(13)^{507} when divided by 9 leaves the remai | vedclass

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(A) We need to find the remainder when $(13)^{507}$ is divided by $9$.$(13)^{507} = (9 + 4)^{507}$.Using the Binomial Theorem,$(9 + 4)^{507} = \sum_{k

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$1$

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(A) We need to find the remainder when $(13)^{507}$ is divided by $9$.$(13)^{507} = (9 + 4)^{507}$.Using the Binomial Theorem,$(9 + 4)^{507} = \sum_{k=0}^{507} \binom{507}{k} 9^{507-k} 4^k$.All terms except the last term (where $k=507$) contain a factor of $9$.Thus,$(13)^{507} \equiv 4^{507} \pmod{9}$.Now,$4^1 \equiv 4 \pmod{9}$,$4^2 = 16 \equiv 7 \pmod{9}$,$4^3 = 64 \equiv 1 \pmod{9}$.Since $4^3 \equiv 1 \pmod{9}$,we have $4^{507} = (4^3)^{169} \equiv (1)^{169} \equiv 1 \pmod{9}$.Therefore,the remainder is $1$.

$(13)^{507}$ when divided by $9$ leaves the remainder :-

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