Integer formula problem

Expert replies

Sak32: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Mon Oct 14, 2013 11:06 am

by Sak32 » Sun Nov 10, 2013 2:37 am

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?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

B is the answer.
Please provide clear explanations for the solution of this problem.

Like this post

Source: Beat The GMAT — Problem Solving | Sun Nov 10, 2013 2:37 am

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sun Nov 10, 2013 3:10 am

Sak32 wrote: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?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

B is the answer.
Please provide clear explanations for the solution of this problem.

Hi Sak32

You Need to find the Sum of Even numbers between 99 and 301
That is 100+102+104+......296+298+300
=> 2(50+51+.....149+150)

Sum of 51 to 150 can be calculated as sum of (0-150) - (0-49)

As it is given,the sum of the first n positive integers equals (n(n+1))/2

So 0-150 is (150*151)/2 =>(3*50*151)/2
0-49 is (49*50)/2
So above Highlighted statement can be written as

2{50/2(3*151-49)} = 20,200.

Hence answer is [spoiler]B[/spoiler]

Hope it helps you.

Regards,
Uva.

Known is a drop Unknown is an Ocean

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Nov 10, 2013 4:00 am

For any positive integer n, the sum of first n positive integer 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

The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, the following can be used to calculate the sum of any set of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (biggest - smallest)/interval + 1

The INTERVAL is the distance between one term and the next.
Since we're adding only the even integers here, the interval is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Like this post

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Sun Nov 10, 2013 4:13 am

Using Direct Formula:

Sum of first EVEN Numbers = (n)(n+1)

Sum from 2 to 300 = (150)(151)

sum from 2 to 98 = (49)(50)

So,

(150)(151) - (49)(50)

50 (3 * 151 - 49)
50 (453 - 49)

50(404)
20200

R A H U L

Like this post

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Nov 10, 2013 8:49 am

Sak32 wrote: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?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

Here's one more approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150 - 50 + 1 = 101)

To evaluate 2(50+51+52+...+149+150), let's add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101 x 200 = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review