Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

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