It’s been a while since we’ve done a puzzle, just for fun. Here we’ll look at two versions of a riddle, about finding children’s ages from a known product, a partially known sum, and a bizarre fact about the oldest. Then we’ll close with an interesting variation.
Product is 36: Just hints
We’ll start with this question from 1997:
Age of Children I have a logic question for you: I meet up with an old friend on the street and ask her how old her three kids are. She says the product of their ages is 36. I say I still don't know how old they are. She says the sum of their ages is the house number across the street. I still don't know how old they are. She then says the oldest one has red hair. I say OH, that's how old they are. How old are the kids and how do I know?
Doctor Mike answered, this time just getting the work started:
This is a great problem! Your first reaction is probably "What the heck does red hair have to do with it?". Let's find out. The special properties of 36 are VERY important here. You need to see that the complete factorization of 36 is 2*2*3*3. You need this to get the most out of these clues. Let's use A, B, and C for the 3 ages, and assume they are in order from younger to older. They could all be different like A = 2, B = 3, C = 6 or some could be the same like A = 3, B = 3, C = 4 (she could have 3 year old twins). Now make a table of ALL the possibilities. Don't forget the ones with the first age being one (1). Also make a column for the sum of the children's ages: A B C sum --- --- --- ----- 1 1 36 38 1 6 6 13 2 3 6 11 3 3 4 10 The table continues on. I got a total of 8 possibilities in my table. Remember A <= B <= C. Now, my first row, with a grown child 36 years old and 1-year-old twins is unlikely, but we have to consider all the possibilities so we don't miss anything. OK?
We’ll let you make the rest of the table.
Now on to clue No. 2. For each line in the table, you should add up the ages and enter that value in the right-hand most column. You will get a different sum of ages A + B + C for most of the rows, BUT two of the rows will come out with the same sum. That must be the house number across the street, because otherwise the second clue would be enough to show which set of ages was right. Before you go on to clue No. 3, be sure you have all eight rows of numbers in the table completely filled out, and that you see which 2 rows have the same sum in the last column.
Because knowing that house number helps figure out the ages, we know that, whatever it is, there must be at least two ways to get that sum. Now we know what the house number must be.
Now on to clue No. 3. This third clue could just as well be "the oldest one has 12 toes". The real clue here is the word "oldest" which eliminates the possibility of one younger child and two older twins. That's because the phrase "the oldest one" implies only one. If you did all the work along with me, you know the answer.
Can you solve it? We’ll be completely solving a similar problem next, so I’m leaving this one for the reader.
If you have an objection to the implication about twins, so do I. Hold on; we’ll get there!
Product is 72: Complete solution
We’ll spend most of our time on this question from 1996, which will include a complete answer to a riddle like our first, and a couple challenges:
Ages of Three Children During a recent census, a man told the census taker that he had three children. When asked their ages, he replied, "The product of their ages is 72. The sum of their ages is the same as my house number." The census taker ran to the door and looked at the house number. "I still can't tell," she complained. The man replied, "Oh that's right, I forgot to tell you that the oldest one likes chocolate pudding." The census taker promptly wrote down the ages of the 3 children. How old are they?
Doctor Tom answered, repeating the ideas we’ve seen before:
Hi Jason, After I know that the ages multiply to 72, here is a complete list of the possibilities: Ages: Sum of ages: 1 1 72 74 1 2 36 39 1 3 24 28 1 4 18 23 1 6 12 19 1 8 9 18 2 2 18 22 2 3 12 17 2 4 9 15 2 6 6 14 ** 3 3 8 14 ** 3 4 6 13 Note that every combination of possible ages which has a product of 72 has its own unique sum of ages - except for 2, 6, 6 and 3, 3, 8, both of which share the sum of 14. Since the census taker can't figure out the ages after looking at the house number, the house number must be 14, because then the ages could be either 2, 6, 6 or 3, 3, 8. Now, the next clue is that the _oldest_ child likes chocolate pudding. This means that there is _one_ oldest child. Well, there is no oldest child if the ages are 2, 6, 6, so the ages of the children must be 3, 3, and 8 years old.
Here we have the complete list, and can see that there is only one sum that would leave the census taker unsure, so that must be the sum. And, as in the first, the uncertainty is resolved by the existence of an oldest, which rules out twins … or does it?
Challenge: a twin can be the oldest!
In 2008, a reader wrote about this:
The fact that you have an oldest child does not mean there cannot be two children that same age. Therefore logically the problem cannot be solved. I have twin daughters and one is 3 minutes older than the other.
Hi, John. I've made the same comment more than once, because I'm in that position myself. Dr. Rick (another Math Doctor) is my twin brother, five minutes older than I am, and I've been aware all my life that he is the oldest! It's not quite as bad as you say, though; in real life, though we shouldn't just assume anything, we can look at the facts and figure out what someone else is thinking, rightly or wrongly. In this case, once you see that the two remaining possibilities are distinguished by the presence of twins, you can see what is intended even if you know it isn't quite right. This is not a useless skill in a world where logic isn't always recognized! My usual comment when I point this out has been, "It's a cute puzzle, but you have to take it with a grain of salt."
So it’s possible that the census taker wrote down the wrong ages, misled by the father; but the fact that the father evidently considered the additional information sufficient to settle the matter argues in favor of the answer being correct.
Many puzzles, like this one, come in versions with varying levels of care in their wording; not all of them were published in our archive.
I've seen several versions of the puzzle. Here is an interesting variation I dealt with earlier this year: Question: A census taker knocks on the door and a woman answers the door. She informs the census taker that she lives in the house with her three sons. The census taker asks for the ages of the boys and the women informs him that the product of their ages equals 36 and the sum of their ages equals the address of the house next door. He finds this rather odd, but walks to the house next door, realizes that he needs more information, and returns. He asks for another hint, and the women tells him that only her oldest son was born in a leap year. How old are the three sons? Explain your answer and show all work!
This is our first version, but with a different punch line that changes things. (Leap years are more relevant to ages than chocolate!)
My answer: The main idea is that knowing the product to be 36 is not enough, even WITH the knowledge of the sum of the ages. If you list possible products that would give 36, you can find a couple sums for which this would be true -- and one for which the added knowledge that the oldest son is the only one born in a certain year would clear up the uncertainty. This version of the puzzle fixes the interesting "error" in the usual formulation in which you are supposed to deduce that the oldest simply EXISTS, and therefore is not a twin. I have a twin brother and a younger brother, and the former is definitely the OLDEST son; his mere existence as such is not enough. The fact that he was born in the same year as I would do the trick. I suppose there is still one small loophole: twins could be born in different years, but still be the same age now. But at least this version shows that someone has noticed the problem!
A twin born just before midnight New Year’s Eve, the other just after, would mess up the puzzle; but that’s very rare.
Counter-challenge: only integer ages are in view
In 2013, yet another reader countered that argument:
Just wanted to point out that the fact that there is an oldest child does not allow us to deduce that there cannot be two of the same age. For example, consider two children both having the same integer age, but one of whom is a few seconds older than the other. If you interpret two children of the same integer age as one being older than the other, then you must interpret their ages as continuous numbers -- e.g., instead of 2 years old, 2.75 years. Any positive number may be expressed as (a + b), where a is a positive integer and b is a positive real number smaller than one. And if we thus interpret their ages as continuous numbers, then the product of the three children's ages will not give an integer, since the product of any three numbers which are all bigger than zero but smaller than one also falls into this range. The only time we can achieve an integer (72) is if we assume that when we say "one is older than the other" that we mean "the integer age of one is older than the integer age of the other." Alternatively, it might imply that all three children are exactly an integer number of years old at the very instant in time. P.S. I am not trying to give criticism, just engaging in some logical thinking, which I enjoy.
It is not quite clear, but the point seems to be that in talking about a whole-number product, we must be thinking about whole numbers of years, and therefore ignoring finer differences in age.
(It doesn’t really matter, but it is wrong to say that you can only get 72 as a product of integers. For instance, the ages could be 2, 4.5, and 8.)
I answered again, asking for clarification while correcting the main idea:
Hi, Ant. As an original contributor to this page (and a twin), I want to find out what you are thinking to see what it adds to the thoughts I wrote. But I confess, I'm having trouble following your reasoning. If I understand you, you are saying that the problem is unambiguous because if two children are the same age, there is no way to distinguish them, so that if there is an oldest child, they can't be the same age. That indistinguishability may be true for an outsider, but not for their parents. As I said on that page, there is no conflict between saying that two children are the same age and saying that one is older. We are just looking at the facts in two slightly different ways in making the two statements, which is normal. No parent who says, "Here is my oldest son; he and his brother are both 7," would be saying that they are both exactly 7; he is saying that their integer ages are both 7, AND one was born before the other, irrespective of their "ages." No claim is made that their birth order can be deduced from what was said about their ages. Can you explain your thinking again?
The reality is that you can know the exact age while reporting a value rounded down.
Ant replied, confirming that the point was to deny any error in the problem:
I am not trying to say that if a parent has 2 children both aged 7, then he would not know who is the eldest. Obviously, he would. I am saying for the purposes of this question that it does not matter, because we must interpret them as having integer ages only, which does not allow them to be distinguished if they are both seven. I am trying to say that the question was fine in the first place and does not need changing. Please do not think I am trying to argue that if you are a twin, then you can't say that you are older than your sibling, or vice versa. My point is based on the assumption that the census taker is able to deduce their ages from the given data. If this is the case, then the eldest must be at least 12 months older than the next oldest sibling; and hence the question works fine as it is.
Fixing the problem by focusing on point of view
I answered, focusing on the last paragraph:
Hi, Ant. Ah! Maybe that's what I've been missing in what you said. Let's look at the problem again and see if you're right that it has a solution even if you allow that a twin can be an oldest child. ... As Dr. Tom said, knowing the product and the sum, she still doesn't know what the ages are, so we know that the sum must be 14, and the ages are either 2, 6, 6, or 3, 3, 8. Now she is told that the oldest child likes chocolate pudding. Your point is that the census taker is now sure that the oldest is not a twin because she is told this, and therefore the oldest must be 8. My point is that she could be WRONG in making this inference, because the father might very well call the older of 6-year-old twins "the oldest one." So we CAN infer that she THINKS the ages are 3, 3, 8, because this is the only reason we can see that she would have thought she knew enough, but we CAN'T be entirely sure that she is right! Nothing in the problem says that she CORRECTLY deduced their ages, which is the basis of your reasoning. So imagine if this were the puzzle's last line, instead: What ages did she write down? This would be a perfectly good puzzle -- and rather ingenious; I'll have to remember this! -- because it gives us permission to take her point of view.
Making the question about the census-taker’s thinking, rather than about the actual ages, fixes the problem.
Actually, I wonder if the father, who clearly is having fun with the census taker, might have deliberately misled her by saying something entirely true (that the older of his 6-year-old twins likes chocolate pudding) that seems to imply something false. My conclusion was, "It's a cute puzzle, but you have to take it with a grain of salt." That is, we read it within its genre (tricky puzzles) and don't expect it to exactly fit real life. There are other issues we could quibble about: the census taker probably needs to write down the sex of each child, too, so she can't just walk away; and she probably would have written down the house number before she walked in if she were really doing her job. But we ignore that and enjoy the puzzle. And I don't take what you're saying as criticism -- just an opportunity to share our enjoyment of tricky puzzles.
Yes, you are right: to account for my point, the question would need to ask something like, "What did the census taker write down?" I'd like to thank you for your response. I definitely understand the problem much better now.
Product is 90, unknown number of kids
We’ll close with a 2002 question that has nothing to do with twins:
Census Taker The census taker says, "I need to know the ages of your children." The mother replies, "I have no one-year-olds. The product of my children's ages is 90, and the sum of their ages is the same as my house number." The census taker replies. "I can see the house number but I still need more information." The mother says, "You're right. You also need to know that the boy across the street is older than my oldest child." The census taker says, "Thank you, I now know the ages of your children." What are the ages of the children, and what is the house number and the age of the boy across the street?
I answered. again just giving a hint:
Hi, Christina. Make a list of all ways to factor 90 (with no ones): 2*3*3*5, 2*3*15, and so on. One of those sets of factors gives the ages. Add the factors in each case; if the house number is that value, the census taker knows that those are the ages - unless there are two possibilities that give the same sum. So the ages must be one of those sets. Which one? Since knowing that the oldest child is younger than the boy across the street tells the census taker the answer, we can presume that the census taker has already been across the street and knows that age. What age will give enough information for him to now determine the answer uniquely?
This time, the key is not the existence of an oldest, but the fact that each possibility for a given sum must involve a different age for the oldest, so that information about that age can be useful.
Can you solve it? And do you see that this, too, can be challenged?