The remainder,when 7^{103} is divided by 23,i | vedclass

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Question

(A) We need to find $7^{103} \pmod{23}$.By Fermat's Little Theorem,since $23$ is a prime number and $\gcd(7, 23) = 1$,we have $7^{23-1} \equiv 1 \pmod

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) We need to find $7^{103} \pmod{23}$.By Fermat's Little Theorem,since $23$ is a prime number and $\gcd(7, 23) = 1$,we have $7^{23-1} \equiv 1 \pmod{23}$,which means $7^{22} \equiv 1 \pmod{23}$.Now,divide the exponent $103$ by $22$: $103 = 22 	imes 4 + 15$.So,$7^{103} = (7^{22})^4 	imes 7^{15} \equiv 1^4 	imes 7^{15} \equiv 7^{15} \pmod{23}$.We calculate powers of $7 \pmod{23}$:$7^1 \equiv 7 \pmod{23}$$7^2 = 49 \equiv 3 \pmod{23}$$7^3 = 7 	imes 3 = 21 \equiv -2 \pmod{23}$$7^6 = (-2)^2 = 4 \pmod{23}$$7^{12} = 4^2 = 16 \equiv -7 \pmod{23}$$7^{15} = 7^{12} 	imes 7^3 \equiv (-7) 	imes (-2) = 14 \pmod{23}$.Thus,the remainder is $14$.

The remainder,when $7^{103}$ is divided by $23$,is equal to:

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