The number 6^12 , when divided by 13, gives remainder:

The correct answer is Option (1) → 1

Working modulo 13, a remainder may be replaced by any number that differs from it by a multiple of 13:

    \( 6^2 = 36 = 2(13) + 10 \equiv 10 \equiv -3\ (\text{mod } 13) \)

    \( 6^4 = (6^2)^2 \equiv (-3)^2 = 9 \equiv -4\ (\text{mod } 13) \)

    \( 6^{12} = (6^4)^3 \equiv (-4)^3 = -64\ (\text{mod } 13) \)

    \( -64 = -5(13) + 1 \equiv 1\ (\text{mod } 13) \)

So the remainder is 1.

Shortcut: By Fermat’s little theorem, if p is prime and does not divide a, then \(a^{p-1}\) leaves remainder 1 on division by p. Here p = 13 and \(6^{12} = 6^{13-1}\).