2021 AMC 8

2021 AMC 8 problems and solutions. The test has not been held, and will never be held.


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Unnamed Theorem

I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).

Claim: Given a set $S=\{1,2,3\cdots n\}$ where $n$ is a positive integer, the number of ways to choose a subset of $S$ then permute said subset is $\lfloor n!\cdot e\rfloor.$

Proof: The number of ways to choose a subset of size $i$ and then permute it is $\binom{n}{i}\cdot i!$. Therefore, the number of ways to choose any subset of $S$ is \[\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.\] This is also equal to $\sum_{i=0}^n \frac{n!}{i!}$ by symmetry across $i=\frac{n}{2}$. This is also $n! \cdot \sum_{i=0}^n \frac{1}{i!}.$ Note that $e$ is defined as $\sum_{i=0}^\infty \frac{1}{i!}$, so our expression becomes \[n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).\] We claim that $\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}$ for all positive integers $n$.

Since the reciprocal of a factorial decreases faster than a geometric series, we have that $\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots$. The right side we can evaluate as $\frac{1}{n(n!)}$, which is always less than or equal to $\frac{1}{n!}$. This means that the terms being subtracted are always strictly less than $\frac{1}{n!}$, so we can simply write it as \[\lfloor n!\cdot e\rfloor.\]

Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?

Solution to example: This is equivalent to the Unnamed Theorem for $n=5$, so our answer is $\lfloor 120e \rfloor=\boxed{326}$.

Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is $\boxed{1}$. Note that this contradicts with the answer given by the Unnamed Theorem.

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