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