HMMT November, 2022

Solution:

We first prove the following lemma, which will be useful later.

Lemma: Let $p$ be a prime and $1 \leq n \leq p-1$ be an integer. Then, $n!(p-1-n)!\equiv(-1)^{n-1}(\bmod p)$.

Proof. Write
$$\begin{aligned} n!(p-n-1)! & =(1 \cdot 2 \cdots n)((p-n-1) \cdots 2 \cdot 1) \\ & \equiv(-1)^{p-n-1}(1 \cdot 2 \cdots n)((n+1) \cdots(p-2)(p-1)) \quad(\bmod p) \\ & =(-1)^{n}(p-1)! \\ & \equiv(-1)^{n-1} \quad(\bmod p) \end{aligned}$$
(where we have used Wilson's theorem). This implies the result.

Now, we begin the solution. Suppose that a prime $p$ divides both $a \cdot n!-1$ and $a \cdot(n+1)!+b$. Then, since
$$-b \equiv a \cdot(n+1)!\equiv(n+1) \cdot(a \cdot n!) \equiv(n+1) \quad(\bmod p)$$
we get that $p \mid n+b+1$. Since we must have $n<p$ (or else $p \mid n!$ ), we get that, for large enough $n$, $n=p-b-1$. However, by the lemma,
$$a(-1)^{b-1} \equiv a \cdot b!(p-1-b)!=a \cdot b!n!\equiv b!\quad(\bmod p)$$
This must hold for infinitely many $p$, so $a=(-1)^{b-1} b!$. This forces all fantastic pairs to be in form $((2 k-1)!, 2 k-1)$.

Now, we prove that these pairs all work. Take $n=p-2 k$ for all large primes $p$. Then, we have
$$\begin{aligned} a \cdot n! & \equiv(2 k-1)!(p-2 k)! \\ & \equiv(-1)^{2 k} \equiv 1 \quad(\bmod p) \\ a \cdot(n+1)! & \equiv(n+1) \cdot(a \cdot n!) \\ & \equiv(p-2 k+1) \cdot 1 \equiv-(2 k-1) \quad(\bmod p) \end{aligned}$$
so $p$ divides the gcd.

The answer is $(1+1)+(6+3)+(120+5)+(5040+7)=5183$.