The timer was ticking: 25:01 ... 25:00 ... 24:59. I was about two and a half hours into the AIME test, having solved questions #1-8, 10, and 12. I had two options: either dive into a sequence of geometry bash problems or settle with the mythical #14 NT problem. I was driven to problem 14 as it offered the prospect of an easy point in the face of some intimidating geometry problems and a ticking timer. I took a deep breath and skimmed through the problem statement. It was something to do with the sigma function, \(\sigma(a)\). Great, that was just what I wanted: a multiplicative function \(\rightarrow\) primes. As Andrew Lin had said in my AIME class, the first instinct with this information should be to examine the prime powers by writing out the prime factorization:
$$a=p_1^{e_1}\cdots p_d^{e_d}.$$
Okay, now what did I have to do - find the smallest \(n\) so that \(2021\;\vert\;\sigma(a^n)-1\). Immediately, I used my factorization to simplify this statement:
$$\sigma(a^n)=\sigma(p_1^{ne_1}\cdots p_d^{ne_d})=\prod_{i=1}^{d}(1+p_i+\ldots+p_i^{ne_i})=\prod_{i=1}^{d}\Big(\frac{p_i^{ne_i+1}-1}{p_i-1}\Big),$$
$$\sigma(a^n)\equiv\prod_{i=1}^{d}\Big(\frac{p_i^{ne_i+1}-1}{p_i-1}\Big)\equiv 1 \pmod {2021}.$$
At first, that did not sound too bad, but then the implication struck me: this condition had to hold for all integers! Oh, so this had to hold for all powers of primes, implying that
$$\frac{p^{na+1}-1}{p-1}\equiv 1 \pmod{2021}\Leftrightarrow \frac{p^{na+1}-1}{p-1}-1\equiv\frac{p(p^{na}-1)}{p-1}\equiv 0 \pmod{2021},$$
where \(p\) could be any prime and \(a\in\mathbb{Z_{>0}}\). Now, the path to the solution was crystal clear. I glanced at the timer to ensure that I was at pace with its sprinting digits: 14:17 ... 14:16. Wow! I was solving this problem as fast as the preliminary problems. Now, I set out to see when the divisbility constraint would hold.
Firstly, \(p\) was never going to be divisible by \(2021\) because primes did not have any factors besides \(1\) and itself. So I rewrote the new condition in bold on the middle of my scratch paper:
$$\mathbf{\frac{p^{na}-1}{p-1} \pmod{2021}}.$$
The biggest enemy in a modulo problem is the denominator. I had to find a way of getting rid of the denominator, so I gave into the dreaded idea of casework:
a. If \(p-1\not\mid 43,47\), then I could ignore it completely (a mathematician's dream):
$$\frac{p^{na}-1}{p-1} \equiv 0\pmod{2021}\Rightarrow p^{na}\equiv 1 \pmod{2021}.$$
A classic problem for Euler's theorem:
$$p^{na}\equiv 1 \pmod{2021}\Leftrightarrow \varphi(2021)=42\cdot46\;\vert\;na.$$
Since \(a\) could be any integer, the above condition eventaully reduces to \(42\cdot46\;\vert\;n\).
b. If \(p-1\;\vert\;43,47\), then I could use LTE. Sweet ;)
$$\nu_{43}(p^{na}-1)=\nu_{43}(p-1)+\nu_{43}(na).$$
$$\nu_{47}(p^{na}-1)=\nu_{47}(p-1)+\nu_{47}(na).$$
For \(p^{na}\equiv 1 \pmod{2021}\), we would need to have that \(\nu_{43}(na)\geq 1\) and \(\nu_{47}(na)\geq 1\). Again, \(a\) could be any positive integer, so \(43\cdot47\;\vert\;n\).
This was it: \(42\cdot46\;\vert\;n\) and \(43\cdot47\;\vert\;n\). The smallest \(n\) that fit these conditions is \(42\cdot43\cdot46\cdot47\). Hence, my final answer is \(42+43+46+47=176\). I typed in my answer and checked back through my work. After skimming my scracth paper a few times, I made sure that I did not misbubble any of the previous answers.
I was now headed down the home stretch with one minute left on the exam time. It was only then that I realised that problem #14 asked me to find the sum of the distinct prime factors of \(n\). \(42\) and \(46\) were not prime! I quickly factored
$$n=2^2\cdot3\cdot7\cdot23\cdot43\cdot47 \text{ and got the right answer of } 2+3+7+23+43+47=\boxed{125}.$$
Phew! That was a narrow escape. 0:02 ... 0:01 ... 0:00. As the timer ran out, it not only marked the end of the AIME test but also the end of my suspicions about the myth of problem #14. Despite its easy outlook, the NT problem at that place in the AIME exam always has the knack of throwing off test-takers. I found it rather ironcial that I started the problem by instinctively thinking about primes and ended up nearly making a silly mistake by overlooking that very same idea of primes. Anyways, I scored 11/15 on the AIME test, and my experince with problem 14 makes this test all the more memorable.