Aerith has an unlimited number of coins of values $5, m-1$, and $m$, and 71 is the greatest value that cannot be made with a combination of these coins. Find all possible values of $m$.
By the Chicken McNugget Theorem, the largest coin value that can't be achieved with 5 and $m-1$ is

$$
5(m-1)-5-(m-1)=4 m-9 \geq 71 \text {, }
$$

so $4 m \geq 80$ and thus $m \geq 20$.

Since 71 is not possible, there cannot be a number less than 71 which is $1(\bmod 5)$ and divisible by $m$ or $m-1$, since then you could add some number of 5 's to it to get 71 . In particular, $m$ and $m-1$ cannot be $1(\bmod 5)$, so $m$ can't be 1 or $2(\bmod$ $5)$.

Since all numbers above 71 are achievable, 76 has to be achievable, and without using any 5 's. If $m \geq 36$, the only way this can happen is if it's either $2 m$ or $2(m-1)$, so $m=38$ or $m=39$. If $m=39$, then there's no way to get 72 . However, if $m=38$, then we can get:

$$
\begin{aligned}
& 72=37+7 \cdot 5, \\
& 73=38+7 \cdot 5, \\
& 74=2 \cdot 37, \\
& 75=37+38, \\
& 76=2 \cdot 38 .
\end{aligned}
$$

We also can't get 71 , so $m=38$ is a solution.

If $m \leq 37$, then $2 m$ also can't be $1(\bmod 5)$, so $m$ can't be $3(\bmod 5)$. Also, $2(m-1)$ can't be $1(\bmod 5)$, so $m$ can't be $4(\bmod 5)$. Thus $m$ has to be a multiple of 5 . But
then $m-1 \equiv 4(\bmod 5)$, so the only way we could get 76 is if it's $4(m-1)$, which gives $m=20$. This works since 71 can't be achieved but

$$
\begin{aligned}
& 72=3 \cdot 19+3 \cdot 5, \\
& 73=2 \cdot 19+20+3 \cdot 5, \\
& 74=19+2 \cdot 20+3 \cdot 5, \\
& 75=15 \cdot 5, \\
& 76=2 \cdot 38 .
\end{aligned}
$$

Thus, the two solutions are $m=20$ and $m=38$.