Here's the puzzle of the month from the April 1960 issue of *Popular Science* [UPDATE: Readers with a better memory than mine have pointed out that I posted this romper la cabeza in 2005.]

Luckily someone told me the answer to this one at the same time as the puzzle. Otherwise –– like the wife of one PS editor –– I might still be working on it:

The census taker, placing one weary foot after another, climbed the steps and rang the bell. “How many people live here?” he asked the person who answered the door.

“Three,” was a prompt answer.

“And their ages?”

The reply startled him: “the product of our ages is 225, while the sum is the same as the house number.”

The census taker looked up to check the number he had already noted on his tally sheet, “Hmm,” he said, “I need to know one more thing: are you the eldest?”

“Why, yes,” the person said. And the census taker, cheered at the novel answers, smiled as he wrote down their ages and walked away.

Puzzle of the Month (PopSci April 1960)

*Mark Frauenfelder is the founder of Boing Boing and the editor-in-chief of MAKE and Cool Tools. Twitter: @frauenfelder. His new book is Maker Dad: Lunch Box Guitars, Antigravity Jars, and 22 Other Incredibly Cool Father-Daughter DIY Projects*

Seems more like an anecdote than a puzzle. Most puzzles I know end with a question.

“What was the census taker’s name?”

Now that’s a puzzler.

Mccrum: Agreed. There is no question mark at the end, therefore this cannot be a puzzle.

Don’t we need to know the house number to solve this?

House number? Probably not.

5 * 5 * 3 * 3 = 225.

Two answers: 25, 3, 3 OR 15, 5, 3. I’d probably go with the answer were there are twins at age 3, so the eldest can be an adult at 25.

(I’m excluding the possibility of a 1 year old, which could give even more answers.)

Also 9, 5, 5.

25, 3, 3 seems most likely, but all of these answers [with no one-year-olds] have an eldest.

I believe calling him “the oldest” as opposed to “the older” cinches that there are 3 ages involved, and only one solution.

The 1 year-old argument, while interesting, leads to remaining “children” whose ages are > 18, and therefore not children.

Those three ages could be 15, 15, and 1. One of the 15-year-olds is older than the other, and the eldest of the three.

“The 1 year-old argument, while interesting, leads to remaining “children” whose ages are > 18, and therefore not children.”

What? Which ones? , ,,,

I would recommend not excluding the 1 year old possibility.

In what situations would the census taker need to ask if they were the oldest?

One of the kids answered, so he’d know which kid was which.

I would recommend not excluding the 1 year old possibility.

In what situations would the census taker need to ask if they were the oldest?

That is why the house number is needed (31), to ensure that there was no one-year-old involved (25, 9, 1; 35 or 45, 5, 1; 51).

Considering that the article was printed in the Mad Men era of 1960, an assumption of a mother who gave birth to twins at the age of 22 (your solution) seems like the only possible answer – teenage pregnancy (the 25, 9, 1 suggests the first child was born when the mother was 16) or pregnancy only after the onset of middle age (45, 5, 1 suggests that) would be too immoral for a puzzle like this back then!

3, 5, and 15? Or 1, 9 and 25?

Or 45, 5 and 1?

I think this is the real answer. Otherwise he wouldn’t have to ask who was the oldest (in the case of twins 25-3-3), and it wouldn’t make sense to have a household of all children (9-5-5, or 15-5-3).

There’s no point asking, if it’s 45; that’s a very distinct age, and there’s no factor _above_ 45 in that sequence going up to 225…

25, 3 and 3?

That’s the right answer

Really? What claim does it have above 25, 9 and 1?

If its the cultural perogative against a 16 year old birth parent, well, its not stated that the person answering the door is a birth parent. They just live together; could be a woman who adopted her elder dead sisters children at any age up to 25.

The reason is: the house number is a clue, and 25, 3 3 has the same sum as 15, 15, 1. This is the only scenario in which an “eldest” lets you disambiguate, so 25, 3, 3 it is.

But why would he ask a 25 year if she were the eldest? She’s clearly not one of the 3 year olds. I think 25, 9, & 1 are the more likely because it lets us assume that by “eldest” he means eldest child–a question that would have no meaning if the children were twins.

This is an old puzzle, there are other variants of it that are more clearly written than this one. Here is one: http://www.teamten.com/lawrence/puzzles/daughters.html

In theory they could be 5, 5, and 9. But yeah, without the house number it seems like there are multiple solutions to the problem. The key is the final question. What sort of solution would knowing that that person is the oldest trigger? Why does he NEED to know that?

It’s reasonable to assume that they’re not 5,5, and 9, though. The person at the door said that 3 people live there, and it’s not likely that the census taker would’ve ignored the fact that 3 children were living alone together in a house, supervised by a 9 year old. :-)

I agree, but this is math, so you know, the real world can go right out the window :-p

15, 15, 1 can’t be right, as then there wouldn’t be an “eldest”…

But 3, 3, and 25, on the other hand, would have an “eldest” AND add up to 31, the same sum as 15, 15, 1.

Shouldn’t it be 15,15,1 ? He wouldn’t ask about the oldest if it was 25,3,3.

Shouldn’t it be 15,15,1 ? He wouldn’t ask about the oldest if it was 25,3,3.

Yes there would. It might be a difference of several months, it might be a difference of minutes, but technically, one would be older than the rest.

Exactly. My objection as well. Clearly the puzzle poser thought that age-rounded-down-to-years alone determines who is the eldest.

Not only are two 15 year olds the exact same age, but they are perfectly spherical, frictionless and in a vacuum as well. This is a math problem after all.

Not only are two 15 year olds the exact same age, but they are perfectly spherical, frictionless and in a vacuum as well. This is a math problem after all.

In all likelihood one twin came out first so there can be an eldest.

In all likelihood one twin came out first so there can be an eldest.

(3*5)+(3*5) + 1 == (5*5) + 3 + 3 == 31. The house number is 31, so he needs to confirm that there is a unique eldest.

This is the correct explanation for what happens.

To rule out 15, 15, and 1? But that doesn’t work because a slightly-older twin would still be the eldest.

Leaving 225, 1, 1; 75, 3, 1; 45, 5, 1; 25, 9, 1; 25, 3, 3; 15, 5, 3; and 9, 5, 5.

And the census-taker can’t choose among these unless he or she can tell people’s ages just by looking at them, which seems unlikely enough in real life and more so in this kind of puzzle.

Possible between the eldest ones in each scenario: 75, 45, 25, 15 and 9…

Old, early-middle-age, adult-youth, teenager, pre-teen; everything is distinct in that set.

We don’t need to know the house number, all we need to know is that it is not sufficient for the census taker to know the answer without asking another question.

There must be (at least) two answers that meet the criteria (product is 225 and sum is x), such that asking the question “Are you the eldest?” distinguishes between them. I got it by running through all the possible combinations. Doesn’t take that long.

Okay so can we find two solutions that add up to the same number? And then see which one requires an eldest?

That’s it. 15,15,1 add to 31 as does 25,3,3. Thus the final question is needed to prove it is indeed 25,3,3

But it doesn’t prove that.

15, 15, 1 has an eldest 15-year-old because one is invariably older than the other.

25, 3, 3 has an eldest 25-year old.

Both have an eldest.

Other solutions -

75, 3, 1

45, 5, 1

25, 3, 3

15, 5, 3

15, 15, 1

I figured the eldest question was to rule out two 15-year-olds. But I’m not very good at logic puzzles!

45, 5, and 1 ? (with house number 51)

Assuming the puzzle is to find out the ages of the residents (or, indeed, the house number), we have all the information we need. Look carefully at the entire puzzle and note that the census taker NEEDED one more piece of information in order to find the correct ages–only one house number could result in that question being required.

We ARE getting stupider, this is proof. Puzzles were a whole lot harder back in the 60′s than the “brain teasers” we get today.

I can’t even figure out what the question is. Am I supposed to figure out the house number? The ages of the inhabitants?

Or are those two questions in fact the same? *Cue twilight zone music*

eldest, thus two 15y olds are ruled out. Sum of 15+15+1=31

age 25+3+3=31

there it is.

You’re assuming they can’t be less than a year apart (permitting duplicate ages). We don’t even know if they’re all related.

“eldest, thus two 15y olds are ruled out.”

does not follow, unless they were born at exactly the same time.

Sounds painful.

The answer is 25, 3, and 3. The sum has to be the same as another set of 3 numbers that multiply to 225 and don’t have a single oldest (in this case 15, 15, 1). Otherwise, knowing the house number would have been sufficient.

Pretty sure it’s 3, 3, 25. Using similar logic to what’s presented here.

http://nielsenhayden.com/makinglight/archives/006557.html

Find the primes, get your list of possible options.

There are only 2 with the same sum 1,15,15 & 3,3,25. There’s only 1 “oldest,” person in the house. So that eliminates 1,15,15. Leaving 3,3,25.

Every twin I’ve ever known will tell you which of their pair is older.

If the census taker can’t tell whether he’s talking to a 15-year-old or 25-year-old, I don’t see how this question resolved the ambiguity.

This doesn’t just count for twins either. I have 2 children who are the same age 2 months of the year.

I think that there was one solution (when the house number is know) and the result was that two people were of the same age and one person was older? By asking this question the census taker had alle the information he needed. But with this piece of information beeing know, the information about the housenumber becomes irrelevant?

I couldn’t agree more:

http://dreadedmonkeygod.net/home/archive/709

Here’s how it works, and the possibilities (I think this is all of them)

75,3,1

45,5,1

25,3,3

15,15,1

15,5,3

9,5,5.

He knows the house number, but that’s insuficient information to figure out their ages. So that means there must be at least two sets of ages that add up to the same number.

25+3+3 = 31 and 15+15+1 = 31. The other sums (79, 51, 23, 19) are unique.

So he asks the follow up question (“Are you the eldest?”) to determine if there -is- an eldest person. Since there is, that rules out 15+15+1, and hence the solution must be 25,3,3.

Just for completeness, in this fantasy 225, 1, 1 is an initial possibility.

The guy can’t tell if the person he’s talking to is 15 or 25?

I think in this riddle the math is the only part that matters. Whether or not the census taker can tell ages apart is irrelevant and distracting to the answer.

Jim,

I did the same analysis and came up with the same answer. There are two things that bug me about this riddle — things that always seem to be glossed over by riddle-makers and riddle-answerers alike: 1) There is an ‘eldest’ in the 15,15,1 case too, as age is not quantized by year (though it is on a census form); 2) any census-taker worth his or her salt would know the difference between a 15 and a 25 year old, so the answer might’ve been known simply with the knowledge that a*b*c=225.

By the way, you left out 25,9,1 and 225,1,1 (arguably, not really a possibility).

Oops. I thought I forgot a few.

But yes, you can immediately rule out 225,1,1, which is why I omitted that one.

Regardless, I still inject a little bit of outside logic into the puzzle and assume that the census taker has to figure there’s at least one adult in the house and that it’s not children living there unsupervised. That really only leaves 75,3,1, 45, 5, 1, and 25,3,3.

At that point, the knowledge that the house number is ambiguous is enough to know that it hs to be 25,3,3 or 15,15,1. And since only one of those combinations has an adult, it’s gotta be 25,3,3 and the question he asked (who’s the eldest?) is moot.

Maybe back in the ’60s twins were less obsessed with knowing who was 30 seconds older? Maybe the reader assumes he wouldn’t be sassing a census taker with such a triviality? Dunno.

I think it’d be a more devious puzzle to use the assumption that there has to be an adult living there to whittle it down. Then you can make the 3rd question any arbitrary thing you want. (“The census taker looked at the house number, then asked the person if he owned a dog/was a communist/watches I Love Lucy/can speak German”). Then the reader focuses on the misdirection of the last question, thinking it’s significant.

Oops. I thought I forgot a few.

But yes, you can immediately rule out 225,1,1, which is why I omitted that one.

Regardless, I still inject a little bit of outside logic into the puzzle and assume that the census taker has to figure there’s at least one adult in the house and that it’s not children living there unsupervised. That really only leaves 75,3,1, 45, 5, 1, and 25,3,3.

At that point, the knowledge that the house number is ambiguous is enough to know that it hs to be 25,3,3 or 15,15,1. And since only one of those combinations has an adult, it’s gotta be 25,3,3 and the question he asked (who’s the eldest?) is moot.

Maybe back in the ’60s twins were less obsessed with knowing who was 30 seconds older? Maybe the reader assumes he wouldn’t be sassing a census taker with such a triviality? Dunno.

I think it’d be a more devious puzzle to use the assumption that there has to be an adult living there to whittle it down. Then you can make the 3rd question any arbitrary thing you want. (“The census taker looked at the house number, then asked the person if he owned a dog/was a communist/watches I Love Lucy/can speak German”). Then the reader focuses on the misdirection of the last question, thinking it’s significant.

You’re assuming he asks that question to disambiguate a ‘same sum’. But there’s another possible reason for the question. If he knows the speaker is the eldest then he knows he can accept it as being supplied by an adult (over 21). It’ll work for 25, 3, 3, but also for 25, 9, 1 and 45, 5, 1.

25, 3, and 3 :)

There’s no reason to ask the “eldest” question for this answer to work, and yet the census taker felt it was necessary.

and he wasn’t asking a 3 year old who could construct advanced word puzzles.

The answer is on page 212: http://books.google.com/books?id=xCoDAAAAMBAJ&lpg=PP1&lr&rview=1&pg=PA212#v=twopage&q&f=false

Hmm, or not. Doh!

Hmm, or not. Doh!

Can’t help myself… Must nit-pick… If there are two fifteen-year olds, there would still be an eldest. Even with twins, there’s one who’s born first. And yes, I know, I need help.

Can’t help myself… Must nit-pick… If there are two fifteen-year olds, there would still be an eldest. Even with twins, there’s one who’s born first. And yes, I know, I need help.

This riddle was posted a few years ago from what I read here :)

By Mark himself, no less. Congrats to being the sort of happy mutant that would snag him being lazy. ;)

By Mark himself, no less. Congrats to being the sort of happy mutant that would snag him being lazy. ;)

This is one of the benefits of having a poor memory – I can entertain myself over and over again with the same material!

Anyone smart enough to solve that in his head ought to have had a better job than census taker in 1960.

But it’s actually not much of a puzzle for the census taker, because he has information (the house number) that the partially blind observer (aka the reader of the puzzle) doesn’t have. I have heard this problem before, though it was with the product 36, not 225. It’s one of my favorites.

Possible ages and house numbers (indicated with #):

1 3 75 # 79

1 15 15 # 31

1 9 25 # 35

3 5 15 # 23

1 5 45 # 51

3 3 25 # 31

5 5 9 # 19

1 1 225 # 227

Possible ages and house numbers (indicated with #):

1 3 75 # 79

1 15 15 # 31

1 9 25 # 35

3 5 15 # 23

1 5 45 # 51

3 3 25 # 31

5 5 9 # 19

1 1 225 # 227

The answer is 3, 3, 25. Since knowing the number of the house wasn’t enough to determine the ages, there must be at least two combinations with that particular sum, whatever it was. In this case, 3, 3, 25 and 1, 15, 15, both add to 31. Once the census taker was informed that he was talking to the eldest, he concluded that there was only person in the house with that age.

“Once the census taker was informed that he was talking to the eldest, he concluded that there was only person in the house with that age.”

Which doesn’t follow. If there are two 15-year-olds, one is always older than the other.

Not necessarily – they don’t have to be related to one another. They could have been born simultaneously to two different mothers :)

Not necessarily – they don’t have to be related to one another. They could have been born simultaneously to two different mothers :)

Mark, it looks like you are repeating yourself. :)

http://www.boingboing.net/2005/08/24/puzzle-of-the-month.html

The answer is actually in the next month’s issue: http://books.google.com/books?id=0yoDAAAAMBAJ&lpg=PA184&dq=popular%20science%20may%201960&pg=PA28#v=onepage&q=popular%20science%20may%201960&f=false

If that census taker can’t determine if the person he is speaking to is 25 or 3, he’s got some pretty big problems.

And the census taker didn’t have to ask the occupant’s occupation, either. Just wrote down ‘wiseass’.

So the puzzle is why a government worker who

just has to write down a few simple numbers from his day’s allotment of houses, hopefully in time to get home before “Gunsmoke” startswould be cheered by some youthful-looking twentysomething that just wants to fuck with him. Right?I missed the “with” first time around in your post and wondered what details I missed in the puzzle.

Isn’t it a requirement to have an legal adult respond to census questions? If so, working through the age/product combinations, wouldn’t that also make the “are you the eldest” question irrelevant?

Asking about the “eldest” does not rule out 15, 15, and 1, since there is still an “eldest” between two 15-year-old twins.

Especially in the case of so-called “Irish twins”.

I grew up in a mostly Catholic town and knew quite a few people with siblings born less than a year apart. Not at all uncommon in the 1950s.

Asking about the “eldest” does not rule out 15, 15, and 1, since there is still an “eldest” between two 15-year-old twins.

45, 10 and 6 months (1/2). It’s a single grandparent with the parent away (college, prison)

I agree that if the problem was based in the real world it would logically make sense that you are more likely dealing with two adults and an infant, rather than a single adult and two twins. However for the census I think infants 0-1 are recorded as such with no discretion of months. And where I live individual house numbers usually end in either a 0 or 5.

Really? Individual house numbers here just end with a number. Not necessarily a whole number either, if the original lot was split up.

Where I live, house numbers are always at least 3 numbers, and almost always 4 numbers.

25, 9 & 1 iff the house number is 35 and no fractional ages.

It is entirely possible to have two children in the family born at precisely the same time. There is no requirement that a family’s children have only one mother; a father can have children by any number of women, limited only by his stamina and persuasiveness, and families are not constructed solely by procreation anyway.

The only answer can be one in which the ages of the two eldest people in the house are close enough to be potentially confusing. Thus, the only possible answer is two fifteen year olds and a one year old

The only answer can be one in which the ages of the two eldest people in the house are close enough to be potentially confusing. Thus, the only possible answer is two fifteen year olds and a one year old

Although, I suspect he must be asking this for some other purpose.

I’d like to know how two fifteen year olds were able to buy a house back then. Kids and their crazy young love, it looks like they can do anything!

I still don’t get what the house number has to do with anything. Is it just to throw you off? The house number can be an arbitrary value, so it tells us nothing

But the census taker knows the house number (standing at the door) but still has to ask another question. So the house number is relevant.

On the census forms the age is reported as your age at your last birthday. Thus, in census-world twins are the same age.

Here’s the “official” answer from the May issue, if anyone wants to see their reasoning (it’s at the very bottom of the page, under “Answer to last month’s P-O-T-M”):

http://books.google.co.uk/books?id=0yoDAAAAMBAJ&lpg=PA184&dq=popular%20science%20may%201960&pg=PA28#v=onepage&q&f=false

Here’s the “official” answer from the May issue, if anyone wants to see their reasoning (it’s at the very bottom of the page, under “Answer to last month’s P-O-T-M”):

http://books.google.co.uk/books?id=0yoDAAAAMBAJ&lpg=PA184&dq=popular%20science%20may%201960&pg=PA28#v=onepage&q&f=false

9, 5, 5 has to be the answer. For any other set of numbers, it would be obvious to the census taker if he was talking to the eldest OR in the case of 15, 15, 1, the answer would have been “no” to his question.

225 1 1 (ridiculous)

75, 3, 1 (obvious)

45, 5, 1 (obvious)

25, 9, 1 (obvious)

25, 3, 3 (obvious)

15, 15, 1 (the answer to the eldest question would be “no”)

9, 5, 5 (the only option left)

The census taker then called Social Services.

Why would the answer be no?

This was the same logic I used. Any solution to this problem seems to require discarding common sense, but of all the explanations I still like this one best — even if a nine-year-old head-of-household is unlikely.

The answer is 25, 3, and 3. There are 7 possibilities of ages, excluding 225 as a possible age, but if you sum up the ages to get the house number, there are only 2 sets of ages that sum to the same house number of 31: 15, 15, 1, and 25, 3, 3. (Here are all the sums: 25+3+3=31, 5+15+3=23, 5+5+9=19, 75+3+1=79, 45+5+1=51, 25+9+1=35, 15+15+1=31.) The answer has to be one of those two, because otherwise by knowing the house number he would know the answer, and wouldn’t have needed to ask that last question about who was eldest. For example, if the house number were 35, he’d know the ages were 25, 9, and 1. So the house number has to be 31, leaving him unsure whether this is a 15, 15, 1 house, or a 25, 3, 3 house. Assuming that one 15 year old can’t be older than another (which is a stupid assumption, but alas), the fact that the door answerer claims to be oldest reveals that the the ages are 25, 3, and 3.

And I just confirmed that this is the same answer and reasoning as in the followup PopSci issue. Hooray.

I hate it when logic puzzlers work on an unstated false premise.

Me too. And this one seems particularly false–I’ve yet to meet a twin who won’t proudly state when asked that s/he is the oldest, even if it’s only by a few seconds. But I imagine asking “Is anyone else in the house the same age as you?” wouldn’t have quite the same ring.

Have you met a twin who wouldn’t proudly state by how many minutes he/she is the eldest? “Why, yes” doesn’t seem to be the answer a twin would give. “By fourteen minutes” does.

So no 15-year-old is older than any other 15-year-old? I think most 15-year-olds would disagree.

So no 15-year-old is older than any other 15-year-old? I think most 15-year-olds would disagree.

I think you folks are forgetting that the census agent had to ask. I honestly don’t think he’d be asking as though it were unclear as to whether the 45 year old person answering the door was the oldest. We have to assume that, visually, the age of the person at the door was unclear enough to not be the oldest.

Regarding the claims that one twin can be ‘the eldest’ – Census-takers record whole ages. For the purpose of the Census, all 15 year olds are the same age.

Regarding the claims that one twin can be ‘the eldest’ – Census-takers record whole ages. For the purpose of the Census, all 15 year olds are the same age.

But do 15 year old twins know that?

Here’s the answer from PopSci:

Answer to P-O-T-M: The Census Taker. You’re looking of three numbers which are all divisors of 225 and which add up to the house number — unknown to you. Now this would be insoluble for a distinct set of numbers, except for the fact that the census taker, who obviously knoew the house number, nevertheless had to ask an additional question: “Are you the eldest?”. He must have had to choose between sets of numbers that not only multiplied to 225, but added to the same sum. There are two such sets: 15, 15, 1 and 25, 3, 3 (Both add to 31). Clearly if there is an eldest, the ages must correspond to the second set.

*headdesk*

“Clearly if there is an eldest, the ages must correspond to the second set.”

So. not. true.

It sounds like something my anti-census relative, on a good day, would tell the census taker.

“I nicked the census man.”

“Now there’s a good boy.”

I’ve read this problem posed in a way that would avoid the problems some people are having. Still a census taker, but instead they’re talking to a mathematician parent and the questions are only about the children:

===

CT: “How many children do you have, and how old are they?”

M: “I have three children. The product of their ages is 72, and the sum of their ages is the same as our house number.”

[CT thinks for a moment]

CT: “I need more information.”

M: “My eldest child likes chocolate ice-cream.”

CT: “Thanks, that’s all I need.”

How old are the children?

===

Since it’s a parent talking about their children and providing an extra piece of information, the whole “but one twin is always older than the other!” objection isn’t as much of an issue.

There is nothing in the clue that requires the occupants to be related, therefore, any of the age combinations that satisfy the arithmetic conditions is correct.

he needs to ask if the person at the door is the eldest because her age must be in more than one solution.

9 appears in more than one answer but as the eldest child in one variant, the middle child in another.

He affirms the solution as 9, 5, 5.

Wolfram Alpha makes finding combos a breeze:

http://www.wolframalpha.com/input/?i=xyz%3D225%2C+x%3E0%2C+y%3E0%2C+z%3E0%2C+z%3Ex%2C+z%3Ey

Looking a bit laterally

What would the answer had been if the person said “no” to the census taker when asked their age?

The other thing is that a one year old baby would often be carried by an adult.

The ages are 15, 15, 1.

The factors of 225 are 5,5,3,3,1 so the the possibles ages are:

A: 15, 15, 1 = 31

B: 25 9 1 = 35

C: 45, 5, 1 = 51

D: 75, 3, 1 = 79

E: 25, 3, 3 = 34

F: 9, 5, 5 = 19

When the person says the the address is that same as the sum then the census taker can rule out B through F because it would be obvious in those cases that the person is the oldest. A is the only case where it is not obvious who is the oldest by minutes or even days because we don’t even know if they are related.