$3$ children riddle, compute the ages based on information given A man has $3$ children such that their ages add up to some number $x$, and whose ages multiply to some number $y$, such that $xy = 756$.  What are the ages of the $3$ children?
Letting the ages be $a$, $b$, and $c$ of the three children, what we know is the following.
$$a+b+c = x$$
$$abc = y$$
$$xy = 756.$$
How can I go about solving this?  I tried just plugging in some numbers and can get semi close such as ages $3,3,7$ which gives an $xy$ value of $819$.
Also I tried working backwards from $756$ to divide thru  by factors and I got $378,189,63,21,7$, which is why I thought one of the ages might be $7$.
 A: If the ages are constrained to integer values only, then prime factorize $756$ as:
$$756=2\times2\times3\times3\times3\times7$$
Thus we find:
$$ (a+b+c)(abc) = 756$$
Then simply use trial and error, keeping in mind that $(a+b+c)$ must be less than $abc$. I found at least one solution:

$$2,3 \text{ and }9.$$

A: Doing in the same method, another solution is
$(18,2,1) $
$(18×2×1)(18+2+1) = 36×21 = 756$
A: I would start going backwards - what is the integer factorization of $756$?
$$756=2^2\times 3^3\times 7$$
Now we should try to distribute these prime numbers between $x$ and $y$ s.t. the first two conditions apply.
$y$ should have at least $3$ prime numbers, since it is the result of $a\times b\times c$.
Also, since a solution with $1$'s, doesn't work, notice that $y>x$ because $x$ is the sum of three integers, and $y$ is the multiplication of that integer with $4$ at least.
With some direct trial an error (starting from a large $y$ and then reducing it) I got an answer. I can leave it in the comments if necessary.
A: By factorizing $756$ into integers, you can choose all possible pairs of subsets of the prime factors and generate all possible combinations of $x$ and $y$. These turn out to be $24$ in total:

$(1,756)$
  $(2,378)$
  $(3,252)$
  $(4,189)$
  $(6,126)$
  $(7,108)$
  $(9,84)$
  $(12,63)$
  $(14,54)$
  $(18,42)$
  $(21,36)$
  $(27,28)$
  $(28,27)$
  $(36,21)$
  $(42,18)$
  $(54,14)$
  $(63,12)$
  $(84,9)$
  $(108,7)$
  $(126,6)$
  $(189,4)$
  $(252,3)$
  $(378,2)$
  $(756,1)$

Now we can repeat the same procedure by looking for the integer factorization of each $y$ candidate into exactly $3$ terms. Similarly, we could look for the integer partition of each $x$ candidate for $3$ terms, but this is computationally unfeasible. This yields a total of $119$ unique triplets, since we have to augment each factorization with a pair of $1$s since $1$ does not influence the product.
Then, by multiplying together the sum and product of each triplet, there are three distinct solutions that satisfy the $xy=756$:

 $(2,3,9),(2,1,18),(3,1,14)$

A: There is also another "reasonable" solution which is the ages are $3.5$, $4$, and $4.5$.  Some children's ages are expressed in "halves" although it is not as common as "wholes".  ($3.5 + 4 + 4.5$) * ($3.5 * 4 * 4.5$) does indeed equal $756$.  It is interesting to note that this is $12 * 63$ and if you were to guess the ages were $4,4,4$, that would give you $12 * 64$ which is very close to the correct answer of $12 * 63$ so that would imply that the ages have to all be very close to $4$ meaning they are likely not all integers.  Ages $3,4,5$ is also a near solution at $12*60=720$.
