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 "2025 AIME II Problems/Problem 2"

(→‎Solution 1)
m (→‎Solution 1)
 
(4 intermediate revisions by 3 users not shown)
Line 17: Line 17:
 
<math>\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z</math>
 
<math>\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z</math>
  
−
Since n+2 is positive,the positive factor of 39 are 1, 3, 13 and 39
+
<math>\Rightarrow \frac{39}{n+2} \in Z</math>
  
−
So x=-1, 1, 11 and 37
+
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>.
  
−
Since x is positive, so x=1, 11 and 37
+
Therefore, <math>n = -1</math>, <math>1</math>, <math>11</math> and <math>37</math>.
  
−
1+11+37=  <math>\framebox{49}</math> is the correct answer
+
Since <math>n</math> is positive, <math>n = 1</math>, <math>11</math> and <math>37</math>.
  
−
～Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]
+
<math>1 + 11 + 37 = \framebox{049}</math> is the correct answer
 +
 
 +
～[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}}

Latest revision as of 13:05, 15 February 2025

Problem

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

Solution 1

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

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

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

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

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

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

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

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

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

～Tonyttian

~ Edited by aoum

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_2&oldid=242772"