Existence of inverses modulo n

Statement

Let $n$ be a positive integer and let $a$ be an integer. Then $a$ has a modular multiplicative inverse modulo $n$ if and only if $a$ and $n$ are coprime.

Remarks

If $\gcd(a, n) = 1$, Bézout's identity gives integers $x$ and $y$ with $a x + n y = 1$. Then $a x \equiv 1 \pmod n$. Conversely, if $a x \equiv 1 \pmod n$, then $a x + n y = 1$ for some integer $y$, so each common divisor of $a$ and $n$ divides $1$. The extended Euclidean algorithm computes the inverse. For example, $4$ has no inverse modulo $6$, because $\gcd(4, 6) = 2$.

Explanations

L1 human

You can "divide by $a$" modulo $n$ exactly when $a$ and $n$ have no common factor. Modulo $10$, the numbers $1, 3, 7, 9$ have inverses, since $1 \cdot 1$, $3 \cdot 7$ and $9 \cdot 9$ all end with the digit $1$. The numbers $2, 4, 5, 6, 8$ have no inverse modulo $10$.