DS, please explain

[sanagesh]

Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only

a finite number of nonzero digits?

(1) P>Q

(2) Q=8

[DWarrior]

Statement 1: Insufficient, if P is 4 and Q is 3, it'll be an infinitely repeating decimal.

Statement 2: We know it's guaranteed to terminate because Q is 8. If any digit is 0, it will not make an integer because the max value the next digit can have is 9, which will only carry over 7 to the 0 place, and there's no way to get an integer that way. I don't think a zero to the left of the decimal counts (so if P=1 and Q=8, the number will be .125, not 0.125, and thus will have all non-zero digits). Sufficient

B

[krusta80]

For any quotient to have a terminating decimal, the denominator of the equation must have 2 and/or 5 as its only prime factor(s).

8 = 2^3, so B is the answer