Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: sum of positive integers [#permalink]
10 Jun 2008, 12:12

D (which is clearly established by now;) )

The way I figured it was this way.

Multipe of 5 is 5, 10, 15, 20, etc. Those not divisible by 2 are 5, 15, 25, 35, etc.

So, there will be 1 number to include for every 10, so 1000 numbers / 10 = 100 numbers to include in the sum.

If you look at 5 + 995 = 1000, 15 + 985 = 1000, 25 + 975=1000, 35 + 965 = 1000...495+505=1000, You do this, then you have 50 pairs that add up to be 1000, so 50 * 1000 = 50,000 (i.e. D)

I tend to see things in patterns rather than formulas. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: sum of positive integers [#permalink]
02 Jul 2008, 03:48

There are 200 no's divisible by 5 upto 1000 and out of this only 100 nos are not divisible by 2. So the sum we need is (5+15+....+995) or 5(1+3+5+....199). The sum of consequtive n integers is n^2. Therefe for sum of 100 odd nos is 100^2.

Re: sum of positive integers [#permalink]
03 Feb 2011, 07:41

puma wrote:

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

a) 10050 b) 5050 c) 5000 d) 50000 e) 55000

My approach:

Step 1: Find out the sum of all integers divisible by 5. Number of such terms= \(\frac{(1000-0)}{5} + 1= 201\). Sum of all such integers= \(201*500\) (using the sum formula of terms in Arithmetic Progression).. (i) Step 2: Find out the sum of all integers divisible by 10: Number of such terms= \(\frac{(1000-0)}{10}= 101\). Sum of all such integers= \(101*500= 101*500\) .. (ii)

Now, the multiples of 10 are the ones which are divisible by both 5 and 2. Therefore, we need to subtract (ii) by (i) to arrive at the terms divisible only by 5 and not by 2

Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
13 Jan 2014, 18:25

100 possibilities, evenly spaced by 10. Since its an evenly spaced set, the median (500) is also the mean. Find the median: will be the average between the 50th observation (495) and the 51st observation (505), which is 500. Average * number of observations == total. 500*100 is 50,0000.

Re: sum of positive integers [#permalink]
05 Feb 2014, 21:17

gmatpapa wrote:

puma wrote:

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

a) 10050 b) 5050 c) 5000 d) 50000 e) 55000

My approach:

Step 1: Find out the sum of all integers divisible by 5. Number of such terms= \(\frac{(1000-0)}{5} + 1= 201\). Sum of all such integers= \(201*500\) (using the sum formula of terms in Arithmetic Progression).. (i) Step 2: Find out the sum of all integers divisible by 10: Number of such terms= \(\frac{(1000-0)}{10}= 101\). Sum of all such integers= \(101*500= 101*500\) .. (ii)

Now, the multiples of 10 are the ones which are divisible by both 5 and 2. Therefore, we need to subtract (ii) by (i) to arrive at the terms divisible only by 5 and not by 2

Re: sum of positive integers [#permalink]
05 Feb 2014, 22:53

Expert's post

1

This post was BOOKMARKED

nishanthadithya wrote:

gmatpapa wrote:

puma wrote:

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

a) 10050 b) 5050 c) 5000 d) 50000 e) 55000

My approach:

Step 1: Find out the sum of all integers divisible by 5. Number of such terms= \(\frac{(1000-0)}{5} + 1= 201\). Sum of all such integers= \(201*500\) (using the sum formula of terms in Arithmetic Progression).. (i) Step 2: Find out the sum of all integers divisible by 10: Number of such terms= \(\frac{(1000-0)}{10}= 101\). Sum of all such integers= \(101*500= 101*500\) .. (ii)

Now, the multiples of 10 are the ones which are divisible by both 5 and 2. Therefore, we need to subtract (ii) by (i) to arrive at the terms divisible only by 5 and not by 2

Can u explain in brief why do you multiply by 500. Thank you

Posted from my mobile device

For every evenly spaced set (a.k.a. arithmetic progression), the sum equals to (mean)*(# of terms). 500 is the mean there: (mean)=(first+last)/2=(5+995)/2=500.

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...