BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

integers

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Mon Mar 02, 2009 4:50 am

integers

by phelps » Fri Jul 23, 2010 1:01 pm
For any positive integer n, the sum of first n positive integers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301.

a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150

Ans b

Please explain
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 1031
Joined: Thu Jul 03, 2008 1:23 pm
Location: Malibu, CA
Thanked: 716 times
Followed by:255 members
GMAT Score:750

by Brian@VeritasPrep » Fri Jul 23, 2010 3:39 pm
Hi Phelps (Michael?):

This is a great question - do you know where it's from? I ask because the first sentence seems out of place although it's a great tip. I haven't seen the GMAT give away a tip like that as part of the question (it almost reads as though it's setting up a sequence problem), and in this case using the tip is probably a lot more cumbersome than anything else.

Sum of consecutive even/odd/overall integers problems have come up fairly commonly in GMAT resources, and it's actually a statistics rule that can help you the most: In evenly-spaced sets, the median of the set will equal the mean. Knowing that means that if you can find the middle number, then you know the average of the set, and can then just multiply the average by the number of terms to get the total:

Average = sum of terms / number of terms

So, using algebra:

Average * number of terms = sum of terms

Here, for even integers, we'll start with 100 and end with 300 (and it's an inclusive set, since both 100 and 300 are within the parameters of 99 and 301), so the middle number is 200.

Now we need to find the number of terms. Because we only care about all of the even numbers, but not the odds, we only care about half of the numbers (every other number). So we can take the range (300-100 = 200) and divide by 2 to get rid of the ones we don't want (the odd ones), giving us a "usable" range of 100 values*.

Because it's an inclusive set, we need to add one to the number of values. It's a rule that you could memorize (inclusive --> add 1; exclusive --> subtract 1), or you could just prove it to yourself with the set 1, 2, 3 (the range is 2, but we know that there are 3 values if we include them all and only 1 if we don't include the endpoints, so you can prove that you add for inclusive and subtract for exclusive).

Adding 1 to the "usable" range of 100 gives us 101 values, and if multiply that by the average of the values, 200, we get:

100*200 + 1 * 200
20000 + 200
20,200 = B


___________________________________________________________________________

*Try this on a number line and it may make more sense. If you only care about, say, the multiples of 3 from 3 to 15, you'd list the numbers as:

3 4 5 6 7 8 9 10 11 12 13 14 15

If you divide by 3, because every third number gives you what you want, you'd have:

1 4/3 5/3 2 7/3 8/3 3 10/3 11/3 4 13/3 14/3 5

The integers that are left are the numbers that are divisible by 3, so you have 5 values, and you can then apply the inclusive/exclusive rules there.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
Top
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 am
Thanked: 135 times
Followed by:7 members

by selango » Fri Jul 23, 2010 8:55 pm
Sum of even integers<=n

If n is odd: (n-1)/2*(n+1)/2

Sum of even intetgers<=99 (98)/2*(100)/2=2450

Sum of even intetgers<=301 (300)/2*(302)/2=22650

Sum of even integers between 99 and 301=22650-2450

20200

Option B
--Anand--
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Fri Jul 23, 2010 9:26 pm
The even integers between 99 and 301 are 100, 102, 104...........300.
Their sum is 100+102+104.......+300 = 2*(50+51+........+150).

First calculate the sum of all integers between 1 and 150 inclusive which is 150*(151)/2 = 11325.

Then calculate the sum of integers from 1 to 49 inclusive which is 49*50/2 = 1225.

The difference of the two sums which is 11325 - 1225 =10100 gives the sum of integers between 50 and 150 inclusive.

Now 2*10100 = 20200 gives the value of 2*(50+51+........+150).

The correct answer is hence b.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Post new topic Post Reply
• Page 1 of 1