The remainder when 3^{91} is divided by 80 is | vedclass

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(D) We need to find the remainder of $3^{91}$ when divided by $80$.First,express $3^{91}$ in terms of powers related to $80$.Note that $3^4 = 81 = 80

Vedclass · Accepted Answer

$27$

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(D) We need to find the remainder of $3^{91}$ when divided by $80$.First,express $3^{91}$ in terms of powers related to $80$.Note that $3^4 = 81 = 80 + 1$.We can write $3^{91} = 3^3 	imes 3^{88} = 27 	imes (3^4)^{22}$.Substituting $3^4 = 80 + 1$,we get $3^{91} = 27 	imes (80 + 1)^{22}$.Using the Binomial Theorem,$(80 + 1)^{22} = \sum_{k=0}^{22} \binom{22}{k} 80^k (1)^{22-k} = 1 + 22(80) + \dots + 80^{22}$.Thus,$(80 + 1)^{22} = 80m + 1$ for some integer $m$.So,$3^{91} = 27(80m + 1) = 27 	imes 80m + 27$.When this expression is divided by $80$,the remainder is $27$.

The remainder when $3^{91}$ is divided by $80$ is:

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