The remainder when 3^{100} times 2^{50} is di | vedclass

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(D) We need to find the remainder of $3^{100} 	imes 2^{50}$ when divided by $5$.First,consider $3^{100} \pmod{5}$:$3^2 = 9 \equiv 4 \equiv -1 \pmod{5

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We need to find the remainder of $3^{100} 	imes 2^{50}$ when divided by $5$.First,consider $3^{100} \pmod{5}$:$3^2 = 9 \equiv 4 \equiv -1 \pmod{5}$.So,$3^{100} = (3^2)^{50} \equiv (-1)^{50} \equiv 1 \pmod{5}$.Next,consider $2^{50} \pmod{5}$:$2^2 = 4 \equiv -1 \pmod{5}$.So,$2^{50} = (2^2)^{25} \equiv (-1)^{25} \equiv -1 \pmod{5}$.Since $-1 \equiv 4 \pmod{5}$,we have $2^{50} \equiv 4 \pmod{5}$.Now,$3^{100} 	imes 2^{50} \equiv 1 	imes 4 \equiv 4 \pmod{5}$.Therefore,the remainder is $4$.

The remainder when $3^{100} \times 2^{50}$ is divided by $5$ is

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