Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

 
 
(5 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math>
+
Let <math>S</math> be the sum of all [[positive integer]]s <math>n</math> such that <math>n^2+12n-2007</math> is a [[perfect square]]. Find the [[remainder]] when <math>S</math> is divided by <math>1000.</math>
 +
==Solution==
 +
 
 +
If <math>n^2 + 12n - 2007 = m^2</math>, we can [[completing the square | complete the square]] on the left-hand side to get <math>n^2 + 12n + 36 = m^2 + 2043</math> so <math>(n+6)^2 = m^2 + 2043</math>.  Subtracting <math>m^2</math> and [[factoring]] the left-hand side, we get <math>(n + m + 6)(n - m + 6) = 2043</math>.  <math>2043 = 3^2 \cdot 227</math>, which can be split into two [[factor]]s in 3 ways, <math>2043 \cdot 1 = 3 \cdot 681 = 227 \cdot 9</math>.  This gives us three pairs of [[equation]]s to solve for <math>n</math>:
 +
 
 +
<math>n + m + 6 = 2043</math> and <math>n - m + 6 = 1</math> give <math>2n + 12 = 2044</math> and <math>n = 1016</math>.
 +
 
 +
<math>n + m + 6 = 681</math> and <math>n - m + 6 = 3</math> give <math>2n + 12 = 684</math> and <math>n = 336</math>.
 +
 
 +
<math>n + m + 6 = 227</math> and <math>n - m + 6 = 9</math> give <math>2n + 12 = 236</math> and <math>n = 112</math>.
 +
 
 +
Finally, <math>1016 + 336 + 112 = 1464</math>, so the answer is <math>464</math>.
 +
 
 +
==See Also==
 +
{{Mock AIME box|year=2006-2007|n=2|num-b=2|num-a=4}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]

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 $464$.

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_3&oldid=46092"