Bunuel wrote:
If x and y are integers, is xy odd?
(1) x^y is odd.
(2) x + y is even.
Great question Bunuel!!
Given: x and y are integers Target question: Is xy odd? Statement 1: x^y is odd There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 1 and y = 1. Notice that 1^1 = 1, and 1 is odd. In this case, xy = (1)(1) = 1, which means the answer to the target question is
YES, xy is oddCase b: x = 2 and y = 0. Notice that 2^0 = 1, and 1 is odd. In this case, xy = (2)(0) = 0, which means the answer to the target question is
NO, xy is not oddSince we can’t answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x + y is evenPro tip: When testing values, always check to see whether you can save some time and reuse values from the previous statement. In this case, we can definitely use them since both pairs of values also satisfy statement 2:
Case a: x = 1 and y = 1. Notice that 1+1 = 2, and 2 is even. In this case, xy = (1)(1) = 1, which means the answer to the target question is
YES, xy is oddCase b: x = 2 and y = 0. Notice that 2+0 = 2, and 2 is even. In this case, xy = (2)(0) = 0, which means the answer to the target question is
NO, xy is not oddSince we can’t answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Notice that I was able to use the
same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: x = 1 and y = 1. In this case, xy = (1)(1) = 1, which means the answer to the target question is
YES, xy is oddCase b: x = 2 and y = 0. In this case, xy = (2)(0) = 0, which means the answer to the target question is
NO, xy is not oddSince we can’t answer the
target question with certainty, the combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent
Thank you very much, Brent for pointing that out to me. After reading your analysis, I see where I made my mistake.