BTGmoderatorDC wrote: ↑

Sun Jan 26, 2020 1:37 am

If \(N\)\(\) is a positive integer, is \(N^3/4\) an integer?

1. \(N^2+3\) is a prime number.
2. \(N\) is the number of odd factors of \(6\).

A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statement (1) and (2) TOGETHER are NOT sufficient.

OA D

Source: e-GMAT

Given that N is a positive integer, for \(N^3/4\) to be an integer, N must be even so that 4 in the denominator could be canceled.

Question rephrased: Is N even?

Let's take each statement one by one.

1. (N^2 + 3) is a prime number.

Given that N is a positive integer, (N^2 + 3) must be a prime number, greater than 3. Again, since (N^2 + 3) is a prime number, and all prime numbers greater than 3 are Odd, N^2 = Odd – 3 = Even. Again, since N is a positive integer, N^2 must be a perfect square; thus, N must be even. Sufficient.

2. N is the number of odd factors of 6.

The factors of 6 are 1, 2, 3, and 6. Out of these, 1 and 3 are odd, so there are 2 odd factors of 6. Thus, N = 2, even. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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