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 19: / Line 19: @@
 <math>\Rightarrow \frac{39}{n+2} \in Z</math>
-Since n+2 is positive,the positive factors of 39 are 1, 3, 13 and 39
+Since <math>n + 2</math> is positive, the positive factors of <math>39</math> are <math>1</math>, <math>3</math>, <math>13</math>, and <math>39</math>.
-So n=-1, 1, 11 and 37
+Therefore, <math>n = -1</math>, <math>1</math>, <math>11</math> and <math>37</math>.
-Since n is positive, so n=1, 11 and 37
+Since <math>n</math> is positive, <math>n = 1</math>, <math>11</math> and <math>37</math>.
-+11+37=  <math>\framebox{49}</math> is the correct answer
+<math>1 + 11 + 37 = \framebox{049}</math> is the correct answer
-～Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]
+～[https://artofproblemsolving.com/wiki/index.php/User:Tonyttian Tonyttian]
+~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
+==See also==
+{{AIME box|year=2025|num-b=1|num-a=3|n=II}}
+{{MAA Notice}}

2025 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions