Q86 / 180 · COMEDK 2012 · Mathematics

The last digit of $583 !+7^{291}$ is

  • A
    1
  • B
    2
  • C
    0
  • Correct answer
    D
    3

Solution

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 $3 .$

Hence, the correct answer is option: 'D'.

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