If x is the sum of all odd numbers between 1001 and 2000, inclusive, and y is the sum of all even numbers between 1001 and 2000, inclusive, what is the value of $x-y$ ?

Answer Choices

Correct answer marked below

-1000

-500

Correct

500

1000

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Comments

ArinSep 6

Wait, is there a quick way to do this without adding all numbers? I started listing odds and evens but that's way too slow. There must be a pattern.

MuskaanSep 14

Yeah, think about pairing them up: each odd and the next even. Like 1001 and 1002, then 1003 and 1004, etc. What's the difference each time?

RandoSep 8

Or use the formula for sum of arithmetic sequence. But the pairing trick is faster.

LokeshSep 21

I got -500 but I'm not sure if 1001 is odd or even? 1001 is odd, right? So the odd sum starts with 1001 and ends with 1999? And even starts with 1002 and ends with 2000? Then x has one more term?

RohitSep 22

Yes, 1001 is odd. The number of odds and evens between 1001 and 2000 inclusive: odds from 1001 to 1999 (500 terms), evens from 1002 to 2000 (500 terms). So equal count.

GiacomoJun 13

This is tricky because you might think x-y is 0 but it's not. I fell for that initially. The pairing method makes it clear: each odd is 1 less than the next even, so for 500 pairs, total difference is -500. So answer is B.

SanyaAug 19

I used the sum formulas: x = 500*(1001+1999)/2 = 500*1500 = 750,000. y = 500*(1002+2000)/2 = 500*1501 = 750,500. Then x-y = -500. Works but takes longer. The pairing is smarter.

VionaAug 29

Nice, I did the same but I like the pairing better. Good to have both.

What this question tests

What it tests

Your command of divisibility, primes, factors, parity, and how the GMAT tests properties of integers.

Common trap

Forgetting that 1 is not prime, that 0 is divisible by every integer, or that negative numbers flip your assumptions.

If x is the sum of all odd numbers between 1001 and 2000, inclusive, and y is the sum of all even numbers between 1001 and 2000, inclusive, what is the value of x−yx-yx−y?