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Bunuel
If p = (n)(5^x)(3^k), is p divisible by 10?

(1) n, x, and k are even.
(2) x > n > k > 0.

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Statement 1: n, x, k are even integers
n=2
x=0
k=0
p = 2 , which is not divisible by 10

n=2
x=2
k=2
p = 450 , which is divisible by 10

Insufficient

Statement 2:
0 < k < n < x
could be 0 < 1/2 < 1 < 2 (p not divisible) or 0 < 2 < 4 < 6 (p is divisible)
Insufficient

Combined, n, x, k is even and 0 < k < n < x
must be 0 < 2 < 4 < 6 or some combination of increasing even integers. so p is always divisible
Sufficient

Answer: C
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Bunuel
If p = (n)(5^x)(3^k), is p divisible by 10?

(1) n, x, and k are even.
(2) x > n > k > 0.

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VERITAS PREP OFFICIAL SOLUTION:

Solution: (C)

Watch out! It’s tempting to think that if n is even and has a factor of 2, the product p will automatically be a multiple of 10. (A number will be divisible by 10 whenever it has factors of 2 and 5.) But this is not the case. Choosing negative values for the exponents here easily shows that the product of these three terms may not be a multiple of 10, even when n is even. So Statement (1) alone is insufficient, and the answer is either (B), (C), or (E). Statement (2) alone doesn’t tell us much more than that the variables take on positive values, and choosing non-integer values will show that this statement is also not sufficient. Combining the two statements, we can conclude that n, x, and k are positive even integers, and in that case the product will indeed always be a multiple of 10, because it will always have at least one factor of 5 and at least one factor of 2. So the correct answer is (C).
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