The remainder when $428^{2024}$ is divided by 21 is __________.
Answer (integer)
1
Solution
<p>$$\begin{aligned}
& 428=21 \times 20+8 \\
\Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\
& 8^2=21 \times 3+1 \\
& 8^{2024}=(21 \times 3+1)^{1012} \\
\Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\
& \equiv 1^{2012}(\bmod 21) \\
& 428^{2024} \equiv 1(\bmod 21)
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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