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Integer Properties

Expert replies
Source: — Problem Solving |

by GMATGuruNY » Sun Jul 06, 2014 3:36 am
[email protected] wrote:What is the value of 10 - 8 + 6 - 4 + ... - (-20) ?



8


10


12


14


16
Look for a PATTERN.
Sum of the first 2 terms = 10 - 8 = 2.
Sum of the first 4 terms = (sum of the first 2 terms) + 6 - 4 = 4.
Sum of the first 6 terms = (sum of the first 4 terms) + 2 - 0 = 6.

Note the pattern exhibited by the values in red.
If n is EVEN, then the sum of the first n terms = n.
From 10 to -20 there are a total of 16 even integers.
The sum of the first 16 terms of the sequence = 16.

The correct answer is E.

On the GMAT, a formula for the sequence would be given.
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by GMATGuruNY » Sun Jul 06, 2014 3:41 am
[email protected] wrote:What is the value of 10 - 8 + 6 - 4 + ... - (-20) ?



8


10


12


14


16
We could also determine the sum of every successive PAIR of terms:
10 - 8 = 2.
6 - 4 = 2.
2 - 0 = 2.
-2 - (-4) = 2.
-6 - (-8) = 2.
-10 - (-12) = 2.
-14 - (-16) = 2.
-18 - (-20) = 2.

Since there a total of 8 pairs, and the sum of each pair is 2, the sum of the entire set = 8*2 = 16.

The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

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by GMATinsight » Sun Jul 06, 2014 6:23 am
What is the value of 10 - 8 + 6 - 4 + ... - (-20) ?

8
10
12
14
16
This is the set of 16 even integers with 8 positive and other 8 negative integers

First alternate integer series

10+6+2-2-6-10-14-18 = -32 (Sum of last two terms because rest cancel out)

Second alternate integer series

-8-4-0+4+8+12+16+20 = 28+20 (Sum of last three terms because rest cancel out)

Sum of them all = -32+28+20 = [spoiler]16 Answer Option E[/spoiler]
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by Matt@VeritasPrep » Sun Jul 06, 2014 6:00 pm
[email protected] wrote:What is the value of 10 - 8 + 6 - 4 + ... - (-20) ?
Wow, three answers and nobody used the method I did ... couldn't resist joining in!

Notice that the first eleven terms of our sum just cancel out (I've written this so it's clear, but you don't need to do this on test day, as you know the series will alternate pos/neg):

10 - 8 + 6 - 4 + 2 - 0 + (-2) - (-4) + (-6) - (-8) + (-10)

So we really have 10 - 10 -8 + 8 + 6 - 6 etc. or a sum of 0.

Hence we only have to sum the last four terms:

12 - 14 + 16 - 18 + 20, or

-2 + -2 + 20, or 16
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