The remainder,when 7^{103} is divided by 17 i | vedclass

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(B) We need to find the remainder of $7^{103} \pmod{17}$.By Fermat's Little Theorem,$a^{p-1} \equiv 1 \pmod{p}$ if $p$ is prime and $p 
mid a$.Here,$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) We need to find the remainder of $7^{103} \pmod{17}$.By Fermat's Little Theorem,$a^{p-1} \equiv 1 \pmod{p}$ if $p$ is prime and $p 
mid a$.Here,$p=17$ and $a=7$,so $7^{16} \equiv 1 \pmod{17}$.Now,$103 = 16 	imes 6 + 7$.Therefore,$7^{103} = (7^{16})^6 	imes 7^7 \equiv 1^6 	imes 7^7 \pmod{17}$.$7^2 = 49 \equiv 15 \equiv -2 \pmod{17}$.$7^4 \equiv (-2)^2 = 4 \pmod{17}$.$7^6 \equiv 4 	imes (-2) = -8 \equiv 9 \pmod{17}$.$7^7 = 7^6 	imes 7 \equiv 9 	imes 7 = 63 \pmod{17}$.Since $63 = 17 	imes 3 + 12$,we have $63 \equiv 12 \pmod{17}$.Thus,the remainder is $12$.

The remainder,when $7^{103}$ is divided by $17$ is $..........$.

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