Q78 / 180 · KCET 2010 · Mathematics

The least positive integer $\mathrm{x}$ satisfying $2^{2010} \equiv 3 x(\bmod 5)$ is

  • Correct answer
    A
    3
  • B
    4
  • C
    1
  • D
    2

Solution

Since, we know that $2^{2} \equiv-1(\bmod 5) \quad\left(\because 2^{2}+1\right.$ divisible by 5$)$ Now, $\quad 2^{2010}=\left(2^{2}\right)^{1005}$ $\equiv(-1)^{1005}(\bmod 5)$ $\Rightarrow \quad 2^{2010} \equiv-1(\bmod 5)$ $\Rightarrow \quad-1 \equiv 2^{2010}(\bmod 5)$ But $\quad 2^{2010} \equiv 3 x(\bmod 5)$ $\Rightarrow \quad-1 \equiv 3 x(\bmod 5)$ (by transitive relation) $\Rightarrow \quad x=3$

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

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