The remainder when 5^{99} is divided by 13 is | vedclass

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(B) We need to find the remainder of $5^{99}$ when divided by $13$.Using Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we

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$8$

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(B) We need to find the remainder of $5^{99}$ when divided by $13$.Using Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we have $5^{13-1} \equiv 1 \pmod{13}$,which means $5^{12} \equiv 1 \pmod{13}$.We can write $5^{99} = 5^{12 	imes 8 + 3} = (5^{12})^8 	imes 5^3$.Since $5^{12} \equiv 1 \pmod{13}$,then $(5^{12})^8 \equiv 1^8 \equiv 1 \pmod{13}$.Thus,$5^{99} \equiv 1 	imes 5^3 \pmod{13}$.$5^3 = 125$.Dividing $125$ by $13$: $125 = 13 	imes 9 + 8$.Therefore,$125 \equiv 8 \pmod{13}$.The remainder is $8$.

The remainder when $5^{99}$ is divided by $13$ is

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