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The remainder when $3^{100} \times 2^{50}$ is divided by 5 is
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Verified Answer
The correct answer is:
4
Now, $\quad 3^{2} \equiv 4(\bmod 5)$
$\Rightarrow \quad\left(3^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(3^{4}\right)^{25} \equiv(1)^{25}(\bmod 5)$
$\Rightarrow \quad 3^{100} \equiv 1(\bmod 5)$
and $\quad 2^{2} \equiv 4(\bmod 5)$
$\Rightarrow \quad\left(2^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(2^{4}\right)^{12} \equiv 1^{12}(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad 2^{48} \cdot 2^{2} \equiv 4(\bmod 5)$
$\therefore \quad 3^{100} \times 2^{50}=1 \times 4(\bmod 5)$
Hence, remainder is 4 .
$\Rightarrow \quad\left(3^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(3^{4}\right)^{25} \equiv(1)^{25}(\bmod 5)$
$\Rightarrow \quad 3^{100} \equiv 1(\bmod 5)$
and $\quad 2^{2} \equiv 4(\bmod 5)$
$\Rightarrow \quad\left(2^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(2^{4}\right)^{12} \equiv 1^{12}(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad 2^{48} \cdot 2^{2} \equiv 4(\bmod 5)$
$\therefore \quad 3^{100} \times 2^{50}=1 \times 4(\bmod 5)$
Hence, remainder is 4 .
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