# Difference between revisions of "2021 AIME I Problems/Problem 14"

## Problem

For any positive integer $a, \sigma(a)$ denotes the sum of the positive integer divisors of $a$. Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$. What is the sum of the prime factors in the prime factorization of $n$?

## Solution

We first claim that $n$ must be divisible by 42. Since $\sigma(a^n)-1$ is divisible by 2021 for all positive integers $a$, we can first consider the special case where $a \neq 0,1 \pmod{43}$.

Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$. In order for this expression to be divisible by $2021=43\times 47$, a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$. By Fermat's Little Theorem, $a^{42} \equiv 1 \pmod{43}$. Moreover, if $a$ is a primitive root modulo 43, then $\text{ord}_{43}(a) = 42$, so $n$ must be divisible by 42.

By similar reasoning, $n$ must be divisible by 46, by considering $a \neq 0,1 \pmod{47}$.

We next claim that $n$ must be divisible by 43 and 47. Consider the case $a=2022$. Then $\sigma(a^n) \equiv n \pmod{2021}$, so $\sigma(2022^n)-1$ is divisible by 2021 if and only if $n$ is divisible by 2021.

Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$, then $\sigma(a^n) - 1$ is divisible by 2021 for all positive integers $a$. The claim is trivially true for $a=1$ so suppose $a>1$. Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$. Since $\sigma$ is multiplicative, we have $$\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.$$ We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$, where $$\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in})$$ where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$, so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$. Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$. If $p_i \neq 1 \pmod{43}$, then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$, and if $p_i \equiv 1 \pmod{43}$, then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of 43, by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$, and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$. Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$.

Then the prime factors of $n$ are $2$, $3$, $7$, $23$, $43$, and $47$, and the answer is $2+3+7+23+43+47 = \boxed{125}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 