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If $$10^{20} +20^{10}$$ is divided with 4 then what would be its remainder?

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Since $$ \begin{eqnarray*} 10^{20}+20^{10} &=&\left( 10^{10}\right) ^{2}+2^{10}10^{10} \\ &=&10^{10}\left( 10^{10}+2^{10}\right) \\ &=&2^{10}5^{10}\left( 2^{10}5^{10}+2^{10}\right) \\ &=&2^{10}2^{10}5^{10}\left( 5^{10}+1\right) \\ &=&4^{10}5^{10}\left( 5^{10}+1\right) \\ &=&4\left( 4^{9}5^{10}\left( 5^{10}+1\right) \right) , \end{eqnarray*} $$

the remainder would be is $0$, because

$$\frac{10^{20}+20^{10}}{4}=4^{9}5^{10}\left( 5^{10}+1\right).$$

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It would be zero.

Namely, $10^{20} = (10^2)^{10}$, and since $4$ divides both $10^2 = 100$ and $20$, it will also divide $100^{10}+20^{10}$.

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As $4$ divides $20 \implies 4$ divides $20^n$ for any integer $n\ge 1$

As $4$ divides $10^2\implies 4$ will divide $10^m$ for any integer $m\ge2$

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