The correct option is C 2,5
{xyis terminating decimal}
⇒xy have to be in the form of x2n.5n, them only it will be terminating
for eg. take n=0⇒x→ Terminating
n=2⇒xy.25=x100→ Terminating
If we have any other no. In deceminator, then it will be x on terminating
∴ In order to be terminating, it should have no prime factors than 2&5∴(C)