Let the number (22)^{2022} + (2022)^{22} leav | vedclass

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(B) Let $N = (22)^{2022} + (2022)^{22}$.For division by $3$:$22 \equiv 1 \pmod{3}$,so $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{3}$.$2022$ is divisi

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$5$

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(B) Let $N = (22)^{2022} + (2022)^{22}$.For division by $3$:$22 \equiv 1 \pmod{3}$,so $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{3}$.$2022$ is divisible by $3$,so $(2022)^{22} \equiv 0^{22} \equiv 0 \pmod{3}$.Thus,$N \equiv 1 + 0 \equiv 1 \pmod{3}$,which means $\alpha = 1$.For division by $7$:$22 \equiv 1 \pmod{7}$,so $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{7}$.$2022 = 7 	imes 288 + 6$,so $2022 \equiv 6 \equiv -1 \pmod{7}$.$(2022)^{22} \equiv (-1)^{22} \equiv 1 \pmod{7}$.Thus,$N \equiv 1 + 1 \equiv 2 \pmod{7}$,which means $\beta = 2$.Therefore,$\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5$.

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by $3$ and $\beta$ when divided by $7$. Then $(\alpha^2 + \beta^2)$ is equal to

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