The remainder on dividing 5^{99} by 11 is | vedclass

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(A) By Fermat's Little Theorem,since $11$ is a prime number and $\gcd(5, 11) = 1$,we have $5^{10} \equiv 1 \pmod{11}$.$5^{99} = 5^{90} \cdot 5^9 = (5^

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) By Fermat's Little Theorem,since $11$ is a prime number and $\gcd(5, 11) = 1$,we have $5^{10} \equiv 1 \pmod{11}$.$5^{99} = 5^{90} \cdot 5^9 = (5^{10})^9 \cdot 5^9$.Since $5^{10} \equiv 1 \pmod{11}$,then $(5^{10})^9 \equiv 1^9 \equiv 1 \pmod{11}$.Now,we calculate $5^9 \pmod{11}$:$5^2 = 25 \equiv 3 \pmod{11}$.$5^4 = (5^2)^2 \equiv 3^2 = 9 \equiv -2 \pmod{11}$.$5^8 = (5^4)^2 \equiv (-2)^2 = 4 \pmod{11}$.$5^9 = 5^8 \cdot 5^1 \equiv 4 \cdot 5 = 20 \equiv 9 \pmod{11}$.Therefore,$5^{99} \equiv 1 \cdot 9 = 9 \pmod{11}$.The remainder is $9$.

The remainder on dividing $5^{99}$ by $11$ is

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