Bunuel wrote:
If x and y are positive integers, what is the remainder when y^x is divided by 2?
(1) y² is an odd integer.
(2) xy is an even integer.
Target question: What is the remainder when y^x is divided by 2?NOTE: If we divide ANY positive integer by 2, the remainder will be EITHER 0 (if the number is even) OR 1 (if the number is odd)
Given: x and y are positive integers Statement 1: y² is an odd integer If y² is odd, then we know that y is ODD
If y is ODD, then y^x will be odd for ANY value of x (since x is a positive integer, y^x = the product of a bunch of y's)
If y^x is odd, then
y^x divided by 2 will leave a remainder of 1Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: xy is an even integer There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 2. In this case,
x^y = 1^2 = 1. So, when we divide x^y by 2, the remainder will be 1Case b: x = 2 and y = 1. In this case,
x^y = 2^1 = 2. So, when we divide x^y by 2, the remainder will be 0Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Answer:
Cheers,
Brent
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Brent Hanneson – Creator of gmatprepnow.com
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