Problem Solving

Hi everyone, I have come across this question from gmathacks.com

Which of the following expresses the sum of the first 50 positive even integers?

A (2)(50)
B (25)(100)
C (25)(102)
D (50)(100)
E (50)(102)


The answer is C. The way I approached this question was under the assumption that zero is a positive integer.

So I figured.

The first even consecutive integer is 0 and the last one is 100. The mean would be (50). I also figured the number of terms would be 50 (since they say this is the number of terms in the question prompt). Therefore, the sum would be (50)(50)

But...The solution says

"the first even consecutive integers between 2 and 100 inclusive. The mean of those 50 umbers is equal to the median. So the mean is 51. We know there are 50 terms and the average is 51, so the sum is (50)(51)...This would then equal (25)(102)"


Question:

I do not understand why we are not counting 0 as an even integer. I though 0 was an even integer. Could someone please let me know if this is a poorly written question or if I just should not assume that 0 is an even integer on the gmat.

Thanks a lot

0 is an even integer, BUT it is NOT a POSITIVE integer, because the definition of "positive" is literally "greater than 0." (Likewise, the definition of negative is "less than 0," so 0 is also not negative.)

One more thing to watch out for -- you didn't actually make a mistake in this regard, because you used the stated fact that we should be dealing with 50 integers, but keep in mind that if we DID list all the even integers from 0 to 100, there would actually be 51 of them (just as if you listed ALL the integers -- even and odd both -- from 0 to 100, there would be 101).

Wow!

Thank you so much for you very very quick response. I just did not read into the question carefully!

Takeway:

O is neither positive or negative.
0 is always an even integer
0 would never be considered a positive even integer or a negative even integer.


Simple enough

Perfect phrasing!

2 + 4 + 6 +...+ 96 + 98 + 100
(adding 2 and 100, 4 and 98, etc. - 25 pairs)
= 102 + 102 + 102 +.... (25 times)
= 102*25

Isn't this essentially the same problem Gauss is said to have solved when he was ten? : )

2 + 4 + 6 + ..+ 100
take 2 common..
2 ( 1 +2 +3 + ...+50)

formula of sum of 1 to n = n*(n+1)/2..
= 2 * (50) * (51) /2
= 102 * 25