Difference between revisions of "2025 AIME II Problems/Problem 2"

Latest revision as of 13:05, 15 February 2025

Find the sum of all positive integers $n$ such that $n + 2$ divides the product $3(n + 3)(n^2 + 9)$ .

$\frac{3(n+3)(n^{2}+9) }{n+2} \in Z$

$\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in Z$

$\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in Z$

$\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in Z$

$\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in Z$

$\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z$

$\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z$

$\Rightarrow \frac{39}{n+2} \in Z$

Since $n + 2$ is positive, the positive factors of $39$ are $1$ , $3$ , $13$ , and $39$ .

Therefore, $n = -1$ , $1$ , $11$ and $37$ .

Since $n$ is positive, $n = 1$ , $11$ and $37$ .

$1 + 11 + 37 = \framebox{049}$ is the correct answer

~ Edited by aoum

@@ Line 25: / Line 25: @@
 Since <math>n</math> is positive, <math>n = 1</math>, <math>11</math> and <math>37</math>.
-<math>1 + 11 + 37 = \framebox{49}</math> is the correct answer
+<math>1 + 11 + 37 = \framebox{049}</math> is the correct answer
 ～[https://artofproblemsolving.com/wiki/index.php/User:Tonyttian Tonyttian]

2025 AIME II (Problems • Answer Key • Resources)
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