M7MBA wrote:Is a positive integer x square of an integer?

A) x has exactly 2 different prime factors
B) x has exactly 12 positive divisors including 1 and x.

The OA is B.

May anyone helps me? How can I prove that the second statement is sufficient? I would be thankful for your answers.

Given: x is a positive integer

We have to determine whether x is a square of an integer.

Question rephrased: Is x a perfect square?

Let's take each statement one by one.

A) x has exactly 2 different prime factors.

Case 1: Say x = 6 (= 2*3). 6 is not a square of an integer. The answer is No.
Case 2: Say x = 36 (= 2^3*3). 36 is a square of an integer. The answer is Yes. No unique answer. Insufficient.

B) x has exactly 12 positive divisors including 1 and x.

Note that an integer is a square if and only if every exponent in its prime factorization is even.

Example: Say x is a perfect square, and its prime factorization gives x = a^m*b^n*c^p; where a, b and c are prime numbers and m, n and p are even integers.

m, n and p must be even integers, else upon taking the square root of x, we cannot get an integer in return.

Note that the number of factors of x = a^m*b^n*c^p is given by (m +1)*(n +1)*(p +1) including 1 and x.

Since m, n and p are even integers, (m +1), (n +1), and (p +1) are odd integers. Thus, the product of (m +1)*(n +1)*(p +1) must be odd. Since (m +1)*(n +1)*(p +1) = 12 ≠ odd, x must not be a perfect square. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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