The least remainder when 17^{30} is divided b | vedclass

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(D) We need to find the remainder of $17^{30}$ when divided by $5$.First,observe that $17 \equiv 2 \pmod{5}$.Therefore,$17^{30} \equiv 2^{30} \pmod{5}

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We need to find the remainder of $17^{30}$ when divided by $5$.First,observe that $17 \equiv 2 \pmod{5}$.Therefore,$17^{30} \equiv 2^{30} \pmod{5}$.By Fermat's Little Theorem,since $5$ is a prime number and $\gcd(2, 5) = 1$,we have $2^{5-1} \equiv 1 \pmod{5}$,which means $2^4 \equiv 1 \pmod{5}$.Now,we can write $2^{30} = (2^4)^7 	imes 2^2$.Substituting the congruence,$2^{30} \equiv (1)^7 	imes 2^2 \pmod{5}$.$2^{30} \equiv 1 	imes 4 \pmod{5}$.$2^{30} \equiv 4 \pmod{5}$.Thus,the least remainder is $4$.

The least remainder when $17^{30}$ is divided by $5$ is

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