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 $c$.
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 $3.5, 4, 4.5$.
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.