The non-negative remainder when 3^200 × 2^50 is divided by 5 is :

The correct answer is Option (4) → 4

Remainders on division by 5 are found by working modulo 5. Both 3 and 2 have fourth powers that leave remainder 1:

    \(3^4 = 81 \equiv 1 \pmod 5\)
    \(2^4 = 16 \equiv 1 \pmod 5\)

So

    \(3^{200} = \left(3^4\right)^{50} \equiv 1^{50} = 1 \pmod 5\)

    \(2^{50} = \left(2^4\right)^{12} \times 2^2 \equiv 1 \times 4 = 4 \pmod 5\)

Multiplying the remainders:

    \(3^{200} \times 2^{50} \equiv 1 \times 4 = 4 \pmod 5\)

The remainder is 4.