Let $a\in\mathbb{Z},n\in\mathbb{N}$. If $a$ has a remainder $r$ when divided by $n$, then $a\equiv r\pmod n$
I've done some of these questions before with modulus and division, but I'm unsure of how to approach it with the addition of a remainder
Let $a\in\mathbb{Z},n\in\mathbb{N}$. If $a$ has a remainder $r$ when divided by $n$, then $a\equiv r\pmod n$
I've done some of these questions before with modulus and division, but I'm unsure of how to approach it with the addition of a remainder
The division algorithm tells us that, when an integer $x$ is divided by a natural number $n$, then $x = qn + r$ for some $q \in \mathbb{Z}$ and $0 \leq r < n$.
Now can you show $n|(x-r)$?
By definition, $x \equiv r \pmod{n} \iff n|(x-r)$.