devinawilliam83 wrote:
If x and y are positive integers is y odd?
(1) \(\frac{(y+2)!}{x!}\) is an odd integer.
(2) \(\frac{(y+2)!}{x!}\) is greater than 2.
Solution:
Statement One Only:(y + 2)!/x! is an odd integer.
If y + 2 = x, then (y + 2)!/x! = 1, which is odd. Since x is 2 more than y, both x and y have the same parity, i.e., they are both odd or they are both even. Since y can be either odd or even, statement one alone is not sufficient.
Statement Two Only: (y + 2)!/x! is greater than 2.
If y + 1 = x, then (y + 2)!/x! > 2. For example, if y = 1 and x = 2, 3!/2! = 3 > 2; if y = 2 and x = 3, 4!/3! = 4 > 2. Since x is 1 more than y, both x and y have the opposite parity, i.e., x is odd and y is even OR x is even and y is odd. Since y can be either odd or even, statement two alone is not sufficient.
Statements One and Two Together:Using the two statements, we have: (y + 2)!/x! is an odd integer greater than 2. In that case, y + 2 > x. However, if y + 2 is at least 2 more than x, i.e., y + 2 ≥ x + 2 or y ≥ x, then (y + 2)!/x! will always be an even integer. For example, if y = 3 and x = 3, 5!/3! = 5 * 4 is even; if y = 5 and x = 4, 7!/4! = 7 * 6 * 5 is even. As you can see, when y ≥ x, there will always be an even factor left after some of the factors of (y + 2)! cancel out with all the factors of x!. Therefore, in order for (y + 2)!/x! to be an odd integer, y + 2 must be 1 more than x, i.e., y + 2 = x + 1 or y + 1 = x. Furthermore, x must be even. For example, if x = 2, y = 1 and (y + 2)!/x! = 3!/2! = 3; if x = 4, y = 3 and (y + 2)!/x! = 5!/4! = 5. As we can see, when x = y + 1 and x is even, only the factor y + 2 is left when all the other factors of (y + 2)! cancel out with all the factors of x!. Since the factor y + 2 is odd, y is also odd. The two statements together are sufficient.
Answer: C _________________
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