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Mock AIME 2 2006-2007 Problems/Problem 3

Revision as of 19:35, 15 September 2006 by JBL (talk | contribs)

Problem

Let $\displaystyle S$ be the sum of all positive integers $\displaystyle n$ such that $\displaystyle n^2+12n-2007$ is a perfect square. Find the remainder when $\displaystyle S$ is divided by $\displaystyle 1000.$

Solution

If $n^2 + 12n - 2007 = m^2$, we can complete the square on the left-hand side to get $n^2 + 12n + 36 = m^2 + 2043$ so $(n+6)^2 = m^2 + 2043$. Subtracting $m^2$ and factoring the left-hand side, we get $(n + m + 6)(n - m + 6) = 2043$. $2043 = 3^2 \cdot 227$, which can be split into two factors in 3 ways, $2043 \cdot 1 = 3 \cdot 681 = 227 \cdot 9$. This gives us three pairs of equations to solve for $n$:

$n + m + 6 = 2043$ and $n - m + 6 = 1$ give $2n + 12 = 2044$ and $n = 1016$.

$n + m + 6 = 681$ and $n - m + 6 = 3$ give $2n + 12 = 684$ and $n = 336$.

$n + m + 6 = 227$ and $n - m + 6 = 9$ give $2n + 12 = 236$ and $n = 112$.

Finally, $1016 + 336 + 112 = 1464$, so the answer is $464$.

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