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

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(C) We need to find $5^{99} \pmod{13}$.By Fermat's Little Theorem,since $13$ is a prime number and $\gcd(5, 13) = 1$,we have $5^{13-1} \equiv 1 \pmod{

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$8$

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(C) We need to find $5^{99} \pmod{13}$.By 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 $99 = 12 	imes 8 + 3$.Therefore,$5^{99} = 5^{12 	imes 8 + 3} = (5^{12})^8 	imes 5^3$.Substituting the congruence,we get $5^{99} \equiv (1)^8 	imes 5^3 \pmod{13}$.$5^3 = 125$.Now,divide $125$ by $13$: $125 = 13 	imes 9 + 8$.Thus,$125 \equiv 8 \pmod{13}$.The remainder is $8$.

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

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