2025 AIME II Problem 2

Attempt Problem 2 of the 2025 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AIME II solutions, or check the answer key.

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2.

Find the sum of all positive integers

n

such that

n + 2

divides the product

3(n + 3)(n^2 + 9).

Answer

Answer: 49

Concepts:modular arithmetic divisibility factor

Difficulty rating: 1890

Small hint:

Modulo

n + 2,

the product

3(n+3)(n^2+9)

is congruent to a constant: substitute

n \equiv -2

Big hint:

You need

n + 2

to divide

3 \cdot 1 \cdot 13 = 39,

with

n + 2 \ge 3

Solution:

Work modulo

n + 2,

where

n \equiv -2.

Then

\begin{gathered} 3(n + 3)(n^2 + 9) \\ \equiv 3 \cdot 1 \cdot (4 + 9) \\ = 39 \pmod{n + 2}, \end{gathered}

so

n + 2

divides

3(n+3)(n^2+9)

exactly when

n + 2

divides

39.

The divisors of

39

that are at least

3

are

3,

13,

and

39,

giving

n = 1,

11,

and

37.

The sum is

1 + 11 + 37 = 49.