The least non-negative remainder when \(3^{128}\) is divided by 7 is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
Since 7 is prime and does not divide 3, Fermat’s little theorem gives:
\(3^{6} \equiv 1 \pmod{7}\)
(Directly: \(3^6 = 729 = 7 \times 104 + 1\).) Now \(128 = 6 \times 21 + 2\), so:
\(3^{128} = (3^{6})^{21} \times 3^{2}\)
\(\equiv 1^{21} \times 9 \pmod{7}\)
\(\equiv 2 \pmod{7}\)
The least non-negative remainder is 2.