The remainder when 428^{2024} is divided by 2 | vedclass

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(A) We need to find the remainder when $428^{2024}$ is divided by $21$. First,express $428$ in terms of $21$: $428 = 21 	imes 20 + 8$. So,$428^{2024}

Vedclass · Accepted Answer

$1$

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(A) We need to find the remainder when $428^{2024}$ is divided by $21$. First,express $428$ in terms of $21$: $428 = 21 	imes 20 + 8$. So,$428^{2024} = (21 	imes 20 + 8)^{2024}$. Using the Binomial Theorem,$(21 	imes 20 + 8)^{2024} = 21k + 8^{2024}$ for some integer $k$. Now,we find the remainder of $8^{2024}$ when divided by $21$: $8^{2024} = (8^2)^{1012} = 64^{1012}$. Since $64 = 21 	imes 3 + 1$,we have $64 \equiv 1 \pmod{21}$. Therefore,$64^{1012} \equiv 1^{1012} \pmod{21}$. $64^{1012} \equiv 1 \pmod{21}$. Thus,the remainder is $1$.

The remainder when $428^{2024}$ is divided by $21$ is ............

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