The least non-negative remainder when 3^128 is divided by 7 is:

The correct answer is Option (1) → 2

Compute $3^{128} \bmod 7$.

Cycle of powers of $3$ modulo $7$:

$3^{1}\equiv 3$

$3^{2}\equiv 9\equiv 2$

$3^{3}\equiv 3\cdot 2=6$

$3^{4}\equiv 3\cdot 6=18\equiv 4$

$3^{5}\equiv 3\cdot 4=12\equiv 5$

$3^{6}\equiv 3\cdot 5=15\equiv 1$

Cycle length is $6$.

$128 \bmod 6 = 2$

So $3^{128} \equiv 3^{2} \equiv 2 \pmod{7}$.

Final answer: $2$