GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Oct 2019, 12:38

### GMAT Club Daily Prep

#### 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

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

# Obtain the sum of all positive integers up to 1000, which

Author Message
TAGS:

### Hide Tags

Manager
Joined: 23 Mar 2008
Posts: 199
Obtain the sum of all positive integers up to 1000, which  [#permalink]

### Show Tags

08 Jun 2008, 12:05
12
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:19) correct 34% (02:27) wrong based on 452 sessions

### HideShow timer Statistics

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
Current Student
Joined: 28 Dec 2004
Posts: 2718
Location: New York City
Schools: Wharton'11 HBS'12
Re: sum of positive integers  [#permalink]

### Show Tags

08 Jun 2008, 12:32
1
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

last term thats not divisble by 10

995=5+(n-1)10

995=5+10n-10
1000=10n
n=100

sum= 100/2(10+(100-1)10)

this give us 50,000.

D it is
Senior Manager
Joined: 01 Jan 2008
Posts: 471
Re: sum of positive integers  [#permalink]

### Show Tags

10 Jun 2008, 13:02
the answer is sum of 5*k - sum of 10*n

5 + ... + 1000 = (5+1000)*200/2 = 1,005*100 = 100,500

10 + .. + 1000 = (10+1000)*100/2 = 1,010*50 = 50,500

the answer is 100,500-50,500 = 50,000 -> D
SVP
Joined: 30 Apr 2008
Posts: 1641
Location: Oklahoma City
Schools: Hard Knocks
Re: sum of positive integers  [#permalink]

### Show Tags

10 Jun 2008, 13:12
D (which is clearly established by now;) )

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.

GMAT Club Premium Membership - big benefits and savings
Senior Manager
Joined: 12 Apr 2008
Posts: 466
Location: Eastern Europe
Schools: Oxford
Re: sum of positive integers  [#permalink]

### Show Tags

10 Jun 2008, 13:57
2
2
Hi!

For those who asked to explain the formula:

For arithmetic progression,

Sn = n*(a1+an)/2

Our series is 5, 15, 25, … , 995. (a1=5, an=995, d=10).

Only, we need to calculate n. In order to find it, we can use the formula an = a1+(n-1)*d – just put in numbers and solve for n.
Intern
Joined: 25 Jun 2008
Posts: 7
Re: sum of positive integers  [#permalink]

### Show Tags

02 Jul 2008, 04:48
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.

Ans = 5 (100^2)=50000 .D
Current Student
Joined: 12 Jun 2008
Posts: 232
Schools: INSEAD Class of July '10
Re: sum of positive integers  [#permalink]

### Show Tags

02 Jul 2008, 05:04
bajaj wrote:
So the sum we need is (5+15+....+995) or 5(1+3+5+....199). The sum of consequtive n integers is n^2.

In your example, 1,3,5,..,199 are not consecutive integers (and I don't see where this results "n^2" come from).

What you can write is that the number we look for is

$$\sum_{k=1 and k is odd}^{200} 5*k = \sum_{i=0}^{99} 5(2i+1) = 5 \left( \sum_{i=0}^{99} 2i + \sum_{i=0}^{99} 1 \right) = 5 \left( 2 \sum_{i=0}^{99} i + 100 \right) = 5 \left( 2 \frac{99*100}{2} + 100 \right) = 5 \left( 9,900 + 100 \right) = 50,000$$
Intern
Joined: 26 May 2008
Posts: 33
Re: sum of positive integers  [#permalink]

### Show Tags

02 Jul 2008, 09:36
5,15,25,.....,995
total terms = 100

sum = n/2 [2a + (n-1)d]

n = 100
a = 5
d = 10

Sum = 50000

Manager
Joined: 27 Jul 2010
Posts: 137
Location: Prague
Schools: University of Economics Prague
Re: sum of positive integers  [#permalink]

### Show Tags

03 Feb 2011, 06:47
2
5+15+25+...+995=
1*5+3*5+5*5+...+199*5=
5*(1+3+5+...+199)

Sum of first n odd numbers: $$n^2$$

#of odd numbers: $$n=\frac{(199-1)}{2}+1 = \frac{199-1+2}{2}=100$$

5*100*100=50000
_________________
You want somethin', go get it. Period!
Senior Manager
Status: Up again.
Joined: 31 Oct 2010
Posts: 464
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
Re: sum of positive integers  [#permalink]

### Show Tags

03 Feb 2011, 08:41
1
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

Step 3: $$201*500 - 101*500$$
= $$500*(201-101)$$
= $$500*100$$
= $$50,000$$. Answer.

_________________
My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Senior Manager
Status: Up again.
Joined: 31 Oct 2010
Posts: 464
Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42
Re: sum of positive integers  [#permalink]

### Show Tags

03 Feb 2011, 08:58
greenoak wrote:
Hi!

For those who asked to explain the formula:

For arithmetic progression,

Sn = n*(a1+an)/2

Our series is 5, 15, 25, … , 995. (a1=5, an=995, d=10).

Only, we need to calculate n. In order to find it, we can use the formula an = a1+(n-1)*d – just put in numbers and solve for n.

This is the quickest approach IMO. you really dont need anything else!
+1
_________________
My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Retired Moderator
Joined: 20 Dec 2010
Posts: 1584
Re: sum of positive integers  [#permalink]

### Show Tags

04 Feb 2011, 08:50
1
Sequence is: 5,15,25,....995

Common difference d = 10
Number of elements: ((995-5)/10)+1=100
Average: (first+last)/2 = (5+995)/2 = 500

Sum= Number of elements * Average = 100 * 500 = 50000

Ans: "D"
_________________
Intern
Joined: 04 Aug 2013
Posts: 28
Concentration: Finance, Real Estate
GMAT 1: 740 Q47 V46
GPA: 3.23
WE: Consulting (Real Estate)
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.
Intern
Joined: 13 Aug 2013
Posts: 3
Re: sum of positive integers  [#permalink]

### Show Tags

05 Feb 2014, 22:17
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

Step 3: $$201*500 - 101*500$$
= $$500*(201-101)$$
= $$500*100$$
= $$50,000$$. Answer.

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

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 58381
Re: sum of positive integers  [#permalink]

### Show Tags

05 Feb 2014, 23:53
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

Step 3: $$201*500 - 101*500$$
= $$500*(201-101)$$
= $$500*100$$
= $$50,000$$. Answer.

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.

Hope it's clear.
_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: Obtain the sum of all positive integers up to 1000, which  [#permalink]

### Show Tags

05 Feb 2016, 07:37
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

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Current Student
Joined: 12 Aug 2015
Posts: 2568
Schools: Boston U '20 (M)
GRE 1: Q169 V154
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
_________________
Intern
Joined: 10 Dec 2015
Posts: 26
GMAT 1: 680 Q45 V38
Obtain the sum of all positive integers up to 1000, which  [#permalink]

### Show Tags

06 Dec 2017, 22:59
Let's first try to figure out what the problem's actually asking us.
Numbers divisible by 5 mean the set : {5,10,15,....,1000}
Numbers divisible by 5 and 2 mean the set : {10,20,30,...,1000}
OUR CRITERION: divisible by 5 and not divisible by 2
So, the set we need to work with is : {5,15,25,35,...995}
This is a generic arithmetic progression problem, where the common difference is 10 and number of numbers is 100*, so we can easily apply the formula,

s=n/2*{2a+(n-1)d} where s=sum of numbers, n=number of numbers a=first term and d=common difference
So, s=100/2*{2*5+(100-1)*10}=50,000 ...OPTION D

*number of numbers is 100 because between 1-10, we have only one number, i.e. 5, now multiply it to our total 1-1000, i.e. 100.
_________________
-Please PM me any doubts which you might have, and I will do my level best to help.
Non-Human User
Joined: 09 Sep 2013
Posts: 13210
Re: Obtain the sum of all positive integers up to 1000, which  [#permalink]

### Show Tags

23 Dec 2018, 05:01
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] 23 Dec 2018, 05:01
Display posts from previous: Sort by