Find the remainder of $a^2-3a+11$ knowing that the remainder of dividing $a$ by $18$ is $5$.

Question

I'd like to check if what I did here is ok just in case.

I'm asked to find the remainder of $a^2-3a+11$ knowing that the remainder of dividing $a$ by $18$ is $5$.

What I did:

The problem states that $a=18q+5$ for some $q$, follows that $$a^2-3a+11 = (18q+5)²-3(18q+5)+11 \\ = 18^2q^2+(10)(18)q-3(18)q+21 \\ = 18^2q^2+(7)(18)q+21 \\=18(18q^2+7q)+21$$.

Could I say that the remainder is $21$?. I'm unsure why, but I have doubts about the last step I made.

Answer 1

You're almost there. The remainder must be smaller than the divisor.

$$18(18q^2+7q) + 21 = 18(18q^2+7q) + 18 + 3 = 18(18q^2+7q+1) + 3$$