The least non-negative remainder, when 5^61 is divided by 7, is

The correct answer is Option (3) → 5

Using congruences modulo 7, a small power of 5 turns out to be \(-1\):

    \(5^2 = 25 \equiv 4 \pmod 7\)
\(5^3 \equiv 4 \times 5 = 20 \equiv 6 \equiv -1 \pmod 7\)

Since \(61 = 3 \times 20 + 1\):

    \(5^{61} = (5^3)^{20} \times 5\)
\(\equiv (-1)^{20} \times 5 \pmod 7\)
\(\equiv 5 \pmod 7\)

So the least non-negative remainder is 5.