<s>[INST] If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
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EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
 [/INST] Sometimes while finding maxima for positive integers you will have to implement some other techniques along with the traditional ones.In your problem you get $xy=\frac{-5(x-10)^2+500}{3}$.Note that $500\equiv 2$(mod 3) and $-5\equiv 1$(mod 3).So we must have $(x-10)^2\equiv 1$(mod 3).So x-10 can't be a multiple of 3.So for minimum value x=11 and thus xy=165
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