M13-13

Official Solution:


Is \((x - y)*(x + y)\) an even integer?

(1) \(x\) and \(y\) are integers.

If both of them are even or both of them are odd, then \((x - y)*(x + y)=even*even=even\) but if one of them is even and another is odd, then \((x - y)*(x + y)=odd*odd=odd\). Not sufficient.

(2) \(x+y\) is even.

If \(x=y=1\) then the answer is YES but if \(x=1.8\) and \(y=0.2\), then the answer is NO, since in this case \((x-y)*(x+y)\) is not an integer at all. Not sufficient.

(1)+(2) Since from (1) \(x\) and \(y\) are integers and from (2) \(x+y\) is even then either \(x\) and \(y\) are both even or both odd. In either case \((x - y)*(x + y)=even*even=even\) is even. Sufficient.


Answer: C
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Can you please explain how x=y=1 gives the equation an even integer... Thanks

If x = y = 1, then (x - y)*(x + y) = 0 = even (recall that 0 is an even integer).
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What is the source of the question. It is the perfect glimpse of what trips us at GMAT.

Source: GMAT Club's Tests.
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This is very good question. My first answer is B as I thought that A and B both must be odd or even and should be enough to answer the question. However, I missed the point that X and Y can be decimal.

I think this is a high-quality question and I agree with explanation.

Why are we not considering negative values of x and y?

The solution shows that the answer is C not mentioning negative or positive nature of x or y. So, (1)+(2) is sufficient no matter whether you consider positive or negative variables.
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I got this wrong since I considered possibility where X and Y can be negative, such as, X=-2 and Y = 4.
Why we are not considering negative number scenarios

(1)+(2) is sufficient no matter whether you consider positive or negative variables.

If x = -2 and y = 4, then (x - y)*(x + y) = -6*2 = -12 = even. The same answer as we got in the solution above.
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I think this is a high-quality question and I agree with explanation. Hi,

Will the GMAT ever ask a question where a negative even integer is a possibility?

Yes.

An even number is an integer which is divisible by 2 without a remainder, so ..., -4, -2, 0, 2, 4, 6, ... are all even numbers.
An odd number is an integer which is NOT divisible by 2 without a remainder, so ..., -3, -1, 1, 3, 5, 7, ... are all odd numbers.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hey Bunel,

we are missing the possibility of X & Y being integers with same value as the question doesn't mention anything about X & Y being distinct. If X and Y are integers with same value for example 2 & 2 Then (X-Y)*(X+Y) will return as 0 which is neither even nor odd.

Hence I think option E is right. What is your take on this?

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).

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