The remainder when $428^{2024}$ is divided by 21 is__________
The remainder when $428^{2024}$ is divided by 21 is__________
Solution
$\begin{aligned}
& (428)^{2024}=(420+8)^{2024} \\
& =(21 \times 20+8)^{2024} \\
& =21 \mathrm{~m}+8^{2024}
\end{aligned}$
Now $8^{2024}=\left(8^2\right)^{1012}$
$\begin{aligned}
& =(64)^{1012} \\
& =(63+1)^{1012} \\
& =(21 \times 3+1)^{1012} \\
& =2 \ln +1
\end{aligned}$
$\Rightarrow$ Remainder is 1.
Asked in: JEE Main 2024 (09 Apr Shift 1)
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