Difference between revisions of "2025 AIME II Problems/Problem 2"

Latest revision as of 00:58, 28 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

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

@@ Line 30: / Line 30: @@
 ~ 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
 ==See also==

2025 AIME II (Problems • Answer Key • Resources)
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