Competition problem — Combinatorics

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Combinatorics Difficulty 5.5 AIME, harder Prove it Ukraine

At the New year's party each boy gave each girl one candy and each girl gave each boy one cake. After that each boy ate two of given cakes and each girl ate three of given candies. It turned out that kids had eaten quarter of all given things. What is the maximal number of kids that could participate in this party?

Solution

Let us denote the number of boys at the party by $x$ and the number of girls by $y$ . Then from the problem statement we can get the following equality: $2xy = 4(3y + 2x)$ , which is equivalent to the equation $(x-6)(y-4) = 24$ . Now it is enough to consider all divisors of $24$ and to get all possible $x$ and $y$ , among which the greatest value of $(x+y)$ is achieved in the two cases:
$\begin{cases} x-6=1 \\ y-4=24 \end{cases} \text{ or } \begin{cases} x-6=24 \\ y-4=1 \end{cases}$
or this greatest value is:
$x+y=7+28=30+5=35.$

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