Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 3"

Latest revision as of 09:50, 4 April 2012

Problem

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

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 $464$ .

@@ Line 13: / Line 13: @@
 Finally, <math>1016 + 336 + 112 = 1464</math>, so the answer is <math>464</math>.
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+==See Also==
+{{Mock AIME box|year=2006-2007|n=2|num-b=2|num-a=4}}
-*[[Mock AIME 2 2006-2007 Problems/Problem 2 | Previous Problem]]
-*[[Mock AIME 2 2006-2007 Problems/Problem 4 | Next Problem]]
-*[[Mock AIME 2 2006-2007]]
 [[Category:Intermediate Number Theory Problems]]

Mock AIME 2 2006-2007 (Problems, Source)
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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 3"

Latest revision as of 09:50, 4 April 2012

Problem

Solution

See Also