Is the positive integer x even? (1) x is NOT divisible by n!, where n

Official Solution:


Is the positive integer \(x\) even?

(1) \(x\) is NOT divisible by \(n!\), where \(n\) is a positive integer.

If \(n=2\), then the above would mean that \(x\) is NOT divisible 2, so in this case \(x\) must be odd;

If \(n=3\), then the above would mean that \(x\) is NOT divisible 6, so in this case \(x\) could be even (say 2) as well as odd (say 1).

Not sufficient.

From (1) we can deduce the following: \(n \neq 1\), because every number is divisible by 1.

(2) \(x^2\) is divisible by \(n!\), where \(n\) is a positive integer.

If \(n > 1\), then \(n!\) is even and if \(x^2\) is divisible by an even number, then that would mean that \(x\) itself must be even;

If \(n=1\), then \(x\) could be any number, even or odd.

Not sufficient.

(1)+(2) Since (1) ruled out \(n=1\) case then from (2) we are left with \(n > 1\) case, which means that \(x\) is even. Sufficient.


Answer: C