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Question

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


Solution

Let $$\displaystyle x=6k_{1}+4\: and\: y=6k_{2}+5$$
$$\Rightarrow x^{2}+y^{2}=(6k_{1}+4)^{2}+(6k_{2}+5)^{2}$$
$$=\displaystyle 36k{_{1}}^{2}+48k_{1}+16+36k{_{2}}^{2}+60k_{2}+25$$
$$=\displaystyle 36k{_{1}}^{2}+48k_{1}+36k{_{2}}^{2}+60k_{2}+41$$
Obviously when this is divided by $$6$$, the remainder will be $$5$$.

Mathematics

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