Source: Official Guide for the GMAT 13th Ed. Problem Solving; #172 Official Guide for the GMAT 2015 14th Ed. Problem Solving; #172

5

For any positive integer n, the sum of

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

4 Explanations

2

SONIKA VASANTH

The formula given in the question n(n+1)/2 help us to solve for sum of odd integers?

Dec 10, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Sonika :)

Yes, this formula can be used to solve for the sum of a series of consecutive odd integers, as well. Note that the prompt indicates that the provided formula works "for any positive integer n." As long as the integers are positive, they can be even or odd.

Hope this helps!

Dec 10, 2016 • Reply

2

RAJESH K SARAOGI

Hi, In this question, why can't we take the sum of all 99 integers and sum of all 301 integers and subtract the sum of 301 from sum of 99 and divide this by 2. We could get an approximate.

Dec 8, 2015 • Comment

Cydney Seigerman, Magoosh Tutor

While you're correct that the quantity that we get using the method you've described is approximately the answer we're looking for, the method requires several steps and is quite calculation-heavy. Since calculators are not permitted on the Quant Section, it's best to look for solutions that avoid complicated or time-consuming calculations. I'd definitely recommend working to understand the method Mike explains. It is quite calculation-friendly and once you have a good grasp of the concepts he uses, you'll be able to apply the method to similar questions about sums of integers :) I hope this helps!

Dec 8, 2015 • Reply

2

Hersh Bhadra

We can use properties of evenly spaced sets.

Sum of elements = (Mean) * (Number of elements)

In this case we are asked to calculate for even numbers so we can leave 99 and 301 out of our calculations

Mean = (Last + First) / 2
= (300 + 100) / 2
= 200
(Note: Mean = Median in evenly spaced sets)

Number of elements = {(Last - First) / Increment} +1
= {(300 - 100) / 2} + 1
= 101

Sum of elements = 200 * 101
= 20,200

Mar 26, 2015 • Comment

Jonathan , Magoosh Tutor

Hi Hersh,
That works. Great job. Thank you for the clear solution :)

Mar 29, 2015 • Reply

1

Gravatar Mike McGarry, Magoosh Tutor

Dec 28, 2013 • Comment

YONGXI PENG

How to calculate the number of even numbers/pairs between 100 and 198?

Nov 3, 2014 • Reply

, Magoosh Tutor

To know how many pairs are in a set of numbers, we have a few formulas we can use. Let me start from the beginning and work to the answer. :D

We know 99 is not even and 301 is not even; so ignore both of those as Mike does.

The actual sequence is

100, 102, 104, 106, 108, ..., 296, 298, 300

In the above sequence, the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences, there is a formula to find the number of elements:

{(Last Number - First Number)/d} + 1
So, we need to find out how many elements are in the sequence:

{(300 - 100)/2} + 1 =

{(200)/2} + 1 =

100 + 1 =

101

Now we are ready to find pairs of numbers. We know that to have pairs we should just divide our number in half. But it is odd so we'll make it even by removing one:

101 - 1 = 100

100/2 = 50 pairs

Nov 6, 2014 • Reply

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