Suppose we have four integers, no two of which are congruent $\pmod 6$. Let $N$ be the product of the four integers.

If $N$ is not a multiple of $6$, then what is the remainder of $N$ when $N$ is divided by $6$?
I know that the remainders of the four integers when divided by $6$ are different from each other, and they could be any of $0, 1, 2, 3, 4, 5$. I can try to pair up the remainders in different ways and see what the product of each pair has as a remainder when divided by $6$. For example, if I pair up $1$ and $2$, their product has a remainder of $2$ when divided by $6$. If I pair up $3$ and $4$, their product has a remainder of $0$ when divided by $6$. If I pair up $4$ and $5$, their product has a remainder of $2$ when divided by $6$. If I pair up $5$ and $1$, their product has a remainder of $5$ when divided by $6$. I notice that the only way to avoid getting a remainder of $0$ when multiplying two pairs of remainders is to multiply $1$ and $2$ with $4$ and $5$. That means the four integers must have remainders of $1, 2, 4, 5$ when divided by $6$, in some order. Now I can multiply the four remainders together and see what the remainder of $N$ is when divided by $6$. The product of the four remainders is $1 \times 2 \times 4 \times 5 = 40$. The remainder of $40$ when divided by $6$ is $4$. So the answer is $4$. 