A man has $3$ children. Their ages are $a,b,$ and $c$. The constraints are that the ages add up to some value d and multiply to some value e such that $de=756$. Also, no child is more than $3$ times the age of any other. Solve for $a,b,$ and $c$.

  • $\begingroup$ ages $2,3,9$ has been proposed but is a failed solution since $9$ breaks the rule of having to be no more than $3$ times any other child's age. $\endgroup$ – David James Apr 7 '16 at 7:03
  • $\begingroup$ Shouldn't this be math-tagged instead of a word-problem-tag. I don't really see a story or situation in this question as the tag-info describes: word-problem: "A puzzle that's stated in words, usually in terms of a story or a situation that dresses the problem up." $\endgroup$ – Kevin Cruijssen Apr 7 '16 at 7:03
  • $\begingroup$ The word-problem tag seems appropriate to me but someone can edit this post to add another tag if it is more appropriate. $\endgroup$ – David James Apr 7 '16 at 7:06
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    $\begingroup$ I got 3 possibilities: (1,2,18), (1,3,14) and (2,3,9), all of which break that rule. $\endgroup$ – Dennis_E Apr 7 '16 at 7:06

The solution I have is that

The children are aged 3.5, 4, 4.5


Note that $756 = 2*2*3*3*3*7$, and $de = abc(a + b + c)$.

At first, I was working under the assumption that all the ages had to be integers. Thus, the tricky part is figuring out if the sum is divisible by 7, or if one of the ages is itself a multiple of 7.

If you assume the sum is a multiple of 7, you can rule out 7 itself and anything 28 or bigger. Problem is, both 14 and 21 don't give valid solutions (I get 9,3,2 and 18,2,1 respectively).

Thus, I assumed that one of the ages must be a multiple of 7. You can rule out 14 pretty quickly because of the "no child is more than 3 times the age of any other" rule; the products are already too big even if you make the other two as young as possible. If you assume one child is 7, then the only sums I found to be practical to try were 12 and 18; any higher than that and the products got too big. Unfortunately, I didn't get any solutions for those sums either.

Therefore, I assumed that the ages weren't integers, and I just happened to guess that at least one of the ages was a "and a half". A minute on WolframAlpha later gave me $3.5, 4, 4.5$.

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    $\begingroup$ Nice, to think out of the box like that. It does mean that some of these children have different mothers, haha. $\endgroup$ – Ivo Beckers Apr 7 '16 at 7:11
  • $\begingroup$ Yes very well done. Most people think the ages of children have to be integers but that is not true. Also some people may think the ages have to be at least 1 year apart if not the same but that is also not true (the kids could be adopted or the true biological kids could be ages 3.5 and 4.5 with the age 4 kid being adopted). $\endgroup$ – David James Apr 7 '16 at 7:11
  • $\begingroup$ Also, considering that this answer is correct, there is most certainly polygamy involved... $\endgroup$ – CodeNewbie Apr 7 '16 at 7:13
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    $\begingroup$ Not necessarily polygamy. The problem stated a man has $3$ kids (meaning in his possession and/or he is the guardian of them). It didn't say they were of his creation. He might just be babysitting them for example. $\endgroup$ – David James Apr 7 '16 at 7:15
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    $\begingroup$ Run a computer simulation and step the ages by $0.01$ (years) and see if any other "reasonable" solutions exist. If none then don't bother with anything finer (like $0.001$ years) since that is not how ages are normally tracked but $1/2$ year is reasonable for the solution. Even $1/100$th of a year is "pushing it". I never heard a kid say he/she is $4.51$ years old for example but I have heard "four and a half". $\endgroup$ – David James Apr 7 '16 at 7:24

A "trick" is to do the prime factoring but then when you see that $7$ cannot be one of the ages, "factor" the $7$ a step farther into $3.5$ x $2$. $3.5$ IS one of the correct ages so from there, it is easier to find the other $2$ ages. Also a good observation is that if all the kids were age $4$, that would be a very close "near" solution at $12*64 = 768$. That implies the ages must all be close to $4$ but with one of them likely being $3.5$ so to "balance" things out, it is reasonable to guess that the oldest/eldest child is age $4.5$. That gives the correct solution of ($3.5,4,4.5$) rather easily.

  • $\begingroup$ "A "trick" is to do the prime factoring but then when you see that 7 cannot be one of the ages, "factor" the 7..." You'd still have to prove that the sum of the ages can't be a multiple of 7. $\endgroup$ – Dennis Meng Apr 7 '16 at 15:23

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