This is one of my all time favorite problems to give to students.
Determine the ages of Alfie's children from analyzing the following conversation.
Alfie: Can you guess the age's of my children? The product of their ages is 36.
Bernice: That's not enough information.
Alfie: The sum of their ages is the same as our street address.
Bernice: I'll need a little more info.
Alfie: My oldest is a girl.
Bernice: Okay, I've got it.
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18 comments:
This is killing me. I'm hoping that the ages must be whole numbers. I've got {36,1,1}, {4,3,3(,1)}, {18,2,1}, or {9,2,2(,1)}. I've ruled out any of the 2 number combos because the suffix '-est' in 'oldest', if I remember correctly, suggests more than 2. I've also ruled out combos like {3,3,2,2} because there's no 'oldest' in that group. There's got to be something about the street address clue that I'm missing. ARG!
This sure looked familiar and I googled key phrases and was led right back to yofx! Didn't figure it out then, couldn't figure it out now. So, annoyed, I kept googling. And am now in possession of the solution. A couple of remarks:
1. a totally unnecessary correction: the word "ages" appears four times in the post and has an apostrophe in one of them when it shouldn't.
2. the site I found tells you how many kids Alfie has; you don't. Knowing this is useful. There is only one solution to this problem and you don't really need to tell us how many kids he has. It's just a little harder not knowing this.
3. this is a very interesting problem. if you're like me, after about 2 minutes thinking about this, both times, I was convinced you were asking us a totally bogus question which could not be solved. such questions are easy to create and are very nasty indeed.
4. this is *not* one of those questions.
Jim identified the crux of the problem. How do you make use of the second clue? That's what makes this problem really good.
Lee, I'm curious where you found the link. Will you post it in the comments in a couple of days?
James - You're close. There are a few triples you're missing, namely {1,6,6}, {2,3,6}, {1,3,12} and {1,4,9}. If you sum the ages in each group you'll find two of the groups have the same sum. Bernice knows Alfie's address.
Possible tricks that I see...
The word "ages" has an apostrophe in one instance and not in any other.
There can be anything short of an infinite number of children that are one year in age. Twins? Triplets? Dodeculplets? Infinity-minus-one-lets?
Could be.
So far I've got these candidates but no answer.
{36,1,1(,1)}
{18,2,1(,1)}
{12,3,1(,1)}
{9,4,1(,1)}
{9,2,2(,1)}
{4,3,3(,1)}
{6,3,2(,1)}
"The sum of their ages is the same as our street address."
Maybe the sum of their ages SOUNDS like the street address?
"OUR street address"?
Hmm.
Pronoun trouble.
Whose street address?
Alfie and the kids?
Do they live in the same place?
Hendree, I've got a team of geeks working on this and we're really angry with you.
** if you don't want any help, don't read my post **
The fact that there is seemingly an indeterminate number of one year olds is really killing me. here are the possible sets after clue 1 and their sum range (whole ints):
∑{36,1(,1...)} >= 37
∑{18,2(,1...)} >=20
∑{12,3(,1...)} >= 16
∑{9,4(,1...)} >= 13
∑{9,2,2(,1...)} >=13
∑{6,3,2(,1...)} >= 11
∑{4,3,3(,1...)} >= 10
∑{3,3,2,2(,1...)} >= 10 (but actually clue 3 removes this possibility)
The first thing is to identify all possibilities which you have done. (And the fact there are an infinite number doesn't actually mean there is no answer.)
Sorry, I forgot the second thing which is to figure out why the information provided to Bernice is decisive amongst the infinite possibilities. She knows things that you don't know that help her figure out the ages, but the fact that you know she knows those things helps you figure them out.
** possible spoiler - do not continue reading if you don't want help **
OK, I still say this question is not solvable, but if it were solvable, the set that fits the criteria is {4,3,3(,1...)}. There is no way to determine if there is or is not a one year old, regardless of whether or not Bernice knows the # of children.
As Tinyc Tim said, there are only 2 pair of sets which have the same sum:
∑{9,4(,1...)} >= 13
∑{9,2,2(,1...)} >=13
and
∑{4,3,3(,1...)} >= 10
∑{3,3,2,2(,1...)} >= 10
Once Bernice knows that their ages sum up to the street address, the only reason she would still need more info is if the answer was in one of these sets. If the street address had been 37, for example, well, there's only one set that solves that criteria, so she wouldn't ask for more info.
The final clue about the oldest being a girl means that the last set above is no longer an option.
That leaves these possibilities:
∑{9,4(,1...)} >= 13
∑{9,2,2(,1...)} >=13
∑{4,3,3(,1...)} >= 10
Again, if Bernice knows the house number, the solution can't be one of the sets which add up to >= 13, because at that point she'd still need more information. but she doesn't. She has the answer now, so that leaves just the final set:
∑{4,3,3(,1...)} >= 10
But the thing is, Bernice still doesn't know for sure how many one year olds there are, and neither do we. All we can say is that the number of one year olds is a constant ... and that's what threw us for a loop for a long time. We forgot that the number of 1yr olds is not in a state of flux when comparing one set to another. So if you say there's one 1yr old, there's one 1yr old for all the sets, and the sets which have the same sum stay the same.
So, I'm going with {4,3,3(,1...)} but still feeling uneasy as I don't see a way to rectify those darn 1's.
Jim and I have resigned ourselves to the following assumptions:
-Bernice MUST know the address and therefore the sum. Without this she cannot ever solve the problem.
Knowing this sum/address, and that there is a single 'oldest' (clue 3) lets her eliminate any possibilities which match the address but have two elder children with the same age. Otherwise she couldn't ever get an answer.
I'd like to preface all this by saying that this is a *very* tricky
problem. This is my best shot at trying to wrap this up.
To summarize what we all now understand:
Before the 3rd clue, Bernice is in doubt. This is due to the fact that two of the possibilities have equal sums. This sum must be >=10 or >=13.
Case: sum is >=10
4,3,3(,1...)
3,3,2,2(,1...)
Only the 1st entry has an oldest kid.
Case: sum is >=13
3,3,2,2,1,1,1(,1...)
4,3,3,1,1,1(,1...)
9,2,2(,1...)
6,6,1(,1...)
9,4(,1...)
Only the 2nd, 3rd and 5th entries have an oldest kid.
*If* the number of the house across the street is 10, 11 or 12, the kids' ages are 4,3,3 or 4,3,3,1 or 4,3,3,1,1 respectively.
*If* the number of the house across the street is >=13, the kids' ages must be 9,4 or 9,2,2(,1...) or 4,3,3,1,1,1(,1...). But ... since the 3rd clue brings Bernice over the top, it must be that the ages are 9,2,2 *and* that she knows how many kids Alfie has and that it's 3. 9,4 and 9,2,2 and 9,2,2,1 and 4,3,3,1,1,1 and 4,3,3,1,1,1,1 contain the same amount of "information." Bernice throws out the 6,6,1 and accepts the 9,2,2 because she knows he has 3 kids.
A *much* simpler problem is one where the number of kids Alfie has is known. The link I found that discusses this problem is at
http://mathforum.org/library/drmath/view/58493.html
@tiny: Wonderful proof! I like the approach you took! Still I have some questions lingering. How can we say anything about the house numbers when the number of one year olds is indeterminate? Your proof uses 2 conditions regarding the houses across the street to reach the conclusion, but if the number of one year olds is indeterminate, Alfie's house number could be anything, and so could the neighbors'. We're using 10 and 13 because they're our low boundaries, but those one year olds spell trouble.
My solution makes the assumption that Bernice knows Alfie's address. Yours makes the assumption that Bernice knows the address, the number of kids, and that the houses on the street use up all numbers in succession (which is not always the case http://en.wikipedia.org/wiki/House_numbering#United_States_and_Canada).
Hendree said that there *is* information that B knows that we don't know, but that knowing she knows leads us to the answer, but I really think we need to know exactly WHAT data she knows without actually being told what that data is in order to solve this problem conclusively.
James -
So much for my "wrap up" being the last word! Thanks for your kind
remarks about my proof. I will return the compliment by saying your and Ken's seeing that 1-year-olds make this problem more difficult was a great insight. I would be interested to know if Hendree or his students considered this possibility.
I don't know where I came up with the "house across the street" idea. "Our street address" clearly means Alfie's house number. In any case, I think you might have misinterpreted me and thought my solution(s) somehow depended on the way houses are numbered on each side of a street. This was not my intention. All I really know is that Bernice knows *some* house number (singular). It doesn't really matter whose house number she knows (but now we all know it is Alfie's).
In any case, I think I want to stick with my conclusions, namely that there are only 4 possible answers; 3 of the possibilities come from Alfie's house being numbered 10, 11 or 12 and the fourth if it's numbered 13.
- Lee
This is the way my students and I thought this problem out. The existence of one year-olds does trouble the problem because it renders the second clue useless. If you can stick a bunch of 1 year-olds on then every sum is going to be non-unique. It kind of ruins the cleverness of the problem. If you do the problem the clever way, clue two means the sums are non-unique so as Jim said it could only be one of the 13's: (9,4) (9,2,2) or one of the 10's: (3,4,3) (2,2,3,3). The third clue tells us it has to be the 10's because the hint that there is an 'oldest' allows Bernice to decide. So the answer would be (3,4,3).
Now when you think about the problem with the 1 year-olds. The second clue losses its power to eliminate options. Essentially it makes every sum higher than 10 non-unique because we can just add on 1 year- olds to (3,4,3). What is still important is that clue three (that there is a single oldest) allows Bernice to decide. That is still true if the sum is 10. Bernice can use the last clue to decide between (4,3,3) and (3,3,2,2)
Let's assume the sum (house number) is higher, say 11. Including 1 year-olds, these are the options
(6,3,2)
(4,3,3,1)
(3,3,2,2,1)
The last clue still allows us to throw out (3,3,2,2,1), but it is no longer “enough” info to decide between (6,3,2) and (4,3,3,1). It is easy to see that for any sum higher than 11 the same is true—hint 3 is no longer enough info to decide. Therefore the house number has to be 10, and the kids ages are (4,3,3). Let me know if that satisfies the armies of angry geeks
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