The remainder when 2^340 is divided by 341 is:

The correct answer is Option (2) → 1

A small power of 2 is just one more than a multiple of 341:

    \(2^{10} = 1024 = 3 \times 341 + 1\)
\(2^{10} \equiv 1 \pmod{341}\)

Raising both sides to the power 34:

    \(2^{340} = \left(2^{10}\right)^{34} \equiv 1^{34} = 1 \pmod{341}\)

So the remainder is 1.

Tip: 341 = 11 × 31 is not prime, so Fermat’s little theorem does not apply to 341; the answer 1 agrees with it here only by coincidence.