Congruence is compatible with multiplication

Statement

Let $n$ be a positive integer, and let $a$, $b$, $c$ and $d$ be integers. If $a \equiv b \pmod n$ and $c \equiv d \pmod n$, then $a c \equiv b d \pmod n$.

Remarks

The proof: $a c - b d = a (c - d) + d (a - b)$, and $n$ divides both terms. With compatibility with addition, this result shows that the multiplication of integers modulo n is well defined. By induction, it gives compatibility with powers.

Explanations

L1 human

To find the remainder of a product, you can multiply the remainders and reduce again. For example, the last digit of $27 \cdot 43$ is the last digit of $7 \cdot 3 = 21$, that is, $1$. This is the case $n = 10$.