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

Revision as of 00:35, 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 $n + 2$ .

$3 \pmod 3 mod n + 2$

$n + 3 \pmod 1 mod n + 2$

$n^2 + 9 \pmod 13 mod n + 2$ since $n^2 - 4 \pmod 0 mode n + 2$ with a quotient $n - 2$

$3(n + 3)(n^2 + 9) \pmod 3(1)(13) mod n + 2 \pmod 39 mod n + 2$

This means 39 = (n + 2)k mod $n + 2$ 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 41: / Line 41: @@
 Let's express the right-hand expression in terms of mod <math>n + 2</math>.
-\equiv 3 mod <math>n + 2</math>
+<math>3 \pmod 3 mod n + 2</math>
-<math>n + 3</math>\equiv 1 mod <math>n + 2</math>
+<math>n + 3 \pmod 1 mod n + 2</math>
-<math>n^2 + 9</math>\equiv 13 mod <math>n + 2</math> since <math>n^2 - 4</math>\equiv 0 mode <math>n + 2</math> with a quotient <math>n - 2</math>
+<math>n^2 + 9 \pmod 13 mod n + 2</math> since <math>n^2 - 4 \pmod 0 mode n + 2</math> with a quotient <math>n - 2</math>
-<math>3(n + 3)(n^2 + 9)</math>\equiv <math>3(1)(13)</math> mod <math>n + 2</math>\equiv 39 mod <math>n + 2</math>
+<math>3(n + 3)(n^2 + 9) \pmod 3(1)(13) mod n + 2 \pmod 39 mod n + 2</math>
 This means 39 = (n + 2)k mod <math>n + 2</math> where k is some integer.
@@ Line 57: / Line 57: @@
 The sum of all possible <math>n</math> is 1 + 11 + 37 = <math>\boxed{049}</math>.
-~Sohcahtoa157 (Note: Someone fix it so that the \equiv becomes a symbol)
+~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

Page

Toolbox

Search