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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 $7$.

David
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5 Answers5

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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:

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

FreezingFire
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    Is there any better way to do this question without using trail and error? – user262291 Apr 07 '16 at 04:51
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    Factorizing helped vastly reduce the space to search by trial and error, that's for sure. – Dan Uznanski Apr 07 '16 at 04:54
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    I cannot say for sure, but there seems to be no general method for such questions. I searched a little and found a page that might point in the right direction: Diophantine equations. Also, on that page, "Hilbert's tenth problem" section says that no algorithm exists for solving a general problem like this. – FreezingFire Apr 07 '16 at 05:07
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Doing in the same method, another solution is

$(18,2,1) $

$(18×2×1)(18+2+1) = 36×21 = 756$

N.S.JOHN
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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.

Roby5
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Yotam D
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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 $xy=756$:

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

Roby5
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maruko
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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 $12*60=720$.

David
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