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.

_________________

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