GMATinsight wrote:

Is Integer x divisible by 25?

(1) \(x^2\) is a multiple of 50

(2) \(x^3\) is a multiple of 625

Source:

http://www.GMATinsight.comIf x is an integer, then its square and cube will also be integers.

(1) x^2 is a multiple of 50. Now 50 = 2 * 5^2

This tells us that prime factorisation of x^2 has at least one 2 and at least two 5's.

If x^2 has at least two 5's, then x will have at least one 5. So x will be divisible by 5, but we cannot be sure if it will be divisible by 25.

Eg., if x^2 = 100, then x = 10, not divisible by 25. But if x^2 = 2500, then x = 50, which is divisible by 25.

So

not sufficient.

(2) x^3 is a multiple of 625. Now 625 = 5^4

So this tells us that cube of x has at least four 5's. So x also must have '5' in its prime factorisation.

But if x has only one 5 in its prime factorisation, then its cube will have only three 5's, and this given condition will not be satisfied.

Case rejected Only if x has

at least two 5's in its prime factorisation, will then its cube have at least six 5's, and

then only will its cube be divisible by 625.

This tells us that x must have at least two 5's in its prime factorisation, making it divisible by 25. So

sufficient.

Hence

B answer