OA from

Veritas:

This problem rewards students who remember that all multiples of the integer x will be x units apart. Statement (1) gives that y is divisible by x−1. This can be proven insufficient with a quick number picking thought experiment. If x=4 is it possible to have a number that is divisible by both x=4 and x−1=3? If you multiply 3 and 4 together to get 12, the answer is yes, there are an infinite number of ways to have a value of y that is divisible by both x and x−1.

Conversely, there are also an infinite number of integers that are divisible by x−1 that aren't divisible by x. In the case of x−1=3, 6, 9, and 21 are all divisible by 3 but aren't divisible by 4. Since this means you can get both a "yes" and a "no" based on the information given in statement (1), statement (1) is insufficient. Eliminate (A) and (D).

Statement (2) gives that y=x!+x−1. This may seem to give information similar to what is given in statement (1), but it is much more sufficient.

Remember that x! can be written as "(x)(x−1)(x−2)(x−3).....(2)(1)".

This means that for any given value for x where x is an integer greater than 1, that x! must be divisible by x−1 as well as x.

If you add x−1 to x!, you get a number that is divisible by x−1 but that cannot be divisible by x. To understand this, it helps to pick numbers.

If x=4, then x!=(4)(3)(2)=24. If you add x−1=3 to that, you get 27, which is still divisible by 3 but that isn't divisible by 4.

This is because multiples of the same number are evenly spaced - all multiples of 4 will be 4 apart, all multiples of 3 will be 3 apart, etc. So for whatever number you pick for x, if you add x−1 you will always get a number that is 1 less than a multiple of x, which means that y can never be divisible by x.

This means that statement (2) gives you a consistent "no" and is sufficient. Answer choice (B) is correct.