The non-negative remainder when $3^{200} \times 2^{50}$ is divided by 5 is:

4

The correct answer is Option (4) → 4

$3^{200} \equiv (-2)^{200} \pmod{5}$

$(-2)^{200} = 2^{200}$

$2^4 \equiv 1 \pmod{5}$

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

$3^{200} \cdot 2^{50} \equiv 1 \cdot 2^{50} \pmod{5}$

$2^{50} = (2^4)^{12} \cdot 2^2 \equiv 1^{12} \cdot 4 \equiv 4 \pmod{5}$

$\text{ therefore }3^{200} \cdot 2^{50} \equiv 4 \pmod{5}$

The non-negative remainder is 4.