China Team Selection Test

For any positive integer $d$, we aim to prove that there are infinitely many positive integers $n$ such that $d(n!) - 1$ is a composite number.

### Case 1: $d = 1$

Assume for the sake of contradiction that for all sufficiently large $n \in \mathbb{N}$, $n! - 1$ is prime. Define $p_n = n! - 1$ to be this prime for all $n \geq N$ for some $N \in \mathbb{N}$. Notice that $p_n > n$. Consider $n = p_n - k_n$ for some $k_n \in \mathbb{N}$.

In $\mathbb{F}_{p_n}$, by Wilson's Theorem, we have:
$$1 = (p_n - k_n)! = \frac{(p_n - 1)!}{(p_n - 1)(p_n - 2) \cdots (p_n - k_n + 1)} = \frac{-1}{(-1)^{k_n - 1} \cdot (k_n - 1)!} = (-1)^{k_n} \cdot \frac{1}{(k_n - 1)!}$$
Thus,
$$(k_n - 1)! \equiv (-1)^{k_n} \pmod{p_n}$$
If we pick $n \equiv 1 \pmod{2}$, then $k_n \equiv p_n - n \equiv 0 \pmod{2}$ as $p_n$ is odd, and therefore $p_n \mid (k_n - 1)! - 1$.

Since $k_n > C$ for all sufficiently large $n \in \mathbb{N}$ for any $C \in \mathbb{N}$, and considering the pigeonhole principle, we conclude that $p_n = (k_n - 1)! - 1$. Hence, $p_n = n! - 1$ implies $p_n - n - 1 = k_n - 1 = n$, which means $p_n = 2n + 1$. This implies $2n + 1 = n! - 1$ for infinitely many $n \in \mathbb{N}$, which is impossible for large $n$. Therefore, $n! - 1$ cannot be prime for all sufficiently large $n$.

### General Case: $d \geq 2$

Assume for the sake of contradiction that there exists some $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}, n \geq N$, $d \cdot n! - 1$ is prime. Define the function $f: \mathbb{N} \setminus [d] \to \mathbb{N}$ as:
$$f(m) = (d-1)! \cdot (d+1)(d+2) \cdots (m-1) \cdot m + (-1)^m$$

Let $p_m$ be the smallest prime divisor of $f(m)$. For $m \geq 2d$, $p_m > m$ because $\gcd(m!, f(m) - 1) \mid d$ but $d \mid f(m)$ implies $\gcd(f(m) - 1, d) = 1$. Thus, $p_m \mid f(m) + (-1)^m$, meaning:
$$0 \equiv d(f(m) + (-1)^m) \equiv m! + d(-1)^m \pmod{p_m}$$

Using Wilson's Theorem in $\mathbb{F}_{p_m}$:
$$(p_m - m - 1)! = \frac{(p_m - 1)!}{(p_m - 1)(p_m - 2) \cdots (p_m - m)} = \frac{-1}{(-1)^m \cdot m!} = \frac{-1}{(-1)^m (-(-1)^m) \cdot d} = \frac{1}{d}$$
Thus,
$$p_m \mid d(p_m - m - 1)! - 1$$

For sufficiently large $m$, $p_m = d \cdot (p_m - m - 1)! - 1$. By the pigeonhole principle, $k_m = p_m - m$ can appear finitely many times. Therefore, $k_m \geq N + 1$ implies $d \cdot (p_m - m - 1)! - 1$ is prime, leading to a contradiction.

Hence, for any positive integer $d$, there are infinitely many positive integers $n$ such that $d(n!) - 1$ is a composite number.

The answer is: $\boxed{\text{There}}$ are infinitely many positive integers } n $\text{}$ such that } d(n!) - 1 $\text{}$ is a composite number.}}