. 2. If either 2 or 3 does NOT evenly divide the number, then 6 must also NOT be a divisor (since if 6 were a divisor, then both 2 and 3 would be too); so neither 2 nor 3 can be one of the two nondivisors. Look for this sort of relationship among other sets of numbers. You might need to make a table or diagram showing these connections, or it may be enough to cross off from a list each number that is excluded. You should soon get a feel for what kind of numbers one of the pair has to be, and then discover what the other one must be.
Doctor Greenie also responded, saying he had trouble understanding his own hint (which is not uncommon, when we look back at our own old answers!), and adding more:
The input that Doctor Peterson gave is important. Where you were looking at the divisibility rules in isolation, the thing you want to be looking at is the relationship between the divisibility rules. For example, a number is divisible by 24 if it is divisible by both 3 and 8; a number is divisible by 14 if it is divisible by both 2 and 7. Note in that first example, it is important to say a number is divisible by 24 if and only if it is divisible by both 8 and 3. It would not be sufficient to say it is divisible by 24 if it is divisible by 4 and 6. 4 times 6 is 24; but 4 and 6 have a common factor of 2. 12 is divisible by 4 and 6; but it is not divisible by 24. So if 24 were one of the two numbers that did not divide the teacher's number, then it would not be divisible by 3 or 8 either. But then you have numbers that are neither consecutive nor divisors of the teacher's number. So 24 can't be one of the two consecutive nondivisors. What about 23? It is prime, so having a number divisible by 23 does not imply divisibility by any smaller number. There is a possibility that 23 is one of the two consecutive nondivisors. But we already know that 24 can't be one of the two nondivisors; and divisibility by 22 implies divisibility by both 2 and 11. So even though 23 had the possibility of being one of the two nondivisors, each of its neighbors can't be the other one. Similarly, we can rule out 22, 21, and 20 as being one of the two consecutive nondivisors, because for each of them divisibility implies divisibility by smaller numbers that are not adjacent: nondivisibility by 22 > nondivisibility by 2 or 11 nondivisibility by 21 > nondivisibility by 3 or 7 nondivisibility by 20 > nondivisibility by 4 or 5 Continue down the list to find the only place where nondivisibility by two consecutive numbers does *not* imply nondivisibility by any smaller numbers. In the process of trying to provide an explanation that could help you solve the problem, I came up with this alternative method, which you might (or might not!) find easier. Start by finding the smallest number (in prime factorization form) of the number that is divisible by all the numbers from 1 to 25. Then search for ways to remove certain prime factors so that two consecutive integers become nondivisors.
Now, this summer, Sarah got interested in solving this, and wrote to us for help:
I found this in the Ask Dr.Math archives, and got curious what the solution is: http://mathforum.org/library/drmath/view/77746.html. After reading the hints, I thought the obvious answer is 2 and 3, since they’re both prime? Is this correct? And how would you find the actual number the teacher wrote? (Out of curiosity and as a check)
We went back and forth on this for four days (partly due to being in distant time zones). I first corrected her attempt:
No, the answer can’t be 2 and 3; as I said in my answer,
If either 2 or 3 does NOT evenly divide the number, then 6 must also NOT be a divisor (since if 6 were a divisor, then both 2 and 3 would be too); so neither 2 nor 3 can be one of the two nondivisors.
What we know is that every integer from 1 through 25 is a divisor of N except for two consecutive numbers. If nothing else, you could just try every pair from 1,2 to 24,25 and think about the implications; along the way, you might get some special insight. But I will tell you that my immediate thought (which could be wrong) is that the pair of nondivisors might more likely be large than small. The 2,3 case shows why, and this may be why Dr. Greenie started his list at 25.
Give it another try. I’ll be thinking about it while you do. (I don’t immediately recall what the answer was.)
She gave it a try that didn’t work, and I responded with further clarification:
Let’s look more closely.
We want to find two consecutive numbers such that our unknown number N can be a multiple of every number from 1 to 25 except those two.
Suppose the answer were 24 and 25. That would mean that N is a multiple of everything up to 23, but not of 24 or 25.
First, consider 25 = 5^2. If N were a multiple of 25, then it would also be a multiple of 5; that doesn’t matter (just now). But if N is not a multiple of 25, it can still be a multiple of 5 (e.g. 35), so there is no problem. None of the other numbers is forced to be a divisor of N by this fact. We might call 25 a “safe” number.
But consider 24 = 2^3 * 3. If N were a multiple of 24, then it would also be a multiple of any divisor of 24 (1, 2, 3, 4, 6, 8, 12). But what matters to us at the moment is the implication if N is not a multiple of 24.
We can split 24 into 3*8, where 3 and 8 have no common factors; then we know that if N is a multiple of both 3 and 8, it must be a multiple of 24. Do you see why? This is fundamental to divisibility checks, and would not be true if we chose a pair of numbers with a common factor, like 4 and 6: 12 is a multiple of both, but not of 24.
Now, the contrapositive of that statement in bold is, if N is not a multiple of 24, then it must not be a multiple of both 3 and 8; that is, it must either not be a multiple of 3, or not be a multiple of 8. But we were supposing that 24 and 25 are the only numbers from 1 to 24 of which N is not a multiple. So 24 can’t be one of the numbers we’re looking for. We might say that 24 is not a “safe” number.
Now, I mentioned some facts that we don’t care about yet (with 24 or 25); they will come into play when we look at smaller numbers. For instance, if N were not a multiple of 2, then it couldn’t be a multiple of any multiple of 2 either. That would give us a lot of other numbers from 1 to 25 that N was not a multiple of, so 2 can’t be one of our numbers: it is not “safe”.
What I did to solve the problem was to look individually at each number from 1 to 25 and do thinking like I just did: what other facts would be implied if N were not a multiple of a given n? If n not being a divisor implies that any other numbers than n1 and n+1 are also not divisors, then n can’t be it. Eventually, you should be able to speed up a bit, by noticing certain kinds of numbers that are “safe”. I’m deliberately avoiding some words that would give you too much of a hint!
My introduction of a madeup term, “safe” (by which I meant that it doesn’t “endanger” other numbers), is a common trick of mathematicians to make it easier to talk about a concept that is useful in a particular problem. I think of this as a “local definition”. A common word that is used is “nice”, which in ordinary usage has only a vague positive meaning, and so, like tofu, can be given whatever specific flavor you want. Here, it lets me state the goal quickly: We want to find two consecutive safe numbers.
Sarah gave me a lot of detail on her next try:
I understand what you said, but I’m not sure I know how to continue. I was probably mixing up factors and multiples, even though I know the difference between the two. I’m guessing that by certain kinds of numbers you meant the primes, but thanks for not giving too many hints; it’s good to do some thinking!
Let me try:
If N is not a multiple of 23 – 23 is prime, so it’s “safe”
If N is not a multiple of 22 – it can’t be a multiple of both 2 and 11 so it’s not “safe”
If N is not a multiple of 21 – it can’t be a multiple of both 3 and 7, so its not “safe”
If N is not a multiple of 20 – it can’t be a multiple of both 4 and 5, or both 2 and 10, so it’s not “safe”
If N is not a multiple of 19 – 19 is prime, so it’s “safe”
If N is not a multiple of 18 – it can’t be a multiple of both 3 and 6, or both 2 and 9, so it’s not “safe”
If N is not a multiple of 17 – 17 is prime, so it’s “safe”
If N is not a multiple of 16 – it can’t be a multiple of both 2 and 8, so it’s not “safe”
If N is not a multiple of 15 – it can’t be a multiple of both 3 and 5, so it’s not “safe”
If N is not a multiple of 14 – it can’t be a multiple of both 2 and 7, and 28 can’t be a multiple, so it’s not “safe”
If N is not a multiple of 13 – 26 can’t be a multiple either
If N is not a multiple of 12 – it can’t be a multiple of both 4 and 6, and 24 can’t be a multiple, so it’s not “safe”
If N is not a multiple of 11 – it can’t be a multiple of 22
If N is not a multiple of 10 – it can’t be a multiple of both 2 and 5
If N is not a multiple of 9 – then 18 cannot be a multiple, and N can’t be a multiple of 3
If N is not a multiple of 8 – it can’t be a multiple of both 2 and 4
If N is not a multiple of 7 – 14 can’t be a multiple, and neither can 21
If N is not a multiple of 6 – it can’t be a multiple of both 2 and 3
If N is not a multiple of 5 – 10, 15, 20, and 25 can’t be multiples either
If N is not a multiple of 4 – 8, 12, 16, 20, and 24 can’t be multiples either
We’ve already ruled out 2 and 3, and 1 obviously can’t be, since every number is divisible by 1.
She was very close now. My reply:
Primes were one of two kinds of numbers I could have mentioned. Certainly they are always “safe”. (Or not? Comment below.)
So you’ve listed these numbers as “safe”: (By the way, I first used that term just on a whim as I was typing, but I now see how good an idea it is!)
17, 19, 23
I’d also listed 25 as safe.
I see that for 14 and 13, you mentioned numbers greater than 25; you might want to reconsider. That will be a significant change in the way you have to think once you get to smaller numbers. But you were still thinking, because you saw that 11, though prime, is not safe, because 22 is less than 25. So we should say that primes greater than 13 are safe.
The main problem is that you didn’t take to heart my comment about the importance of “factors with common factors”. For 18, you say “it can’t be a multiple of both 3 and 6, or both 2 and 9, so it’s not “safe”. That’s correct for 2 and 9, which are relatively prime (I can use that word now), but not for 3 and 6. Since these have a common factor, 3, a number can be a multiple of both but not be a multiple of their product: 12 is a multiple of 3 and of 6, but not of 18. So you can’t say that if N is not a multiple of 18, then it can’t be a multiple of both 3 and 6. (This was part of what Dr. Greenie had in mind about divisibility rules.)
The main thing you need to do is to work through things a little more slowly, and perhaps do as I had not done and write explicit statements like “primes greater than 13 are safe” so you can check them out. But I’ll give you one more specific hint: you know why that latter fact is true (well, check it out!); what is it that makes 25 safe? It isn’t prime! Is it just being the highest number in the list, or is there more to it?
Sarah’s next attempt:
This is very interesting! I also like the idea of defining safe numbers!
My first thought about 25 is that it is safe because it is a square number, so I’m thinking the solution is 16 (square number) and 17 (prime greater than 13). Let’s check it out:
16 is 2 * 8, and 2 is a common factor, so I can’t say that if N is not a multiple of 16, it can’t be a multiple of both 2 and 8. (In this case, I can’t find a number that is a multiple of both 2 and 8, but not of 16). If N is not a multiple of 16, it can be a multiple of 4 though.
We’ve already said 17 is a safe number.
I’ve just reread what you wrote about 25, and realised that you wrote “consider 25 = 5^2”, so it was there in front of me! Nice hint
But let’s check another example, to be sure:
9 – if N is not a multiple of 9, it can be a multiple of 3 (example 6)
So think the answer is 16 and 17.
Also, I don’t think 18 is safe, since it can’t be a multiple of both 2 and 9. Otherwise, the answer would have been there already – 17 and 18.
The reference to 25 = 5^2 was not intended as a hint; I was showing the factors of everything. But it did make the key idea visible.
I answered:
Elaborating on what you said about 16, the general idea is to look at the prime factorization of a number, in this case 2^4. With only one prime factor, there is no pair of factors (other than 1 and 16) that are relatively prime; so the only conclusion regarding smaller numbers is that if 16 is a divisor of N, then every lower power of 2 is a divisor of N (2, 4, 8). Given that 16 is not a divisor of N, we can’t conclude that any lower number must not be a divisor of N. Also, since 2*16> 25, we can’t conclude that any larger number must not be a divisor of N. Thus we can conclude that 16 is safe. (You said, “I can’t find a number that is a multiple of both 2 and 8, but not of 16”; did you try 8?)
Your answer is correct.
Now, notice how much more has to be said than I said explicitly last time:
We define a safe number as a positive integer n such that, if n is not a divisor of some number N, then no other number in {1, 2, .., 25} is necessarily not a divisor of N.
Whether a number is safe depends on two things: whether there are any smaller numbers that are forced not to be divisors, and whether there are any larger numbers up to 25 that are forced not to be divisors. These work in different ways, and must both be considered.
No number less than 13 can be safe, because doubling it will yield a number less than 26.
Any prime n greater than or equal to 13 is safe, because no smaller number (except 1) is a factor of n, and no larger number through 25 is a multiple of n.
Any prime power n greater than or equal to 13 is safe, because there are no pairs of relatively prime numbers that are both factors of n.
After this, we discussed a similar problem I had run across with some interesting differences. But this post is already over my selfimposed size limit …
]]>A couple weeks ago, in discussing the value of estimates, I included one example of a (very simple) Fermi problem: one in which it is necessary to invent the data as well as the method of solution. Today, I will examine one answer in which we dug deeper into a much more elaborate question, and see what it says about estimation.
Here is the initial question, from Kenneth in 2002:
Estimation and Fermi Questions I'm currently learning about Estimation techniques similar to those used by the famous scientist Enrico Fermi, who proposed the question, "How many piano tuners are there in Chicago?" The question I invented was the following: If there are 12,000 students who attend a certain college, how many professors are employed by the college?
For the benefit of future readers, I gave several references, of which the only one that is still accessible is this nice introduction to Fermi Problems, by Austin Gleeson via Eric Smith, including the classic Piano Tuner Problem. (Today, of course, you can also just look it up in Wikipedia.) Quoting from that page,
Fermi was uncontestably one of the most important research physicists of this century, and a great many of the working tools of the modern physicist were invented by him. He was also for many years a professor at the University of Chicago, who had a reputation for asking his students outrageous and seemingly impossible questions, and then showing them that they had the necessary knowledge and tools to answer them, which is why this kind of problem came to be named Fermi problems.
The point of Fermi problems is in part to show how much use can be made of commonly available knowledge by the person willing to be resourceful and make approximate simple calculations, but it is more to illustrate the difference between estimation and guessing. The fabled canonical Fermi problem was the question “how many piano tuners are there in Chicago?”. Faced with such a question without warning in a physics lecture hall, one response would simply be to declare that you don’t know and cannot know, and if forced to produce an answer to simply guess at one that is “plausible”. The primary disadvantage of this guessing is not that it yields imprecise answers, because any question answered with incomplete or notwell measured information necessarily yields imprecise answers. And in fact, even guesses about common experiences are often plausible, precisely because they are constrained by an intuitive touch with reality. One would not even guess that the number of piano tuners in Chicago is something comparable to the number of people living in Chicago, because that would violate the experience that most people one meets are not piano tuners.
The problem with guessing is that one does not know how much confidence to place in the answer, because the constraints from which it follows have not been clearly identified, and so even if estimations have been made at an intuitive level, the degree of imprecision in the estimations has no way to be identified.
Kenneth went on to show his own work on the question, for which he guessed both what information would be useful to solve it, and what values would be reasonable:
I came up with the following estimation. Can you tell me if my reasoning is reasonable? Thanks very much! 1) The average professor teaches about 1 hour a day, so in one week (Monday  Friday) he teaches 5 hours. 2) Each class takes about an hour, so in one week he teaches 5 classes. 3) But he doesn't teach 5 different classes in one week; the same classes are held 23 times a week (either MondayWednesdayFriday classes or TuesdayThursday classes). To make it simpler, let's say a professor teaches each class 2 times a week (assume only TuesdayThursday classes exist). Therefore, he sees the same class 2 times a week, meaning every half week he sees the same students, but that also means every week he only sees the same students because the classes repeat. 4) If a professor teaches for 2.5 hours per half week (5 hours per week), where each class takes about an hour, and assuming that a typical class consists of 50 students, then he sees 2.5 classes x 50 students = 125 students per week. 5) Since there are 12,000 students for all professors to lecture, then there are probably 12,000 / 125 = close to 100 professors on campus. From a scale of 110, how would you rate this estimation? Any suggestions or comments are greatly appreciated.
There are, of course, two ways to judge his work: the logic, and the data. At the time, I focused only on the former, which is appropriate in our context; today, having become an adjunct professor at a community college two years after this question, I have to laugh at the numbers he used. Of course, colleges differ, and he is imagining one at the other end of the spectrum from mine, probably a major university. Our fulltime professors typically teach at least 5 courses (34 hours each), not just 5 hours, per week, because our emphasis is on teaching rather than research; we have around 12,000 students (depending on how you count them), with 295 fulltime and 710 parttime instructors according to the website. Looking at the other end of the scale, I find that Yale University, near where I grew up, also has about 12,000 students, with 4,410 faculty members (not differentiated as full or parttime, but including many who don’t teach). Clearly Kenneth has underestimated; we’ll see at the end if I got it any better!
Looking only at the reasoning, I immediately saw a gap:
It seems to me that you have left out one important factor: each student attends more than one class. This can be tricky to describe clearly: in your model each class has 50 students, and each student has, say, 5 classes. So is it 50 students per class, or 5 classes per student? One way to avoid this trouble is to think of a specific name for the relation of a student to a class. One I've thought of is "seat." Each class has 50 seats (students in the class); each student has 5 seats (in different classes). You can diagram this: Student > Seat < Class < Prof 1 5 50 1 2.5 1
I’ll explain the source of this diagram later; I think the idea of “seats” is standard in describing a college, as “bed” is in a hospital. (The terms aren’t meant to dehumanize, but to make it possible to talk about such ideas!) But as I’ve said elsewhere, a key to problemsolving is to find a good representation so that you can see relationships, and this is it.
Kenneth asked for clarification:
Each student has 5 seats. There are 50 seats in one class. But does that mean each class holds only 10 students? I'm not sure I understand this concept. And I thought that the total number of students (12,000) attending the college would matter in the estimation. Can you explain this further?
To which I replied, showing the actual work:
Maybe you can get a better understanding if you try to tell me what role the number of classes each student takes should play in your estimate. It takes a bit of wrestling on your own before you can quite pin this idea down. Obviously I didn't say that each class has only 10 students; but if each student took only one class of ten students, you would need the same number of professors, so in a sense it is equivalent to that. Or, instead of 12,000 students taking 5 courses each, you could have 60,000 students taking one course each; that would require the same number of professors, which is five times as many as you estimated. Does that help? And I didn't say the number of students doesn't matter; my diagram (a variety of Entity Relationhip Diagram) only shows relative numbers, not absolute numbers. It says that for each student there will be 5 seats (in different classes), each of which is 1/50 of a class, each of which needs 1/2.5 of a professor, so 12,000 students will need 5 seats 1 class 1 professor 12,000 students *  *  *  = 480 professors 1 student 50 seats 2.5 classes There's a lot of useful math and logic in this question!
So there’s my answer (based on his assumptions). It’s actually not too far off the number for my own school; if I change the assumptions to 25 students per class and 5 classes per professor, according to my own experience, we get the very same result!
5 seats 1 class 1 professor 12,000 students *  *  *  = 480 professors 1 student 25 seats 5 classes
That’s a bit low if I consider an adjunct equivalent to a little less than half a fulltimer, so that we have the equivalent of 650 faculty. But it’s not bad for a rough estimate.
On the other hand, the estimate is extremely low for Yale. I suspect that if we drop nonteaching faculty (listed as “research”) and count parttime faculty appropriately, the number may be more like 2000 rather than 4000, but it’s still a lot more than 480. To increase the number, we’d need smaller classes and fewer classes per professor; dropping it to 1 class each, we get 1200, which is still low. I imagine some are counted as teaching faculty even though they only work with a couple individual students. But that’s not the main point of our discussion!
Kenneth had a last question: How does one learn to think this way?
Thanks very much, I understand your solution. One last general question on how I can sort of think the way you do, so to speak. What you said about how I missed a link between the students and the class, where we're not sure if each student has 5 classes or if each class has 50 students, makes me ponder about how you realized and picked that up. Have you worked with problems such as these repeatedly, so you know all the nitty gritty details? Or is it because I'm illogical or have less ability to reason that I totally neglected that fact. If so, is there a particular math course I can take to help open up my mind to incorporate more reason and logic into my mathematical thinking? I feel incompetent when I see other students being able to answer such questions while I'm struggling. Can you give some advice?
I had a little special knowledge, but mostly just experience:
An interesting question! I suspect a lot comes from experience  and that experience is probably what Fermi was trying to develop with his questions. Logical ability is not just something you are born with (though it may be that some of us naturally gravitate toward it, and therefore develop the skills without having to be forced into it); I think everyone has to develop it by practice. In this case, as with the piano tuner problem, a lot of the thinking needed depends on specific knowledge of the subject matter. You have to picture a university and know that there are classes and professors and so on, or picture a piano tuner's job and see that he will do more than one piano a day, and they will be in different homes, and so on. In this case, I just thought about what factors would play a role, saw that each student would be in several classes, and expected to see that somewhere in your analysis. When I didn't, a red flag went up. No logic, just visualization. But I probably would have noticed it anyway, by going through your presentation in order and falling off the end when you didn't mention what each student does. Maybe you can call that "followthrough"  you can't stop your logic just because you've made contact with the goal, but have to keep thinking until you can't think any further. You did fine until then; you just didn't take it all the way. Assuming that "problemdomain" knowledge, you have to be able to think through the problem, both in a straight line (each professor teaches N classes; in each class there are N students; ...), and also sometimes coming at it from all sides, just brainstorming to think of all the relevant factors. The former requires the ability to stay focused and think in an orderly way; the latter requires defocusing and letting wild ideas come in. Both have their place, and some of us are probably better at one than the other. In this case, to find my own answer, I just went through it sequentially, starting at the student (since I knew the hard part was at that end). Now I happen to have an advantage over you in doing a Fermi problem, one that I haven't seen mentioned in discussions of them. I am a computer programmer, and part of my work involves designing relational databases, where you might have one table listing all the students, another listing all the classes, and so on. That's where my Entity Relationship Diagram came from  it's a tool used to see how these tables relate to one another, and design additional tables that capture the information needed for these relations. While I was thinking about how to explain the "5 classes per student and 50 students per class" problem, that method popped into my mind  I suppose that is an example of the nonlinear type of thinking, pulling a tool out of my toolbox because the kind of thinking I was doing reminded me of it, not because it was the next thing to think of. That's not essential for this kind of problem, but it can always be handy. Speaking of databases, if I had actually been trying to design one, I would have based my thinking on the paperwork involved in a university. Each of the 12,000 students has a course schedule, listing his or her 5 subjects. Each of those 60,000 subject lines corresponds to one seat in one class; 50 of them together form one of 1200 classes. Each professor has a schedule listing 2 or 3 courses, among which the 1200 classes will be found. In the database, you would add a table containing the information from all the students' schedules, which relates students to classes. (That's why it's called a "relational" database.) This image of actual pieces of paper can often make the abstract ideas of numerical ratios more concrete, and help in logical thinking. Where can you build these skills? Many math classes will incorporate them implicitly, in geometric proofs or word problems, for example. Other fields need it too, from Fermi's physics to law school. There are all sorts of puzzles you can find in books or elsewhere.
The great thing about Fermi problems is that they are challenging puzzles that make you bring together all sorts of ideas. We have discussed a number of such questions over the years, not always using the term “Fermi problem”. Here are a few pages of interest:
Estimation in 3rd Grade (Doctor Tom, 1996, listing Fermi problems) Estimating Seating Capacity (Doctor Jesse, 2000) Billions Taste the Rainbow Thousands of Billions of Billions of Times (Doctor Ian, 2014)]]>
Two of these methods, Binary Search and Newton’s Method, were discussed in their more general form last week when we discussed numerical methods of approximation. I also showed a student’s discovery of linear interpolation for finding square roots.
The simplest thing to do is just to guess a square root, check it by squaring, and then make a new guess, higher or lower, based on the new information you have, but still within known bounds. This is not mere random guessing, but orderly guessing!
The key idea is that if \(a^2 < n < b^2\), then \(a < \sqrt{n} < b\). For example, since 20 is between 16 (\(4^2\)) and 25 (\(5^2\)), its square root must be between 4 and 5.
Here are the steps to estimate the square root of a number n:
Example 1: Estimate the square root of 683 to the nearest tenth by guessandcheck.
We first look for two consecutive perfect squares that 683 is between:
We know that \(20^2=400\) and \(30^2=900\), so \(\sqrt{683}\) must be between 20 and 30.
Try 25: \(25^2=625\) which is less than 683, so the square root is greater than 25.
Try 26: \(26^2=676\) which is less than 683, so the square root is greater than 26.
Try 27: \(27^2=729\) which is greater than 683, so \(\sqrt{683}\) is between 26 and 27.
Now we look for the tenths that \(\sqrt{683}\) lies between:
Since 683 is considerably closer to \(26^2\) than to \(27^2\), we should start on the low side.
Try 26.2: \(26.2^2=686.44\), which is high, so our square root is a little lower than 26.2.
Try 26.1: \(26.1^2=681.21\), which is smaller than 683, so we know that \(\sqrt{683}\) is between 26.1 and 26.2.
To make the final decision which is closer, we can check the number halfway between, 26.15:
\(26.15^2=683.8225\) which is high, so we know that \(\sqrt{683}\) is closer to 26.1.
Here is one way to record our work, listing tens, then ones, then tenths and marking which our root is between:
The circled number is our answer. A calculator shows that \(\sqrt{683}=26.134268…\), so this is a good estimate.
The main difficulty with this method is that it may take several tries to get each decimal place; but the only arithmetic we have to do is multiplication, which can be done by hand without much trouble.
This sort of trialanderror, and the closely related method of binary search, are described here:
What is the Square Root of 8? (Doctor Tom, 1996) Finding square roots (Doctor Dave, 1996; Doctor Jerry, next method) Method to Find Square Root of Pi (Doctor Steve, 1997: binary search)
This ancient method is similar, in that we make a series of approximations, but this time we don’t merely guess; a formula tells us what number to use next. It is based on the fact that if \(ab=n\) and \(a\) is less than \(\sqrt{n}\), then \(b\) has to be greater than \(\sqrt{n}\). So if we start with a guess \(x\) that is on one side of \(\sqrt{n}\), then \(\frac{n}{x}\) will be on the other side of \(\sqrt{n}\), and their average will be closer to \(\sqrt{n}\). Here is the method:
Example 2: Estimate the square root of 683 to the nearest tenth by divideandaverage.
Since, as before, \(\sqrt{683}\) is between 20 and 30, let’s start with 25.
The next guess is \(\displaystyle\frac{25+\frac{683}{25}}{2}=\frac{25+27.32}{2}=26.16\).
The next is \(\displaystyle\frac{26.16+\frac{683}{26.16}}{2}=\frac{26.16+26.109}{2}=26.135\). (I am rounding to 3 decimal places.)
These are the same in the tenths place, so our answer is 26.1.
Here is one way to record our work:
The trouble with this method is that we have to divide by numbers with many decimal places, rather than merely multiply. To do it by hand, you have to decide ahead of time how many decimal places to use — I recommend rounding to two places beyond your goal, as I did above (three places). On the other hand, this method improves its estimate very fast, doubling the number of correct digits at each step. If your original guess has 1 digit correct, in two steps you will have about 4 correct digits, as in this example.
Another problem is that this method theoretically keeps going forever, even if the answer can be written exactly; it gets closer to the exact value, but can never reach it. Commonly, however, it will stop because of rounding, and you can tell that you have the exact answer. Here is an example:
Example 3: Estimate the square root of 68.89 to the nearest tenth.
Since 68.89 is between 64 and 81, let’s start with 8.
The method is actually a special case of Newton’s method for approximating a zero of any function, applied to the function \(f(x) = x^2 – n\). Newton’s method is based on calculus; from a first guess, it makes a succession of guesses calculated as \(\displaystyle x_1 = x_0 – \frac{f(x_0)}{f'(x_0)}\), where \(f’\) is the derivative of \(f\). In our case, \(f'(x) = 2x\), so the formula gives \(\displaystyle x_1 = x_0 – \frac{x_0^2 – n}{2x_0} = \frac{x_0^2 + n}{2x_0} = \frac{x_0 + \frac{n}{x_0}}{2}\).
This method was described here:
Square roots by hand: Divide and Average (Doctor Jerry, 1996) Square Roots: Estimate, Divide, Average (Doctor Mitteldorf, 1997, also longhand) Newton's Method and Square Roots (Doctor Jerry, 1998, Newton's method) Heron's Method for Finding Square Roots by Hand (Doctor Paul, 2001, also guessandcheck)
It is extended to higher roots here:
Cube roots (Doctor Jerry, 1997) Calculating 4th Roots, 5th Roots... (Doctor Jerry, 1999) Formula for Finding Mth Root (Doctor Rob, 2001; Doctor Jacques, 2004)
It is also the primary method taught in our FAQ on the topic:
Square Roots Without a Calculator
This method is very different from the others. It looks like long division, and gives us the answer exactly, one digit at a time. However, it requires skill to guess each digit efficiently. Here is the method:
Example 4: Calculate the square root of 683 to the nearest tenth by longhand.
Start at the decimal point and make pairs of digits (including one pair of zeros to the right to make room for tenths in the answer).  
The first digit is 2 because \(2^2\) is the greatest perfect square less than 6.
Subtract \(2^2=4\) from 2, and bring down the next pair. We’re working on 283. Double the root we have so far, 2, to get 4, and write a blank next to it. 

Our goal is to fill that blank so that \(4\_\cdot\_\le 283\), as close as possible.
For a first guess, we can divide 283 by 40, which gives about 7. Trying 7 gives \(4\underline{7}\cdot \underline{7} = 329\), which is too big. For a next try, divide 283 by 47, which is about 6 (or just subtract 1 from the first guess). We find that \(4\underline{6}\cdot 6 = 276\), which is just a bit less than 283, so we use 6. Write 6 in the answer, and subtract 276 from 283. Bring down the next pair. Double the root we have so far, 26, to get 52, and write a blank next to it. 

Now we want to fill this blank so that \(52\_\cdot\_\le 700\).
The obvious answer is 1: \(52\underline{1}\cdot \underline{1} = 521\). Put 1 in the answer (after the decimal point), and subtract 521 from 700. We now have a square root accurate to the tenths place; if we want to round to the nearest tenth, we can test to see whether the next digit is 5 or more: \(522\underline{5}\cdot \underline{5} = 26125\) is too large, so the next digit is less than 5 and we don’t have to round up. 
This method is based on the fact that \((10a+b)^2 = 100a^2 + 20ab + b^2 = (10a)^2 + (20a+b)b\). For a detailed explanation of the reasoning, see
Finding Square Roots (Doctor Peterson, 1998)
Example 5: Calculate the square root of 68.89 exactly. (Compare Example 3 above.)
This method tells you directly when the answer is exact, by giving a remainder of 0.
This method is described here (in addition to the fuller explanation cited above):
Square Roots Without a Calculator (Doctor Robert, 1996) Longhand Square Roots (Doctor Rob, 1998)
Here is the corresponding way to find cube roots, and a detailed explanation of why it works:
Cube Root by Hand (Doctor Rob, 1998) Cube Root Calculation, Explained (Doctor Peterson, 2002)
A fourth method, described as Newton’s Method but really linear interpolation, is described here:
Manual Method for Finding Square Roots (Doctor Ethan, 1995)
Yet another ancient method is discussed here, showing how it is related to Newton’s method or Divideandaverage:
Bakhshali Formula (Doctor Peterson, 2002)
Which method is “best”? It depends on what you are looking for. Let’s explore the question by using each method for another, somewhat larger problem.
Example 6: Use each method to find the square root of 5678 to two decimal places.
(a) Guessandcheck:
This took ten multiplications (because I mostly guessed well).
(b) Divideandaverage:
This took 6 divisions (including dividing by 2), most of them easy but two quite hard to do by hand.
(c) Longhand:
This took three multiplications and three fairly easy divisions, assuming I guessed everything right on the first try.
(d) Calculator: \(\sqrt{5678} = 75.3525049… \approx 75.35\)
Clearly they all give the same answer; see which one fits your style best. I find that, although we tend to push the last two, which seem more “mathematical”, the guessandcheck method may often involve less work for actual hand calculations. All are capable of identifying when an answer is exact.
]]>In many areas of math, an answer can come in several forms, which can make it hard to know if you are right when you compare your answer to the answer in the back of the book. Even worse is when the problem is multiplechoice, and your answer has a different form than the choices given. This can happen very easily in trigonometry. What do you do?
This happened to a student recently, and he wrote to us to help unravel it. Here is his question:
Here is the problem I am working on:
The general solution of the equation \(\tan^2\alpha + 2\sqrt{3} \tan\alpha = 1\) is given by …
(A) \(\alpha = \frac{n\pi}{2} (n\in I)\)
(B) \(\alpha = (2n+1)\frac{\pi}{2} (n\in I)\)
(C) \(\alpha = (6n+1)\frac{\pi}{12} (n\in I)\)
(D) \(\alpha = \frac{n\pi}{12} (n\in I)\)I have tried solving it this way:
Firstly I applied the Quadratic Formula to get,
\(\tan(\alpha) = (2+\sqrt{3}) \text{ or } (\sqrt{3}+2)\)
Now we have two cases,
CASE 1: When \(\tan(\alpha) =\tan(\pi/12) = 2\sqrt{3}\).
So General Formula here will be, \(\alpha =n\pi + \pi/12\).
Now, CASE 2: When \(\tan(\alpha) =\tan(5\pi/12) = (2+\sqrt{3})\).
So General Formula here will be \(\alpha = n\pi – 5\pi/12\).
I do not know what should I do next to get the answer? Please tell me how to proceed further.
The answer given in the key is the option (C).
I will be thankful for any help!
He’s done good work, solving the quadratic equation to find two possible values for \(\tan(\alpha)\), and then finding all values of \(\alpha\) for each case. The difficulty is that the choices are given in ways that blend all solutions into a single formula using an integer parameter, while Navneet has two separate formulas, neither of which looks quite like any of the choices, though there is a clear similarity. How do you blend them? Or is there another way to decide?
After checking his work, I answered:
You have the solution; now you just have to put it in the desired form.
For this sort of multiplechoice problem, all you need to do is to compare your solution to each choice. Your solution, written as a pair of lists, is (taking n=0, 1, 2, …)
α = π/12, 13π/12, 25π/12, … or 5π/12, 7π/12, 19π/12, …
That is, putting these together,
α = …, 5π/12, π/12, 7π/12, 13π/12, 19π/12, 25π/12, …
There are two directions to go from here. Either do what I just did with each of the choices, and find which one matches; or observe that successive solutions actually differ by 6π/12 = π/2, so that the solution in simpler form is a single arithmetic sequence,
α = π/12 + n π/2.
Now you can see that two of the choices have π/12 factored out, so you can do that to your solution:
α = π/12 (1 + 6n)
and we have the answer.
Of course, if this were not multiplechoice, your original answer would have been acceptable, though you might want to do the same work I did here just to find a somewhat neater form.
The fact that the question was multiplechoice provided a hint at the desired form (a single combined formula), which would not have been present in an openended question (where we would have faced the similar issue of determining whether the answer in the book agreed with ours). This is similar to the difference between a problem that asks us to simplify an expression (which might have many “simpler” forms, all equally valid in the absence of specific instructions), versus one that asks us to prove a trig identity (where we are in essence told what the answer is, and just have to get there). Seeing the form of the choices, we had some ideas for directions to take our work.
(We could instead have used the basic strategy for multiplechoice problems, not actually solving at all but just checking each proposed solution. A and D both make 0 a solution; clearly that doesn’t work in the equation. B makes π/2 a solution, which is wrong (the tangent is undefined). So only C is left. But I can’t stand doing that; it works against the whole idea of learning, testing an entirely different skill than what is supposed to be learned. A betterdesigned question might at least make this method harder, and force you to do a little actual math.)
Four days later, the same student asked a related question that was not in a multiplechoice context and did not involve merging two sets of solutions, but was actually more interesting:
I am again facing a problem in solving a question of this type.
See question number 3 in the attachments:
Problem 3:
Solve the equality: \(2 \sin 11x + \cos 3x + \sqrt{3}\sin 3x = 0\)
Answer:
\(\displaystyle x = \frac{n\pi}{7} – \frac{\pi}{84} \text{ or } x = \frac{n\pi}{4} + \frac{7\pi}{48},\ n \in I\)
This is my solution:
One of the solutions is matching with the solution given in my textbook but the other solution is not matching.
Why isn’t my solution matching with the solution given in my textbook?
I will be thankful for help!
This is an openended problem, where our difficulty is not in choosing an answer, but in deciding whether we are right. Navneet’s first answer should agree with the book’s second, but looks quite different. In fact, if we try choosing a value for n and seeing whether the answers agree, it looks at first as if it must be wrong.
I had to think a bit before I answered:
At first I thought this might be an actual arithmetic error, but then I saw that it is just a more complicated version of the previous kind of issue.
First, I was impressed by the way you wrote a single expression initially for all angles whose sine is the same as for 11x; but then I saw that you turned that into what I would have written first, giving two separate cases (for n odd and even). I presume the first form is one you were taught, that I have not seen.
All your work is excellent, and very clearly written.
But your answer seems different from theirs in the odd case. Why? To see what they probably did, try replacing 2n+1 with 2n1 in your work, which is equally valid, and see what you get!
It turns out that they, again, just made a different choice than you, which led to their n having a different meaning in that case. Looking at the problematic half of the answer, they have n pi/4 + 7 pi/48 [= (12n + 7)pi/48], while you have n pi/4 – 5 pi/48 [= (12n 5)pi/48]. Here are the values for some values of n:
n theirs yours
2 17pi/48 19pi/48
1 5pi/48 7pi/48
0 7pi/48 5pi/48
1 19pi/48 17pi/48
2 31pi/48 29pi/48These are, in fact, the same solutions, but in reverse order and starting at different places! As before, doing this at least can reassure you that your solution is correct, even though it has a different form.
When you do what I suggested above, you should see why.
Interestingly, I recently wrote a blog post in part about a similar issue that occurs in calculus, where two solutions can look entirely different, and the difference turns out to be absorbed in an arbitrary constant. Here, the difference is “absorbed” by an index that can start in different places or move in different directions.
Let’s “do what I suggested above” to see what I meant. Here is Navneet’s work, using \(2n+1\):
\(3x+\pi/6 = (2n+1)\pi + 11x\)
\(3x+\pi/6 = 2n\pi +\pi + 11x\)
\(3x – 11x = 2n\pi + \pi – \pi/6\)
\(8x = 2n\pi + 5\pi/6\)
\(\displaystyle x = \frac{n\pi}{4} – \frac{5\pi}{48}\)
Here is my work, using \(2n1\):
\(3x+\pi/6 = (2n1)\pi + 11x\)
\(3x+\pi/6 = 2n\pi \pi + 11x\)
\(3x – 11x = 2n\pi – \pi – \pi/6\)
\(8x = 2n\pi – 7\pi/6\)
\(\displaystyle x = \frac{n\pi}{4} + \frac{7\pi}{48}\)
This is their solution!
As we can see, when the book’s answer looks different from ours, we can (a) “spot check” some values to see that it doesn’t really disagree; (b) write both answers in a common form, such as a list of values, to see that they actually agree; or (c) take the form of their answer as an addendum to the question, suggesting the form we should try, and proceed in that direction. With experience, we can learn how an answer can be disguised (here, by reversing the direction of the index n), and not be too surprised.
Or, we can ask the Math Doctors!
]]>Last time, as part of our series on estimation, we looked at some numerical methods for solving equations approximately. I mentioned the Method of False Position, but when I looked for more detailed expositions in our archive, I realized that in a sense it is really two different things, and we primarily cover the form that is not a method of approximation. Today I will focus on two answers from the archive, one related to each aspect of the method, in order to dig more deeply into both.
This method goes back to the ancient Egyptians or Babylonians; the name comes from the fact that we start by supposing (positing) a solution (though we know it is false), and using the information obtained to get a new answer — which, for some (linear) problems, will in fact be the exact answer, but for others (nonlinear) will just be a better approximation. In the latter form, it is similar to the Bisection Method I discussed last time, but rather than getting a mere “up or down” indication from the function, we have more information, so we can move faster.
The ancient form of the method (for linear problems) came up in this question from 2004:
The Method of False Position There is a quantity such that 2/3 of it, 1/2 of it, and 1/7 of it added together becomes 33. What is the quantity? Solve the problem by the method of false position. I know that using the method of false position, we are looking for the root of an equation, and need a way of making a guess that is better than our previous guess.
This is a classic problem of the Egyptian type, though Lynne didn’t mention such a source; it may be a direct translation from an Egyptian original, or a problem from the 19th century (see below), which explains its awkward wording. If you think about it, you will realize that it is really a simple algebra problem, from our perspective. The hard part is that it is written in terms of a sum of fractions; this is the only way the Egyptians could write fractions, as their notation only allowed for “numerators” of 1, apart from a special case for 2/3 or 3/4. Today, we would just combine the fractions and multiply by the reciprocal. There is no need for a numerical approximation, and in fact this method quickly gives the exact answer for such a problem.
Doctor Douglas answered:
Let's let 'x' be the unknown quantity, and we can write the following equation: x*(2/3) + x*(1/2) + x*(1/7) = 33 Assume x = 42. This is a convenient first guess because it is divisible by all of the denominators {3,2,7}, which makes our life a bit easier on the first step. 42*(2/3) + 42*(1/2) + 42*(1/7) = 28 + 21 + 6 = 55. So our first guess of 42 is approximately too big by a factor of 55/33 = 5/3. Our next guess is therefore 42*(3/5) = 126/5 or 25.2. We plug this, our second guess, in for x in the equation above to obtain (126/5)*(2/3) + (126/5)(1/2) + (126/5)(1/7) = 84/5 + 63/5 + 18/5 = 165/5 = 33 which is exactly what we want it to be, and we've therefore found the value of x for which the equation is true. There is of course a more direct method to solve the original equation using algebra techniques, but I think that this problem shows you how the method of false position works so that you can also apply it to cases where you cannot simply solve for x.
Both Lynne and Doctor Douglas appear to know False Position primarily as an approximation method, despite this being a classic problem for the linear form of the method. I added a reference to the following page, which I had written the previous year, about its (presumed) Egyptian origin and significance:
The Egyptians' Method of False Position
What we have used here is called “Simple False Position”, and applies specifically to problems of the form \(ax = b\) (direct proportion). In it, we make one guess, and adjust the guess proportionally to the error in the result. As I see it, what was done above could just as well be described as multiplying by a common denominator to simplify the equation, and then dividing by the remaining coefficient. If so, then it is not all that different from modern methods. Their main difficulty was the lack of notation, which forced them to say everything in words.
The more general method, used to solve equations of the form \(ax + b = c\), is called “Double False Position”, and is the precursor of the full approximation method we’ll look at below. In it, we make two guesses, and use a complicated formula to find the new “guess” lying between them, which for a linear equation will be the final answer. I’ll discuss this formula after we see it in an example of approximation.
The best discussion I find of this is in the following question from 2006, where Jacob essentially discovered the core of the method himself:
A Way to Estimate the Square Root of Any Number Today in math class, we learned about finding the square root of certain numbers, such as the square root of 36 = 6. I found a way to find the approximate square root of any number, and I need to know why it works. Example: the square root of 63. To get the square root of 63 first you take the nearest two perfect squares, 49 and 64. Take the square roots of those numbers, which are 7 and 8. Set it up like this: 49 63 64 7 ? 8 Next we need to make a fraction. Subtract 49 from 64 to get your denominator, 15. To get your numerator you subtract the first two numbers, 63  49 = 14. That leaves you with 14 over 15 or .93. Since the square root of 63 is higher than 7 but less than 8 your whole number is 7. Add .93 to 7 and get 7.93, close to the real square root of 63. In my math class we always question why. So, why does this work?
I love his original title for the question: “Wow! I have found it, now why does it work?” I think he is in a wonderful class.
I explained what he had done in terms of the graph:
What you've discovered is called "linear interpolation". It amounts to approximating a point on a graph by using a straight line that is close to the graph. Draw the graph of the square root, y = sqrt(x). It passes through the points (49,7) and (64,8), since the square roots of 49 and 64 are 7 and 8 respectively. Between them, the graph is slightly curved. If you draw a line between those two points, you'll find that the line is quite close to the graph of the square root. Here's a closeup of the line: 8  o /  o  /  1 / h  /   7  o++    49 63 64 \_________/ 6349=14 \_______________/ 6449=15 Since 63 is 14/15 of the way from 49 to 64, h is 14/15 of the way from 7 to 8. That means the value of y, when x is 63, is 7 14/15. That's your estimate of the square root. This method is often used for quick estimates; back before calculators, we were taught to use it to find values from trigonometric tables when we needed extra accuracy.
I was not satisfied with the rough “graph” I’d made, so I sent a better one:
Here's a picture of the full graph that I talked about in case you'd like to see it, with the triangle I drew in red. You can see that the straight line is very close to the actual square root graph: Did you discover this on your own, or did you have some hints that led you in this direction? It's a very nice thing to have found!
Jacob replied,
I discovered it on my own and my teacher was amazed at the discovery. She had never heard of figuring it out in that way. Thanks for your reply. That makes sense.
Now, as I said, what he did was a common way to make an estimate from a table (it’s still used in countries where tables are used for trigonometry or logarithms); Jacob just did it once, so it is not yet an iterative method. Let’s look a little deeper to see how this could be turned into (a) a formula for Double False Position, and then (b) a method for iterative approximation.
First, what is the formula? In looking for the square root of 63, Jacob first made two “false guesses”, 7 and 8, knowing that \(7^2 < 63 < 8^2\), and then calculated that 63 is 14/15 of the way from 49 to 64, so that (if the curve were a straight line), x should be 14/15 of the way from 7 to 8. So, in general, to find the square root of n, we make two guesses a and b, and estimate that x should be \(\displaystyle a + \frac{n – a^2}{b^2 – a^2}\cdot (b – a)\).
More generally, suppose we were searching for a zero of the function \(f(x)\), and we find two numbers \(x_1\) and \(x_2\) that bracket the zero we want. Let’s call \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\). Then the xintercept is \(0 – y_1\) out of \(y_2 – y_1\) of the way from the first point to the second vertically, so our approximation will be \(\displaystyle \frac{y_1}{y_2 – y_1}\) of the way from \(x_1\) to \(x_2\). Our approximation is therefore \(\displaystyle x_1 + \frac{y_1}{y_2 – y_1}\cdot (x_2 – x_1) = \frac{x_1 y_2 – x_1 y_1 – y_1 x_2 + y_1 x_1}{y_2 – y_1} = \frac{x_1 y_2 – y_1 x_2}{y_2 – y_1}\). This is our formula. If the goal is not zero, the y‘s are taken to be errors (difference from the goal).
Ancient people (starting at least with the Chinese, Indian, and Islamic worlds, and, in fact, all the way into the 19th century!) used this as a rote method for solving linear problems by using two guesses; this was called Double False Position, or just Double Position. (The word “position”, here, means something like “supposition”: we are twice “positing” a possible but false solution.) In my cursory examination of old books discussing this, I see some literally crossmultiplying (arranging the guesses x and the errors y at the ends of an X), some using something like our unsimplified formula instead. Some modified the rule when the two guesses were on the same side of the solution, rather than use signed numbers. Here is one example from 1827:
Here’s another, from 1836:
A course of mathematics, Charles Hutton
It seems incredible today that such a cumbersome method should have been commonly taught; but algebra was considered too advanced for most students. This is just arithmetic!
Again, these versions of the method apply specifically to linear problems, and will give an exact answer that can be checked. But if we apply it repeatedly to a nonlinear problem, we will get a better approximation at each step. Let’s take Jacob’s example and do that.
His work, again, used the formula
\(\displaystyle a + \frac{n – a^2}{b^2 – a^2}\cdot (b – a)\);
applied to his example with n = 63, a = 7, b = 8, we had
\(\displaystyle 7 + \frac{63 – 7^2}{8^2 – 7^2}\cdot (8 – 7) = 7 + \frac{14}{15} = 7.9333\).
Checking, we find that \(7.9333^2 = 62.9372\), which is a little low, so the solution lies between 7.9333 and 8. Repeating using these two guesses, we have
\(\displaystyle 7.9333 + \frac{63 – 7.9333^2}{8^2 – 7.9333^2}\cdot (8 – 7.9333) = 7.9333 + \frac{63 – 62.9372}{64 – 62.9372}\cdot (8 – 7.9333)\) \(\displaystyle = 7.9333 + \frac{0.0628}{1.0628}\cdot (0.0628) = 7.9333 + .0037 = 7.9370\)
as our new approximation; checking, we see that \(7.9370^2 = 62.996\). That is quite close to 63. (A calculator gives the value 7.9372539… .)
More generally, we would repeat our formula \(\displaystyle x_{n+1} = \frac{x_{n1} y_{n} – y_{n1} x_{n}}{y_{n} – y_{n1}}\), or any equivalent form, and then pair this with the last guess for which y had the opposite sign.
This is, in essence, the method of False Position as used today as an approximation method. There are a number of variations on it.
Next week, I will discuss finding square roots by other methods that are more useful by hand.
]]>First, an overview, based on a question from Joanne in 2005:
How to Solve Equations with No Analytic Solution Method One of my friends gave me this question several days ago: If x(e^x) = 3, find the value of x. I realized that even if I changed the base to "ln" form, it would not help as the two x's are in different levels: x(e^x)= 3 ln x + x = ln 3 x = ln 3  ln x x = ln (3/x) Using "ln", "e", or any sort of combination seems fruitless for this case. Please help.
It’s a good exercise to try solving this as Joanne did; but eventually you will agree that it is fruitless. When I look at the problem, I know ahead of time that it probably can’t be solved algebraically, because the variable occurs both inside and outside of an exponent. Doctor Fenton took this, showing some of the basic methods that can be used to get an approximate solution:
Thanks for writing to Dr. Math. Simplelooking problems can be difficult to solve. This equation can't be solved analytically, that is, by finding a formula and substituting 3 into it. Instead, you have to use an iterative method, which makes an initial estimate ("guess"), and then finds improved estimates by some algorithm.
Unlike algebraic solutions, you will not get an exact answer. But the more you repeat the algorithm, the closer you will come to the exact value. In the real world, that is enough.
There are many such algorithms, some better (faster) than others. The simplest, which always works when the problem has a solution, but which is usually the slowest, is bisection. You first write the problem in the form f(x) = 0, so that you are looking for a root of the function f. Then you need to find two numbers a and b such that f(a) and f(b) have different signs. (I am assuming that f is a continuous function, so that it must cross the axis between two values where it has different signs.) Next, you evaluate f at the midpoint m=(a+b)/2 of the interval. If this should happen to be 0, you are done. Otherwise, f(m) will have the same sign as one of the old endpoints. Replace that old endpoint with m, and you have a new interval on which f has different signs at the endpoints, and the new interval is half as large as the original interval. Then you repeat the above process with the new interval. After 10 steps, the interval will be 1/1024 as large as the original interval; after 20, less than 1/1000000 of the original length.
This method is typically slow, as he mentions; but is by far the easiest to explain! We’ll see an example later.
"False Position" is a method which takes the straight line through the two points (a,f(a)) and (b,f(b)) and finds where it crosses the axis, say, at x = c. Evaluate f(c). Again, unless f(c) = 0 (possible but extremely unlikely), you get a new, smaller interval on which which f has opposite signs at the endpoints. It generally finds an approximate root faster than bisection.
This method follows the same basic procedure as bisection, but finds a different “midpoint” that is better suited to a particular problem. I will examine it more closely in a later post.
The best method is usually Newton's method (although it sometimes doesn't work), and it works extremely well here. It takes calculus to derive the formulas, but for this problem, you make an initial estimate x_0. Given an estimate x_n, you make an improved estimate x_(n+1) by using xe^x  3 next x = x   . (x+1)e^x The function f(x) = xe^x  3, and f(1) < 0, while f(2) > 0 . Using a spreadsheet, with x_0 = 1, I got 10 decimal place accuracy in 5 steps. With x_0 = 2, I got 8 decimal place accuracy in 5 steps. Newton's method generally doubles the number of correct decimals in each step, once you are "close" to the root.
Here are the results of applying each of the three methods to this particular problem; you can see how much more quickly Newton’s method reaches a good answer:
First, bisection:
a  b  mid  f(a)  f(b)  f(mid) 
1  2  1.5  0.28172  11.77811  3.722534 
1  1.5  1.25  0.28172  3.722534  1.362929 
1  1.25  1.125  0.28172  1.362929  0.465244 
1  1.125  1.0625  0.28172  0.465244  0.074446 
1  1.0625  1.03125  0.28172  0.074446  0.10779 
1.03125  1.0625  1.046875  0.10779  0.074446  0.01773 
1.046875  1.0625  1.054688  0.01773  0.074446  0.02809 
1.046875  1.054688  1.050781  0.01773  0.02809  0.005113 
After 8 steps, our answer is 1.050781, where \(xe^x = 3.005112\).
Next, false position (where c is the new value for x, which replaces one of the previous values):
a  b  f(a)  f(b)  c  f(c) 
1  2  0.28172  11.77811  1.02336  0.15247 
1.02336  2  0.15247  11.77811  1.035841  0.08154 
1.035841  2  0.08154  11.77811  1.042471  0.04333 
1.042471  2  0.04333  11.77811  1.04598  0.02294 
1.04598  2  0.02294  11.77811  1.047835  0.01213 
1.047835  2  0.01213  11.77811  1.048815  0.0064 
1.048815  2  0.0064  11.77811  1.049332  0.00338 
1.049332  2  0.00338  11.77811  1.049604  0.00178 
After 8 steps, we have 1.049604, where \(xe^x = 2.99821\), just a little closer but on the other side. (A modified version of the method could have been much faster.)
Now, Newton’s method:
Newton’s method  
x_n  f(x_n) 
1  0.28172 
1.051819  0.011205 
1.049912  0.0000159 
1.049909  0.000000000032 
Only 4 steps take us to 1.049909, where \(xe^x = 3.000000615\). Not bad!
Now let’s look at more detailed explanations, with different examples, for two of these methods.
A student in 1998 asked specifically about this method in the context of polynomials, though it can be used with any continuous function:
Roots and the Bisection Method What is the Bisectional Method? How would I solve 2x^3+x^2+5x+2 = 0 or any other polynomial? Can the method be used by something greater than the 3rd degree?
Doctor Benway answered, starting with a familiar example in a guessing game:
The bisection method is one of my favorite surefire methods for finding roots of polynomials. The great thing about this method is that it works for everything, at least everything normal that you're likely to run into at this point in your mathematical education. Have you ever watched the game show "The Price is Right"? That show actually uses a form of the bisection method, in a way. For example: Bob Barker: This lovely toaster oven/can opener from General Electric can be yours if you can guess the price within thirty seconds. Start the clock. Contestant: Ummm....$20 Bob Barker: Higher Contestant: $100 Bob Barker: Lower Contestant: $60 Bob Barker: Lower Contestant: $40 Bob Barker: Higher Contestant: $50 Bob Barker: Correct! You win! Contestant: Wheeeeeeeee! As you notice, the contestant made two initial guesses, $20 and $100, which were too low and too high respectively. This told her that the price was between these two guesses. She then guessed halfway between, $60, and found that was too high, so it must be between $20 and $60. Her next guess, $40, was too low, so she knew it was between $40 and $60. She guessed $50, halfway between, and she was correct!
The method essentially consists of a series of guesses, narrowing down the range.
This example turns out to be very similar to how the bisection method works. The usual method is to first plot a function to get an idea about where the roots are. Say f(x) has a root somewhere between 2 and 3 when you graph it. Your goal is to guess numbers as above until you get a number that turns out to be zero when you plug it in. Your initial "guesses" are 2 and 3. To find out whether you need to go "higher" or "lower," you plug the numbers you are guessing into the function. In this case, f(x) being positive or negative will tell you whether you need to go higher or lower. The best way to do this is to look at the graph to tell which way you should go. If you have an increasing function on that interval, a negative number means you guessed too low, and a positive number means you guessed too high (see below): / + / too high ___________*__________ /  / too / low / If the function is decreasing, then the opposite rule applies.
This is like searching for a hidden object while a friend tells you when you are getting “warmer” or “colder” — but here the friend is the function itself.
Where "bisection" comes in is that the best way to do this is to pick the midpoint of the interval you know the root to lie in, just as the contestant picked midway between the price range she knew. Eventually you will come closer and closer to the answer, and when you feel you have an x where f(x) is close enough to being zero, then you can put it down as an approximation for the root. This works for all continuous functions. It's not the best method, but it is certainly the easiest. If you take a class such as Numerical Analysis, you'll learn better methods for approximating roots.
Of course, the midpoint is not the “best” next point to try; but it is always going to be closer than one of our previous guesses, and it is easy to do. When I look at False Position next time, it will be clear that that method is essentially Bisection with a better way to make the next guess.
This is the method we most often recommend, for those who have some knowledge of calculus, or who can just use a formula we provide. In the following example from 2006, a student wants to know how to solve a transcendental equation, and Doctor Jerry demonstrates this method:
Solving Trig Equations with Newton's Method What is the technique to solve a trigonometric equation like this? Do I need to use calculus? I've studied a bit of that. 3sin(x) = x + 1 (0 < x < 2pi) What if the above equation is in quadratic or some higher order, what is the technique to do that?
Here again we have the x both inside and outside a “transcendental function”, this time a trigonometric function. Doctor Jerry first pointed out the need for an approximation method:
This equation is of "higher order;" it is called a "transcendental equation." The needed technique is either numerical or graphical. I'll assume that you can do the graphical with the help of a calculator. Actually, most "scientific" calculators these days can also solve this kind of equation numerically.
The typical graphing calculator can both show you the graph (as he does below), and find a solution more accurately than you can by reading the graph. It likely uses Newton’s method or something like it. Since Sam mentioned knowing some calculus, this was the right method to explain here.
If you want to do it "by hand," you can use something like Newton's Method, which uses calculus. Take a look at this figure: Notice that I've graphed in the left figure the two original graphs. In the right figure I've graphed the function f(x) = 3sin(x)  (x+1). We want to find its two zeros. I'll concentrate on the left one, near 0.53.
Newton’s Method is designed specifically to find zeros of a function; that is, it finds a value of x such that f(x) = 0. This is why he has defined f as the difference of the two functions that are to be equal. The method also depends on an initial guess near the desired solution, which makes an initial (rough) graph useful. We can see that one solution is near 0.5; in order to make the process more visible, Doctor Jerry chose to start a little farther away.
In Newton's Method, one makes a guess as to where the root is. In this case, I'll guess that the root is near 0.75. Call this x1. We work out a sequence of guesses, which usually improve and converge to the root. To find x2 we do this: From (x1,0) we go vertically until we hit the graph of f. At that point we draw a tangent line. Call the xintercept of this tangent line x2. Once we have x2, we repeat this procedure. It's not difficult to show that x2 = x1  f(x1)/f'(x1) x3 = x2  f(x2)/f'(x2) and so on.
So what we have is a formula to find the next “guess”, \(x_{n+1}\), given the current “guess”, \(x_{n}\). The formula in general is \(\displaystyle x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)}\), where \(f'(x)\) is the derivative of \(f(x)\).
For this particular function, \(f(x) = 3\sin(x) – (x+1)\), the derivative is \(f'(x) = 3\cos(x) – 1\), so the formula becomes \(\displaystyle x_{n+1} = x_n – \frac{3\sin(x_n) – (x_n+1))}{3\cos(x_n) – 1}\).
The results are:
If x1 = 0.75, then x2 = 0.503222 x3 = 0.537907 x4 = 0.53847 We may be reasonably confident that the root is near 0.538. We can check this by calculating f(0.538). We find f(0.538) = 0.000741359. We can continue generating new approximations until we achieve the accuracy we want.
Note that the last two guesses both rounded to 0.538, so the solution will, also. In four steps, we found a solution to three decimal places.
For another example, explained in great detail, see:
Using Newton's Method to Solve an Implicit Equation
Next time, I’ll discuss the Method of False Position, which has two sides to discuss, only one of which really fits the present discussion, but both of which are interesting.
]]>A recent question about an old puzzle leads to multiple references to our archive.
Here is the question, from a student in a nonEnglishspeaking country:
At finals of campuswide Brain Bowl Tournament, your team was asked to measure out exactly one gallon of water using only a 5gallon and a 3gallon container, without wasting any water. Your team completed the task. How did they do it?
My attempt:
I think … this is hard. I don’t know where to start, they didn’t even have initial volume of water or measurement of water so I can’t manipulate anything. But here’s my idea: fill the 3 gallon water container until it is full, then estimate where the 1/3 of the water then whoosh … I know it is wrong but at least I gave it a try.
The initial difficulty with this common puzzle is to get a clear sense of what it means. What are you starting with, and what are you allowed to do? The wording in this version leaves some facts unstated; this happens often when a problem is so wellknown that an author doesn’t consider how a newcomer will read it. We have seen several variations in which the rules differ; in particular, the phrase “without wasting any water” is not usual.
Doctor Jacques answered, providing mostly just an explanation of the implied rules:
You must measure one gallon exactly: estimation is not allowed.
There are three things you can do at any stage:
 Fill one container from the tap
 Empty one container into the drain
 Pour the content of one container into the other one until either the source container is empty or the destination container is full.
Initially, the only thing you can do is (1): fill one of the containers. It turns out that either choice will lead you to the solution, although one choice is faster.
If you try to play with this, you will notice that, after the first operation, there is only one sensible thing to do at each stage, without undoing what you did in the previous stage.
I will let you experiment with these ideas, feel free to write back if you get stuck.
That last statement is one that many students miss: in order to understand an unfamiliar problem, you need to experiment with it, trying things out to get a feel for it.
In this case, the clarification and some experimentation was all that was needed:
OMG OMG!!! I’ve got an idea!! I will fill the 3gallon container first and pour all of it in the 5gallon container, then I will fill the 3gallon container then pour it into the 5gallon container until it is full, then the remaining water in the 3gallon container is exactly one gallon!!!!! Yaaaayy!!
Would you mind if I ask the other strategy doing this problem?
This turns out to be a rather straightforward version of the puzzle, in which nothing has to be wasted and the path to the answer is fairly direct. Below, I will discuss some variants we’ve seen, and different approaches. But first, Doctor Jacques answered the student’s question, showing the other way while also demonstrating one way to organize your work:
That is correct. The other strategy would be to start with the 5gallon container. If we call A the 5gallon container and B the other one, the operations would be:
A B Fill A 5 0 A → B (3) 2 3 Empty B 2 0 A → B (2) 0 2 Fill A 5 2 A → B (1) 4 3 Empty B 4 0 A → B (3) 1 3 and you are left with 1 gallon in container A.
You may notice that this solution not only takes more steps; it also requires dumping a total of 6 gallons of water, which is not acceptable given the statement of the puzzle.
For harder versions of the puzzle, you may want more help than this student needed. We have collected some of our past discussions of questions of this type on the Selected Answers page:
Measuring with Two Containers
The most detailed of these is this one, from 2002:
4 and 9liter Pails: How to Measure 7 Liters I have a problem that needs to be solved. I tried using cups to do this problem but without the ability to measure I just can't seem to think it out properly. Can you help? Here it is: You are trapped in a cave that has an endless supply of water. You have to create an explosive device to get out. You need to add exactly 7 liters of water to a premade mixture. You only have a 4liter pail and a 9liter pail. Neither pail has markings on it and they are not symmetrical. You do not have the use of any kind of marking device. How do you get out?
This is a considerably more difficult version (that is, it will take more steps, and therefore more thinking), but it is the more typical type, explicitly stating that there is an endless supply of water, so that we can always get more, and won’t mind “wasting” some by dumping it out. Note the additional comments that the pails are unmarked and not symmetrical (so that you can’t tip one and be sure it is exactly half full, for example). This makes it clearer what the rules are.
I started my answer by referring to two previous answers to such puzzles, and then stating, as Doctor Jacques did above, what steps are allowed. I also suggested a similar notation, making a table to keep track of what has been done:
What you want (and these pages don't give) is a strategy for solving this sort of puzzle. Let me suggest one. One thing we need is a way to keep track of what we are doing. I like to use a chart that shows how much is in each container: (4) (9)   0 0 Now at each step we can either fill a pail, empty a pail, or pour from one to the other until we either empty the former or fill the latter. We can indicate on the chart what we did and what the result was: 0 9 fill (4) 4 5 pour into (4) 0 5 empty (4) You might just draw arrows to show what happened; that's harder to do in print.
What I did here differs from Doctor Jacques’ only in the way I label the columns (using the sizes as a reminder rather than an arbitrary letter) and my use of words rather than symbols in the explanations (which was largely because of the limitations of typing text).
Next, I suggested an overall strategy. Here, I was using the puzzle as a model for general problemsolving: first we can invent a notation, then a way to think about the problem:
Now we have to figure out how to choose the steps to take. There are only a few things to do at each step, so you might just make several charts starting with different first steps, and see where each one takes you; whenever you have a choice to make, start a new chart at that point if your first choice doesn't work out.
This is an orderly search of the “state space” (the set of possible states for the pails); in other problems it might be far too timeconsuming, but it is often a good way to start — this is the experimentation I referred to.
Another common strategy is to work backwards from the end of the process, as well as forward from the beginning:
It will also be helpful to know the goal. You should try working backward from the end of the chart, to see what the next to last step should look like. You want to end with 7 liters in the 9liter pail. How could you get to that state? 0 7 You might either have poured 4 into (9): 4 3 0 7 pour into (9) or poured 2 from (9) into (4): 2 9 4 7 pour into (4) So you want to see if you can get amounts of 3 or 2, knowing that if you can get either of those, you can get 7. (That is, 7 = 92 and 7 = 3+4.) This should make the work a lot easier. You might find it interesting to keep a list of what amounts you are able to get in either bucket, and how many steps it takes.
At this point, I left it for Cynthia to apply these ideas. I also gave a link to a page that goes much deeper into the puzzle:
2 Pails Puzzle  CuttheKnot, Alexander Bogomolny
This also links to a further page on the “Three Jugs Problem”. Unfortunately, these depend on Java applets, which are often not supported today.
On one occasion, in 2009, I wrote up an (unarchived) explanation of a puzzle of this type based on the graphical representation of the state space that Bogomolny uses:
Bogomolny shows a way to represent the problem on a grid, which allows you to solve it almost automatically once you get used to it. For your version, you want to get 6 cups from 7 and 4, so you'd make a grid 7 by 4: 4 ++++++++         3 ++++++++         2 ++++++++         1 ++++++++         0 A+++++B+ 0 1 2 3 4 5 6 7 Here, the horizontal coordinate represents the amount in the 7cup container, and the vertical coordinate is the amount in the 4cup container. We want to get from A(0,0), where both are empty, to B(6,0), where there are six cups in the larger one. If you fill a container from the source, or dump it out, you move vertically or horizontally as far as possible in the grid. If you pour from one into the other, you move at a slope of 1 to the left or right until you reach an edge of the grid. (Along such a line the sum of the amounts is constant.) You're looking for a combination of those motions that will get you from A to B. An example of a first few steps might be to follow the numbers in order as shown (on paper I'd draw all the lines showing where I've been): 4 1+++3+++  \     \    3 +\+++\++   \     \   2 ++\+++\+    \     \  1 +++\+++4     \     0 A+++2+B+ 0 1 2 3 4 5 6 7 That means I filled the 4, then poured it into the 7, then filled the 4 again and poured as much as possible into the 7, leaving 1 in the 4 and the 7 full. We're awfully close, but we can't simultaneously reduce each by 1 cup; all we can do here is to empty the full one and keep going. So we continue: 4 1713+39++  \  \  \   \  \   3 +\\\+\\+   \  \  \   \  \  2 11+\\\+\10  \   \  \  \   \  1 5\+\\\+4  \  \   \  \  \   0 A612+28B+ 0 1 2 3 4 5 6 7 So we get to the goal, B, on the 14th step! Here is a shorter solution, which starts by filling the 7 rather than the 4: 4 +++2++6+     \    \  3 4+++\++5  \     \    2 +\+++\++   \     \   1 ++\+++\+    \     \  0 A++3++B1 0 1 2 3 4 5 6 7 One more step takes us to B, but since we already have 6 cups in the 7cup container, we can say we're done after these 6 steps.
I also find another unarchived explanation by Doctor Jacques from 2008, which relates the problem to some concepts in algebra and number theory, before doing essentially what we have been doing. This time, we have 5 and 11 gallon buckets, and need to make 7:
Let A be the 11gallon bucket and B be the 5gallon bucket. As we have to make 7 gallons using multiples of 11 and 5, we try to find integers x and y such that: 11*x  5*y = 7 (1) As 11 and 5 are relatively prime, this is always possible. In fact, there is a systematic way of finding x and y, based on the Extended Euclidean algorithm; however, given your age, I'm not sure you already know about it. If this is the case, you can proceed by trial and error: list the multiples of 11 until you find one that is 7 more than a multiple of 5. You will easily find the solution x = 2, y = 3, i.e.: 11*2  5*3 = 7 Now, if we had a third large enough container, the solution would be easy : pour 22 gallons into it (using A) and remove 15 gallons (using B). The catch is that we have no such container: everything must take place using only A and B; obviously, the 7 gallons must end up in A. Well, let us not despair, and look at what happens. As (1) shows, we must try to add 11 gallons twice, and remove 5 gallons three times; the important point is that we can do these operations in any order. We can start by filling A once (11 gallons) and removing 5 gallons twice (using B, and emptying B after each operation). This gives the following sequence: A B 11 0 fill A 6 5 pour form A to B 6 0 empty B 1 5 pour from A to B 1 0 empty B We are left with one gallon in A and nothing in B. We have added 11 once (to A) and removed 5 twice. According to (1), we still need to add 11 once and remove 5 once (indeed, 1 + 11  5 = 7). The problem is that we can do neither directly: we cannot add 11, because A is not empty, and we cannot remove 5, because A does not contain enough water. The trick is to combine both operations. We can pour the remaining gallon into B (without emptying B), fill A with another 11 gallons, and then pour the remaining 4 gallons into B to fill it. As B is now full, we have effectively removed 5 gallons and added 11: A B 1 0 (previous state) 0 1 pour from A to B 11 1 fill A 7 5 pour from A to B (4 gallons needed to fill B) and we are left with 7 gallons in A, as required. Note that, in total, we have indeed added 11 gallons 2 times and removed 5 gallons 3 times (although the last operation was done in two steps). Note that we used A to add and B to subtract. There is another solution, which is, in this particular case, longer. Instead of (1), we can look for a solution to: 5*x  11*y = 7 (2) and, by trial and error (or otherwise), we find the solution x = 8, y = 3, i.e.: 5*8  11*3 = 7 In this case, we would use B to add and A to subtract. The first steps are: A B 0 5 fill B 5 0 pour from B to A 5 5 fill B 10 0 pour from B to A At this point, we can neither add 5 to A nor remove 11 from it. As before, we combine the operations: A B 10 5 fill B 11 4 pour from B to A (1 gallon needed to fill A) 0 4 empty A 4 0 pour from B to A and the net effect is that we have added 5 and subtracted 11. You can then repeat the process until you get 7 gallons in B. In this case, the solution is longer, but, for other capacities, this second method would be faster. Note that we don't really need to compute x and y  I did it only to show what happens. In practice, you wold start by filling one container, and repeat the above operations until you get the desired result (this will always happen if the capacities are relatively prime). Note that, once you have filled one of the containers as the initial step, there is only one sensible thing to do at each subsequent step (without undoing what you already have done).
Let’s close with an example of the threecontainer version, in which we don’t have an unlimited supply, and we can’t waste anything. (In effect, the third container is both our supply and our dumping place, and that is enough.) In 1997, Doctor Rob answered this question:
The Three Canteens Two men wandering in the desert come upon a trail. In hopes of finding help sooner, they decide to split up, each taking half their remaining water, and set off in opposite directions along the trail. They have only one 14cup canteen full of water, and two empty canteens that will hold nine and five cups respectively. The only way for them to measure water is by pouring water from one canteen to another until the first is empty or the second is full. How can they measure seven cups into each of the two larger canteens? Please send me a solution with the least number of "pourings."
Doctor Rob had a very different representation of the problem:
At each step you have either: (1) just emptied a canteen, so all you can do is pour either of the other two into it, or pour either of the other two into each other; or (2) just filled a canteen, so all you can do is pour from it into either of the other two, or pour either of the other two into each other. In either case, there are at most four possibilities. You can draw a diagram with dots representing the contents of the three canteens and arrows indicating that you can get from one state to another by filling or emptying a canteen into another. The initial state is 1400; there are only two possible next states: 905 (fill the small canteen) or 590 (fill the medium canteen. From the former state, you can get to 1400 (empty the small canteen into the big one), 950 (empty the small canteen into the middle one), or 095 (empty the large canteen into the middle one). From 590 you can get to 1400 (empty the middle canteen into the large one), 095 (empty the large canteen into the small one), or 545 (fill the small canteen from the middle one). This part of the diagram looks like this: O 1400 ^ ^ / \ 14>9/ \14>5 / \ v v O 590 O 905 ^ ^ ^ ^ / \ / \ 9>5/ \ / \5>9 / 14>5\ /14>9 \ v v v v 545 O > O 095 O 950 / \ 5>9 / \ / \ / \ As you draw the diagram, you will discover that two of the arrows leading out of any dot will lead to two of the four dots labeled 1400, 905, 590, and 095. These you can ignore, since they lead back to the beginning of the diagram. Furthermore, another arrow will lead back up to a dot just visited (making a twoheaded arrow), which can also be ignored. That leaves just one arrow leading to a new dot. There are two short paths in this diagram that lead from 1400 to 770. One passes through 545, and the other through 950. I leave it to you to find the single pourings which lead out of each of these two dots to new, notyetvisited dots. Then from each of these, there is a single pouring leading to a new dot, and so on. Eventually each will lead to 770. One is shorter than the other, and that is your answer.]]>
Last time I looked at reasons for learning to estimate. In searching for answers on that topic, I ran across a question that touches not just on reasons for estimation, but on other ways to check an answer, and on some of the specific ideas we will be looking at later.
The question comes from Stephanie, in 2002:
Checking When Rounding This is what I do understand: To check Decimal ADDITION: 13.3 + 26.5 = 39.8 To check: Look at the tens, and round: 13. + 27. = 40. Then look at the addition answer and round it: 39.8 is rounded to 40. Answer is correct What I DON'T understand: To check Decimal SUBTRACTION: 7.6  1.4 = 6.2 OR 86.8  43.9 = 42.9 To round and then check, I don't understand what I need to round, and do I add to check?
We love it when students tell us what they do understand, and then clearly show what they don’t! That makes it much easier for us to see how we can help.
Stephanie has learned to check an addition by rounding each addend to the same place (to the nearest integer, here — she probably meant “look at the tenths” as part of that process), using the standard method of rounding to the nearest. The result she gets is just what she gets by rounding the claimed answer, she concludes that it is correct. I’ll have some comments on that, but on the whole it is well done.
For subtraction, the basic concept is the same, so we will have to find out what it is that is stopping her from doing it. But the last question, “Do I add to check?” suggests that her issue is that she has learned to check subtraction by adding, and she is wondering if that is what “check” means here.
I first looked at the addition, pointing out that she made too strong a conclusion:
It's worth noting that this check does not tell you that the answer is correct; you just know that it makes sense  it is not too far off. But if the correct answer were 39.9, you would not know. Also, the answer you get by rounding and adding will not always be the same as what you get by adding and then rounding. For example, 12.5 + 23.6 = 36.1; but 13 + 24 = 37, and 36.1 does not round to 37. What has happened is that the errors introduced by rounding accumulated when they were added together, so that the result is more than .5 away from the exact answer. But since 36.1 and 37 are reasonably close, the answer is reasonable.
This is an example of “rounding error”: the result after rounding the addends is not always the same as the result of rounding the sum, so we shouldn’t expect that to happen always. Stephanie’s example is atypical, and gives too optimistic an expectation. Knowing what to expect allows you to decide whether the error you do get is small enough to trust the answer.
So on one hand, you shouldn’t call an answer wrong when the estimate is a little off; but on the other hand, you can’t be sure the answer is right just because the estimate looks good!
Next, I looked at the issue of subtraction:
To check subtraction by rounding, do the same thing: round and subtract, and see if the answer is close to the answer you got. The way the rounding test works is simply that you replace a detailed operation (adding decimal numbers) with the same operation on simpler numbers. So for the second problem, you would subtract 87  44 and compare that with 42.9.
I’ll be examining the first example later; it raises an issue I didn’t want to cover at this point. This second example works as neatly as her addition: We round 86.8 – 43.9 to 87 – 44 = 43, which is very close to 42.9, so it is likely to be correct (or at least is not clearly wrong).
I didn’t yet comment on other ways to check subtraction, but left that for her to bring up if she wanted to.
Her reply dealt with two other issues: first, why we would bother rounding; and, second, as we’ll see below, what’s going on in her first example. Here is the first part of her answer:
I understand your directions, and thank you. But I want to know if this is one of those weird Math things that isn't really needed? What's the use of checking, if you are only going to get a ballpark answer? That seems like a waste of time.
First, I dealt with different ways to check a subtraction, and why they are all useful:
Checking is important, but each kind of check has a different value. Checking by estimation is especially important when you use a calculator, since you know it won't make mistakes on the details but if you fail to type in a decimal point or a digit, it can make big errors. So if you did a calculation that said you needed, say, 1.5 tons of concrete to make a bridge strong enough, and your estimate said it should be about 150, you would go back and do the calculation again! Another kind of check, "casting out nines," is unaffected by the size of the answer, but would show if some one digit somewhere was wrong. And that check, in turn, is unaffected by the common error of transposing two digits, so you might prefer another check that would reveal that error.
So we have several tools in our checking toolkit, and each will catch a different kind of error. (In the same way, doctors have different ways to prevent infection, from washing their hands to a variety of antibiotics, and they may use all of them rather than depend on one.) If you are not familiar with “casting out nines”, you can search for that term on our site or elsewhere, to see both the method and its limitations. Here is one place to read about the latter, with a link to an explanation:
When Casting Out Nines Fails
I didn’t mention a third method, which Stephanie had mentioned in her initial question: You can just add in reverse. In her first example, we have
7.6 6.2  1.4 + 1.4  is equivalent to  6.2 7.6
Since the addition is correct, the subtraction is too. And this is a very convenient check, because we can literally add upwards, just reading from the bottom as if it looked like this (but without actually rewriting anything):
7.6 ^   + 1.4  6.2 
This check will catch any error, as long as you do it correctly. But the other checking methods will catch errors you make going both ways.
But she had another question, wondering if something was wrong with rounding for her first example. This is probably the very reason I had left that example until now.
Also, can you please show me how to check the first problem I gave you when I asked about subtraction? I am wondering if my book has explained this wrong. This is what it says: "To round a decimal number to the nearest whole number, look at the tenths digit. If the digit is 04, the ones digit remains the same and all the digits to the right are dropped. If the tenths digit is 59, the ones digit is raised one and all the digits to the right are dropped. This is called a CONVENTION."
Clearly, Stephanie tried rounding and something went wrong, and she thinks the rounding method she is using may be incorrect. The book emphasizes that the rule they teach is not the only way to do it, but is a “convention” — that is, a definition that has been agreed upon, rather than being mathematically necessary. This is so because when a number ends in 5 in the tenths place, there is no one whole number that is “nearest”, and we have to make a choice. But this convention, and even the very choice to round to the nearest whole number, is not always the best, as we’ll see.
So I demonstrated how to check by rounding with her first subtraction example:
Let's look at 7.6  1.4 = 6.2 To estimate the answer, you can round 7.6 up to 8 (since 6 >= 5 ), and round 1.4 down to 1 (since 4 < 5 ); 81 = 7. That doesn't mean there's an error, because rounding can introduce an error this large. If the estimate were, say, 70, you would know something was wrong.
Here I just did what she has been taught (following the convention she quoted), rounding to the nearest whole number. As I mentioned earlier, when you check by rounding, you have to develop a sense of how big an error to expect, in order to make an appropriate judgment of the result. This is probably Stephanie’s main difficulty.
But, as discussed previously when we talked about rounding, this can be improved by not following the convention:
One way to improve the estimate when you do this is to think about how subtraction works. We added something to 7.6, and subtracted something from 1.4; both changes will have increased the answer. (Do you see why?) So we know the real answer is LESS than 7. That makes our check valid. Also, since I know about this problem, I prefer to round both numbers in the SAME DIRECTION when I subtract (and in opposite directions when I add). Here, the .6 wants to go up and the .4 wants to go down, and neither is more persuasive than the other (both are .4 away from the nearest whole number). So I would arbitrarily choose, say, to round both numbers up: 7.6  1.4 ~ 8  2 = 6 That gives a more accurate estimate. But even this way, it won't necessarily be exact.
Here I have shown two ways to improve the check: either think of it as an inequality, telling us that the correct answer is less than 7, which is true of 6.2 so that it is valid; or modify the way we round in order to minimize the error. (Note that if I rounded both numbers down, I would get exactly the same result, so I didn’t spend much time deciding which way to go! This is another thing you learn with experience.)
In conclusion:
You can find several discussions of rounding in our site (try the FAQ first); you'll see more about different conventions for rounding when you are exactly between two numbers. But the important thing here is to realize that rounding is only a tool, and considerations apart from the rule for rounding a single number can lead us to depart from that rule when the goal is to estimate the result of a calculation involving several numbers. The interactions among numbers can make a big difference.]]>
We’ll start with a very broad question, from 2002:
Why Estimate? Why do we use estimation?
Doctor Ian took this one:
The _point_ of estimating things is that estimating will save you space, time, or effort, and in return for those savings, you agree to pay a price in accuracy. There are lots of different ways to estimate things: you can eliminate digits from a number (truncation or rounding), you can replace a decimal number with a fraction (which, technically, is the same thing), you can replace any number with a 'nearby' number that's easier to work with, you can use a graph to replace a list of individual values, you can use a simple function in place of a more complicated function, and so on. In each case, you're making a tradeoff: "It's not going to hurt much if my answer is off by a little bit, especially if I can get that answer ten times more quickly, or by storing a thousand times less information." The _art_ of estimating things properly requires two different sets of skills. The easier one is learning the various techniques of estimation: rounding, graphing, and so on. The harder one is learning to recognize how much accuracy you can afford to throw away in a given situation, and how much benefit you'll get from doing it.
This is estimation in a nutshell: giving up some accuracy in order to gain some sort of efficiency; and deciding how much of the former is a worthwhile price for the latter. Estimating well is not just a matter of technique, but of understanding the application. It is a judgment call (and therefore hard to grade objectively, another reason teachers may avoid it or teach it ineffectively).
Many readers will not recognize what Doctor Ian means by using a simpler function, because this is not seen at lower levels. But, for example, complicated formulas (or even formulas that can’t be worked out at all) are often replaced with approximate polynomial formulas, which can be easily worked with. This is one of the reasons students learn so much about polynomials: not because reallife problems typically have polynomial solutions, but because pretending that they are polynomials makes it possible to solve them approximately. And that is good enough for practical purposes.
A month earlier, we had a specific question about estimation, which served as an example of why we do it:
Why Estimate? Estimate to the nearest thousand a solution to Dave's subtraction problem. The numbers are: 6,107 2,980  4,127 I just don't know how to estimate.
I noticed that the problem could have been written without showing a result for the subtraction. Clearly, the problem is meant to teach something more than how to estimate. I answered:
Assuming that 4,127 is "Dave's" answer to the problem, it illustrates why it can be very important to estimate  it's wrong, it's an easy mistake to make, and it would have a huge effect on whatever he uses the number for, from spending money to building a house. And estimating makes it easy to see the error!
So, the reason we might estimate the result here, even though someone has already claimed to have an exact answer, is to check it. In fact, it is quite easy with a calculator to think you have an exact answer, when it is really very wrong. Estimation is an essential companion to calculators: calculators bypass our own thinking, while estimation forces us to think. We’ll have more to say on this below.
Now I went through the process of estimating as instructed, leaving the actual work to Stephanie:
To estimate to the nearest thousand, the natural thing to do is to round the numbers you are adding to the nearest thousand. Which of these numbers is closest to 6,107? 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Which of those numbers is closest to 2,980? Now subtract those much simpler numbers, and the result is your answer. Sometimes you can get a better estimate by thinking about how much you increased or decreased each number when you rounded them, and how that will affect the result; but in this case the difference of the rounded numbers is good enough.
Notice that I didn’t teach a routine method for rounding to the nearest thousand, but emphasized the meaning of what we are doing. (I later gave links to pages with more detail on the routine process.) Carrying out the work, we see that the “thousand” nearest to 6107 is 6000 (because it is less than the halfway point between 6000 and 7000); and the “thousand” nearest to 2980 is 3000 (because it is beyond the halfway point between 2000 and 3000). We were told to estimate by rounding to the nearest thousand, so we replace the original numbers with the estimates, and do the subtraction: 6000 – 3000 = 3000. That is the answer they want. Since the given answer, 4127, is quite far from this (in fact, farther than the largest possible error introduced by rounding), it must be wrong.
In my post on March 12 about rounding, I discussed how in the context of estimation it is often better not to “round to the nearest”. I just hinted at that here. You may want to try using some of those ideas; there are several things to be learned from this example.
The correct answer is 3127, not 4127; “Dave” made an error in the thousands digit. Note that if he was right there, but wrong in, say, the tens digit, estimation would not catch the error. We would just compare, say, 3217 to 3000 and say it looks good.
So estimates are a good way to become suspicious of an answer, but they don’t catch every error, and can never prove it is correct.
Let’s move on to a very different question, from 2004:
Why Is Mental Math Important? How can I convince a 14 year old girl who is in 8th grade the importance of mental math? I tried to point out some real life situations but her teacher told her it is not really important since there are calculators and computers. I also pointed out that if she does not learn such things it could make her unsuccesful further down the road in life. I think that things like adding 2 digit numbers together using techniques such as adding the tens place first and then adding the ones place and then adding the results are critical skills. Similar skills include doing subtraction mentally when using 2 digit numbers and being able to mentally estimate multiplying 2 digit large numbers.
John is asking about the importance of two different things: “mental math” (getting an accurate answer for simple problems without paper or calculator), and “estimation” (getting approximate answers easily). Two of us answered. Doctor Link started with a pedagogical reason:
Mental math, irrespective of its applications, is an excellent way to stimulate one's mind. Not only does it generally stimulate your mind, but it also helps one get a better "number sense." In other words, one becomes more familiar with how numbers interactthis is very important, because as you know, math is something that builds on itself. If you don't have a good grasp on how numbers interact, then more complicated math will seem like more of a challenge.
Working only with a calculator insulates a student from the numbers, making her just a copyist rather than a participant. Then he moved on to a social reason:
I wasn't going to mention it, but since you brought it up I will. If you can't add and subtract without the help of a calculator, it can certainly reflect poorly on you. Even if you really are a very intelligent person otherwise, this inability may give people a different idea. For better or for worse, our society puts a lot of stock into appearances. If one doesn't appear to be educated or "smart" at first glance, society will often just label that person as such.
Being able to do basic math suggests intelligence, and in some settings that will be important.
There is also a simple, practical reason : habitual use of a calculator can be a hindrance; sometimes it is actually quicker to do the mental math:
Also, the fact is that calculators are most useful in a setting that requires either a) things that are essentially impossible for humans to compute at any reasonable speed (like evaluating the cosine of a number) or b) for calculations with rather large numbers. For two digit addition, not using a calculator should actually be faster, given that you have learned the techniques well.
In working with students face to face, it is far too common to find them picking up a calculator to multiply 1 times 3, or 34 divided by 34! This makes it clear that they are completely bypassing their minds, turning to the calculator when it is absolutely not needed. As Doctor Link said, this reflects poorly on the student, and it also means they don’t gain the experience of thinking. I often tell such students, “Your calculator doesn’t need the practice — you do!”
There are a myriad of other reasons, from being able to dazzle one's friends and future coworkers to giving one confidence about their abilities. Ultimately though, I am very surprised to hear that your daughter's teacher fueled this. When I was growing up (which wasn't too long ago), students constantly questioned the usefulness of mental math. My teachers always maintained that it was very important to be able to do these calculations ourselves. After all, thinking is what makes us humannot dependence on machines to do our thinking for us!
Finally, he touched on a specific practical reason for at least estimating:
My brother also reminded me of one other important thing. What if you are getting your change, and the cashier has either accidentally or purposely given you less change then they should have. Without the ability to do mental math, you may overlook thisso there's your real life scenario for you.
I picked up from here, emphasizing estimation:
In the modern world of computers and calculators, mental estimation skills are MORE important, rather than less! That's because, although calculators and cash registers make it unnecessary to do exact calculations in many instances, it can be easy to trust them too much. People can make mistakes in using them, entering the wrong amount or doing the wrong thing with it, and if we just blindly accept the answer it gives without checking, we'll lose out a lot. One major kind of error is putting the decimal in the wrong place (or failing to enter it at all), another is to miss a key so that what should have been 1234 is taken as 124, off by a factor of ten! Or how about hitting "percent" instead of "divide"! These kinds of mistakes are easily caught by estimating the correct answer. Estimation won't catch errors of pennies, but it will catch dollars, which are more important anyway. You can probably find other examples where this will be important in her life, but money is a good motivator! It's worth noting that estimation uses the same skills as exact mental arithmetic, and more. It also builds number sense by making you aware of what matters most. A teacher who says arithmetic skills are unimportant is depriving students of an essential life skill.
John wrote back that the social argument might be the most persuasive; but money can be, too.
Another question from 2004 about a very different kind of estimating (related to “Fermi problems”, which I will be discussing later in this series) suggests that number sense can take on bigger dimensions than the little problems we have been discussing:
The Importance of Number Sense and Estimating Answers I don't understand how to solve problems using number sense. For example, if an employee at the deli counter slices a 2 foot long salami, about how many slices will he get? How can I figure that out? Why is it important?
Our previous estimation problems involved approximating given numbers. Here, we aren’t even given all the numbers we need! This takes us beyond math proper, into the real world, where our “number sense” has to fill in numbers, and give us confidence that we don’t need accurate numbers at all in order to find sensible answers. This sort of question often arises in real life, where you may need to decide if a project is feasible, without knowing all the facts.
Doctor Ian took this; again, I will only excerpt parts that are relevant to my topic, so you will want to read the whole answer.
One thing that makes this confusing is that there isn't enough information to come up with a definite answer. That is, the number of slices he'll get will depend on how thick each slice is. ... So where does number sense come into this? Basically, you want to make a reasonable assumption about the width of a slice, and use that to get a ballpark figure for how many slices we might end up with. I'd probably say that 1/10 of an inch is a reasonable width for a slice, and that would give me 24/(1/10) = 24 * 10 = 240 slices. Maybe there are more, maybe there are fewer, but this is a _reasonable_ number.
So when we don’t have a number, we can make one up based on experience.
Why do you care? Well, suppose someone works the problem and tells you that the answer is 3, or 150,000. Would you know right away that these are ridiculous answers? The way you'd check them is by making some reasonable assumptions and using those to come up with an answer of your own. So if someone says "About 500 slices", we could say: "Okay, that's about twice what we got, so he's making his slices half as thick. That's still reasonable." Or if someone says "About 100 slices", we could conclude that he's making his slices twice as thick. Which is pretty thick, but not unrealistic.
So, as in our calculator problems, we can recognize a nonsense answer, or validate one that is not far off.
Again, why do you care? Well, this is a madeup problem, but as you get out in the world, you're going to be hearing people throwing all kinds of numbers around, and you'll need to be able to get some sense of whether they're realistic.
Doctor Ian then discusses a typical claim one might hear repeated (which happens far more now, with social media and “fake news”!). On the surface, it doesn’t sound unreasonable — that’s why it might be passed on by so many people without being checked. Read the original for this long example.
The point is that people throw around all kinds of numbers in efforts to convince other people to pass laws, raise or lower taxes, donate money, change lifestyles, declare war, and make other major decisions. Quite frankly, a lot of these numbers are ridiculous. But if you can't figure out for yourself which ones make sense and which ones don't, then you just have to believe what you're told, and hope for the best. And cynicism aside, much of what goes on in business and engineering involves making predictionse.g., estimating how many burgers people will probably buy next week, or calculating how much electrical current will be running through a wire in a given situation. People make mistakes (especially where computers or calculators are involved!) and sometimes they come up with ridiculous predictions. Number sensei.e., being able to quickly come up with ballpark estimatescan be crucial in detecting these mistakes early on, when they can still be corrected easily.
These can be very practical matters, and are good reasons for adults to make such estimates all the time. But too many don’t, because they haven’t been trained to do so.
So, how do you develop this mysterious number sense? In a word: practice. One good way to get this practice is to make a point of never using a calculator unless it's absolutely necessary. When I was trying to convert "one person every 30 seconds" to an equivalent number of people per year, I could have used a calculator, but I deliberately chose not to, because I knew that whatever number I came up with would be approximate, instead of exact; and I knew from experience that I would be able to change the numbers around a little to end up with an easy calculation that would be "close enough for government work".
Be sure to read the original, where you will see some of the ways you can make a problem easier by adjusting the numbers, when you know you don’t need an exact number anyway. Doctor Ian’s choice not to use a calculator is much like choosing to park farther from the mall to give yourself exercise. You do what you need to do to keep both mind and body strong.
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