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)
(→‎Solution 1)
 
(17 intermediate revisions by 7 users not shown)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
−
<math>\frac{3(n+3)(n^{2}+9) }{n+2} \in Z</math>
+
<math>\frac{3(n+3)(n^{2}+9) }{n+2} \in \mathbb{Z}</math>
  
−
<math>\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in Z</math>
+
<math>\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in \mathbb{Z}</math>
  
−
<math>\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in Z</math>
+
<math>\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in \mathbb{Z}</math>
  
−
<math>\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in Z</math>  
+
<math>\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in \mathbb{Z}</math>  
  
−
<math>\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in Z</math>
+
<math>\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in \mathbb{Z}</math>
  
−
<math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z</math>
+
<math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in \mathbb{Z}</math>
  
−
<math>\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z</math>
+
<math>\Rightarrow 3(n-2)+\frac{39}{n+2} \in \mathbb{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 \mathbb{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]
 +
 
 +
==Solution 2==
 +
We observe that <math>n+2</math> and <math>n+3</math> share no common prime factor, so <math>n+2</math> divides <math>3(n+3)(n^2+9)</math> if and only if <math>n+2</math> divides <math>3(n^2+9)</math>.
 +
 
 +
By dividing <math>\frac{3(n^2+9)}{n+2}</math> either with long division or synthetic division, one obtains <math>3n-6+\frac{39}{n+2}</math>. This quantity is an integer if and only if <math>\frac{39}{n+2}</math> is an integer, so <math>n+2</math> must be a factor of 39. As in Solution 1, <math>n \in \{1,11,37\}</math> and the sum is <math>\boxed{049}</math>.
 +
 
 +
~scrabbler94
 +
 
 +
==Solution 3 (modular arithmetic)==
 +
Let's express the right-hand expression in terms of mod <math>n + 2</math>.
 +
 
 +
<math>3 \equiv 3 \pmod{n + 2}</math>.
 +
 
 +
<math>n + 3 \equiv 1 \pmod{n + 2}</math>.
 +
 
 +
<math>n^2 + 9 \equiv 13 \pmod{n + 2}</math> since <math>n^2 - 4 \equiv 0 \pmod{n + 2}</math> with a quotient <math>n - 2</math>
 +
 
 +
<math>3(n + 3)(n^2 + 9) \equiv 3(1)(13) \pmod{n + 2} \equiv 39 \pmod{n + 2}</math>
 +
 
 +
This means <math>39 = (n + 2)k \pmod{n + 2}</math> where k is some integer.
 +
 
 +
Note that <math>n + 2</math> is positive, meaning <math>n + 2 \geq 3</math>.
 +
 
 +
<math>n + 2</math> is one of the factors of 39, so <math>n + 2 = 3, 13,</math> or <math>39</math>, so <math>n = 1, 11,</math> or <math>37</math>.
 +
 
 +
The sum of all possible <math>n</math> is <math>1 + 11 + 37 = \boxed{049}</math>.
 +
 
 +
~Sohcahtoa157
 +
 
 +
 
 +
==Video Solution by Mathletes Corner==
 +
 
 +
https://www.youtube.com/watch?v=WYyDe_VyvvQ
 +
 
 +
~GP102
 +
 
 +
 
 +
==See also==
 +
{{AIME box|year=2025|num-b=1|num-a=3|n=II}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 09:37, 20 May 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

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

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

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

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

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

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

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

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

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

Solution 2

We observe that $n+2$ and $n+3$ share no common prime factor, so $n+2$ divides $3(n+3)(n^2+9)$ if and only if $n+2$ divides $3(n^2+9)$.

By dividing $\frac{3(n^2+9)}{n+2}$ either with long division or synthetic division, one obtains $3n-6+\frac{39}{n+2}$. This quantity is an integer if and only if $\frac{39}{n+2}$ is an integer, so $n+2$ must be a factor of 39. As in Solution 1, $n \in \{1,11,37\}$ and the sum is $\boxed{049}$.

~scrabbler94

Solution 3 (modular arithmetic)

Let's express the right-hand expression in terms of mod $n + 2$.

$3 \equiv 3 \pmod{n + 2}$.

$n + 3 \equiv 1 \pmod{n + 2}$.

$n^2 + 9 \equiv 13 \pmod{n + 2}$ since $n^2 - 4 \equiv 0 \pmod{n + 2}$ with a quotient $n - 2$

$3(n + 3)(n^2 + 9) \equiv 3(1)(13) \pmod{n + 2} \equiv 39 \pmod{n + 2}$

This means $39 = (n + 2)k \pmod{n + 2}$ where k is some integer.

Note that $n + 2$ is positive, meaning $n + 2 \geq 3$.

$n + 2$ is one of the factors of 39, so $n + 2 = 3, 13,$ or $39$, so $n = 1, 11,$ or $37$.

The sum of all possible $n$ is $1 + 11 + 37 = \boxed{049}$.

~Sohcahtoa157


Video Solution by Mathletes Corner

https://www.youtube.com/watch?v=WYyDe_VyvvQ

~GP102


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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_2&oldid=248822"