The remainder when $a+qn$ is divided by $n$ is equal to the remainder of $a$ divided by $n$ where $a,q,n$ are integers

Question

I can't think of a way to prove this, what I've done so far is take the remainder of $a+qn$ when divided by $n$ again, but is there a better way?

Answer 1

I take it by remainder you mean the part not wholly divisible by n.

Consider that

$$\frac{a + qn}{n} = \frac{a}{n} + q$$

Hence any remainder is determined by the remaining part after wholly dividing $a$ by $n$.

Answer 2

If you write $a+qn=r_1+q_1n$ and $a=r_2+q_2n$, where $r_1,q_1,r_2,q_2\in \mathbb{Z}$ and $0\le r_1,r_2 < n$, then $$qn = (a+qn)-a = (r_1-r_2)+(q_1-q_2)n.$$ Then $n|r_1-r_2$. But $-n < r_1-r_2 < n$, so $r_1=r_2$. Therefore $a$ and $a+qn$ have the same remainder.

Answer 3

Proof:

Let $a=qn+r$

where q is the quotient and r is the remainder. This implies that:

$a \equiv r\mod n $ $\hspace{10cm}$ (1)

Looking at $a+qn$ , it follows that: $\hspace{1cm}$(since $qn \equiv 0 \mod n$) $\hspace{2.1cm}$

$a+qn \equiv a \equiv r \mod n $ $\hspace{8.1cm}$ (2)

Thus the remainder is $r$ in both (1) and (2) and we are done.

Source: https://en.wikipedia.org/wiki/Modular_arithmetic#Properties

Answer 4

If $x$ and $y$ are integers, with $y\ne0$, it can be proved that there exist unique integers $p$ and $r$ such that $$ x=py+r\quad\text{and}\quad 0\le r<|y| $$ Thus we can speak of the remainder $r$ and the quotient $p$.

Now if we divide $a$ by $n$, we have $a=pn+r$ with $0\le r<|n|$. Since $$ a+qn=(p+q)n+r $$ we see that $p+q$ and $r$ satisfy the requirements for being the quotient and the remainder of the division of $a+qn$ by $n$. So $r$ is the remainder of the division of $a+qn$ by $n$.

Answer 5

Hint $\, $ The remainder of $\,a\div n\,$ is the least natural in $\,a+n\Bbb Z$

But $\ a\!+\!nq+n\Bbb Z = a+n(\color{#c00}{q+\Bbb Z}) = a+n\color{#c00}{\Bbb Z},\ $ so it has the same least natural.