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2025 AIME II Problems/Problem 2: Difference between revisions

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add solution 2, which disregards n+3 since gcd(n+2,n+3)=1
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~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
~ 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


==See also==
==See also==

Revision as of 19:13, 23 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

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

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

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