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 500,000 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:

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$$.

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.

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]

Show Tags

13 Jan 2014, 19:25

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.

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

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.

Re: Obtain the sum of all positive integers up to 1000, which [#permalink]

Show Tags

05 Feb 2016, 04:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Obtain the sum of all positive integers up to 1000, which [#permalink]

Show Tags

05 Feb 2016, 07:37

Expert's post

puma wrote:

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

A. 10,050 B. 5,050 C. 5,000 D. 50,000 E. 55,000

Total Multiples of 5 from 1 through 1000 = 1000/5 = 200

Half of these 200 multiples of 5 will be even (i.e. divisible by 2) and remaining half will be odd multiples of 5

i.e. Question : 5+15+25+......+955=?

Since it's an Arithmetic Progression (In which difference between any two consecutive terms remain constant) in which sum of first and last term = Sum of second and Second last term = and so on...

100 such numbers i.e.e 50 such pairs and sum of each pair = 5+955 = 1000

i.e. Sum of all pairs = (100/2)*(5+955) = 50*1000 = 50,000

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: Obtain the sum of all positive integers up to 1000, which [#permalink]

Show Tags

14 Mar 2016, 00:59

The question is basically asking us the sum of the AP series 5+15+......995 Number of terms can be calculated from A(n)= A+(n-1)D So N=100 Sum =100/2 * [5+995] = 50,000 hence D _________________

Give me a hell yeah ...!!!!!

gmatclubot

Re: Obtain the sum of all positive integers up to 1000, which
[#permalink]
14 Mar 2016, 00:59

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...