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2025 AIME II Problems/Problem 2

Revision as of 00:58, 28 February 2025 by Sohcahtoa157 (talk | contribs) (→‎Solution 3 (modular arithmetic))
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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

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

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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