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The last digit of $583 !+7^{291}$ is
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Verified Answer
The correct answer is:
3
We have, $583 ! \equiv 0(\bmod 10)$
and $\quad 7^{2}=-1(\bmod 10)$
So, $\left(7^{2}\right)^{145} \cdot 7^{1}=(-1)^{145} \cdot 7^{1}(\bmod 10)$
$\Rightarrow \quad 7^{291}=-7(\bmod 10) \equiv 3(\bmod 10)$
So, $583 !+7^{291}=3(\bmod 10)$
Hence, last digit is $3 .$
and $\quad 7^{2}=-1(\bmod 10)$
So, $\left(7^{2}\right)^{145} \cdot 7^{1}=(-1)^{145} \cdot 7^{1}(\bmod 10)$
$\Rightarrow \quad 7^{291}=-7(\bmod 10) \equiv 3(\bmod 10)$
So, $583 !+7^{291}=3(\bmod 10)$
Hence, last digit is $3 .$
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