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Disclaimer: This is not a homework problem. I stumbled on this puzzle on internet and I also have the answer. However I am not able to figure out whats the method to be used to arrive at the answer.

The puzzle is as below:

The product of the ages of David's children is the square of the sum of their ages. David has less than eight children. None of his children have the same age. None of his children is more than 14 years old. All of his children is at least two years old. How many children does David have, and what are their ages?

The answer happens to be 2,4,6,12.

Please suggest ways to solve this problem systematically.

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And how is this is of interest to research mathematicians? I'm voting to close, however amusing the puzzle might be. There are other sites for mathematics puzzles and they can be found in the FAQ. –  José Figueroa-O'Farrill Aug 24 '10 at 14:40
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For each prime p, how many children must have an age which is a multiple of p, given all the other conditions must also hold? Gerhard "Ask Me About System Design" Paseman, 2010.08.24 –  Gerhard Paseman Aug 24 '10 at 14:43
    
The question has been closed, for reasons that José has explained above. –  Pete L. Clark Aug 24 '10 at 14:44
    
@Jose Why shouldn't it be of interest for Researchers? Just because I have used the word puzzle it need not mean it's an casual problem. For example above problem can possibly have a geometric interpretation that might yield the answer very easily Or can be framed as an optimization problem based on constraint? –  Akshar Prabhu Desai Aug 24 '10 at 15:22
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@Akshar, I suppose a school arithmetic problem could have an interpretation that leads to a proof of the Riemann Hypothesis, but I think it would be up to the proposer to convince MO that there is a link. In the absence of any genuine effort from the proposer to make the link to RH (and ungrounded speculation isn't genuine effort) the arithmetic problem would be closed as off-topic. –  Gerry Myerson Aug 25 '10 at 0:31
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closed as off topic by Gjergji Zaimi, Andrew Stacey, Simon Thomas, José Figueroa-O'Farrill, Pete L. Clark Aug 24 '10 at 14:43

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1 Answer

up vote 2 down vote accepted

Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$.

If there are $7$ or more children, then the product is at least $8!>7056$, so there are at most $6$ children.

Furthermore, if there are $6$ children, the sum is at most $84-8-7=69$, so the product is at most $69^2=4761$, but the product is at least $7!=5040$, so there cannot be $6$ children.

Let $S$ denote the sum and $P$ the product. By the AM-GM inequality, we have $\frac{S}{n} \ge \sqrt[n]{P} = \sqrt[n]{S^2}$, so $\frac{S^n}{n^n} \ge S^2$, or $S^{n-2} \ge n^n$, where $n$ is the number of children. This means that $n \ge 3$, since $n^n \ge 1$, which would contradict $n=2$.

I'm not sure there's much else you can do without getting into some messy casework. To rule out $3$ and $5$, you can divide into cases like "Suppose at least two children are older than 11," etc, and use similar arguments regarding sums and products as above. To then find the result given that $n=4$, you'll need to use some divisibility arguments and, yes, a little bit of casework.

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Thanks. I wanted to know if there is any way other than casework. –  Akshar Prabhu Desai Aug 24 '10 at 15:24
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You can reduce the case work somewhat by the hint I gave in a comment to the question. Namely, note that 11 and 13 cannot appear as ages, and refine Davidac897's analysis above to show a little more quickly that there are at most five children. There may be a nice argument to eliminate 7 and 5 from consideration. A quicker approach, if you can find it, is to show every age is a multiple of 2. That would leave very few cases to check. Gerhard "Ask Me About System Design" Paseman, 2010.08.25 –  Gerhard Paseman Aug 24 '10 at 17:51
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