The remainder when 64^{32^{32}} is divided by | vedclass

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(D) We need to find the remainder of $64^{32^{32}}$ when divided by $9$.First,note that $64 = 9 	imes 7 + 1$,so $64 \equiv 1 \pmod{9}$.Using the prop

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$1$

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(D) We need to find the remainder of $64^{32^{32}}$ when divided by $9$.First,note that $64 = 9 	imes 7 + 1$,so $64 \equiv 1 \pmod{9}$.Using the property of modular arithmetic,$a^n \equiv b^n \pmod{m}$ if $a \equiv b \pmod{m}$.Therefore,$64^{32^{32}} \equiv 1^{32^{32}} \pmod{9}$.Since $1$ raised to any positive integer power is $1$,we have $1^{32^{32}} = 1$.Thus,$64^{32^{32}} \equiv 1 \pmod{9}$.The remainder is $1$.

The remainder when $64^{32^{32}}$ is divided by $9$ is equal to .........................

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