The remainder, when 2^140 is divided by 11, is

The correct answer is Option (2) → 1

Since \( 2^{5}=32=33-1 \), the remainder of \( 2^{5} \) on division by 11 behaves like −1:

    \( \displaystyle 2^{5}\equiv-1\pmod{11} \)

    \( \displaystyle 2^{10}\equiv(-1)^{2}=1\pmod{11} \)

Since 140 = 10 × 14:

    \( \displaystyle 2^{140}=\left(2^{10}\right)^{14}\equiv 1^{14}=1\pmod{11} \)

So the remainder is 1.