The remainder left when 8^{2n} - 62^{2n+1} is | vedclass

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(A) We need to find the remainder when $8^{2n} - 62^{2n+1}$ is divided by $9$.Note that $8 \equiv -1 \pmod{9}$ and $62 \equiv 8 \equiv -1 \pmod{9}$.Su

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) We need to find the remainder when $8^{2n} - 62^{2n+1}$ is divided by $9$.Note that $8 \equiv -1 \pmod{9}$ and $62 \equiv 8 \equiv -1 \pmod{9}$.Substituting these into the expression:$8^{2n} - 62^{2n+1} \equiv (-1)^{2n} - (-1)^{2n+1} \pmod{9}$.Since $2n$ is an even integer,$(-1)^{2n} = 1$.Since $2n+1$ is an odd integer,$(-1)^{2n+1} = -1$.Therefore,$1 - (-1) = 1 + 1 = 2$.Thus,the remainder is $2$.

The remainder left when $8^{2n} - 62^{2n+1}$ is divided by $9$ is:

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