Question

Two numbers $x$ and $y$ are such that when divided by $6$ they leave remainders $4$ and $5$, respectively. Find the remainder when $(x^2+y^2)$ is divided by $6$.

Answer 1

Let $x=6k_1+4andy=6k_2+5$
$\Rightarrow x^2+y^2=(6k_1+4{)}^2+(6k_2+5{)}^2$
$=36k{{}_1}^2+48k_1+16+36k{{}_2}^2+60k_2+25$
$=36k{{}_1}^2+48k_1+36k{{}_2}^2+60k_2+41$
Obviously when this is divided by $6$, the remainder will be $5$.