Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 23 Mar 2008
Posts: 218

Obtain the sum of all positive integers up to 1000, which [#permalink]
Show Tags
08 Jun 2008, 12:05
6
This post was BOOKMARKED
Question Stats:
68% (02:40) correct
32% (02:08) wrong based on 306 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
Official Answer and Stats are available only to registered users. Register/ Login.



Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

Re: sum of positive integers [#permalink]
Show Tags
08 Jun 2008, 12:32
1
This post was BOOKMARKED
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+(n1)10 995=5+10n10 1000=10n n=100 sum= 100/2(10+(1001)10) this give us 50,000. D it is



Manager
Joined: 23 Mar 2008
Posts: 218

Re: sum of positive integers [#permalink]
Show Tags
10 Jun 2008, 12:55
OA is D
Could you please explain the following calculation
sum= 100/2(10+(1001)10)



Director
Joined: 01 Jan 2008
Posts: 622

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 100,50050,500 = 50,000 > D



SVP
Joined: 30 Apr 2008
Posts: 1874
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;) ) 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$$.
GMAT Club Premium Membership  big benefits and savings



Senior Manager
Joined: 12 Apr 2008
Posts: 499
Location: Eastern Europe
Schools: Oxford

Re: sum of positive integers [#permalink]
Show Tags
10 Jun 2008, 13:57
2
This post received KUDOS
2
This post was BOOKMARKED
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+(n1)*d – just put in numbers and solve for n.



Intern
Joined: 25 Jun 2008
Posts: 13

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: 287
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: 40

Re: sum of positive integers [#permalink]
Show Tags
02 Jul 2008, 09:36
5,15,25,.....,995 total terms = 100
sum = n/2 [2a + (n1)d]
n = 100 a = 5 d = 10
Sum = 50000
D is the answer.



Manager
Joined: 27 Jul 2010
Posts: 194
Location: Prague
Schools: University of Economics Prague

Re: sum of positive integers [#permalink]
Show Tags
03 Feb 2011, 06:47
1
This post received KUDOS
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: \(n=\frac{(1991)}{2}+1 = \frac{1991+2}{2}=100\) 5*100*100=50000
_________________
You want somethin', go get it. Period!



Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 533
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
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= \(\frac{(10000)}{5} + 1= 201\). 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= \(\frac{(10000)}{10}= 101\). 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: \(201*500  101*500\) = \(500*(201101)\) = \(500*100\) = \(50,000\). Answer. Thus, the answer is D.
_________________
My GMAT debrief: http://gmatclub.com/forum/from620to710mygmatjourney114437.html



Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 533
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+(n1)*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/from620to710mygmatjourney114437.html



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: sum of positive integers [#permalink]
Show Tags
04 Feb 2011, 08:50
1
This post received KUDOS
Sequence is: 5,15,25,....995 Common difference d = 10 Number of elements: ((9955)/10)+1=100 Average: (first+last)/2 = (5+995)/2 = 500 Sum= Number of elements * Average = 100 * 500 = 50000 Ans: "D"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Current Student
Joined: 04 Aug 2013
Posts: 30
Concentration: Finance, Real Estate
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= \(\frac{(10000)}{5} + 1= 201\). 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= \(\frac{(10000)}{10}= 101\). 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: \(201*500  101*500\) = \(500*(201101)\) = \(500*100\) = \(50,000\). Answer. Thus, the answer is D. 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: 39708

Re: sum of positive integers [#permalink]
Show Tags
05 Feb 2014, 23:53
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= \(\frac{(10000)}{5} + 1= 201\). 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= \(\frac{(10000)}{10}= 101\). 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: \(201*500  101*500\) = \(500*(201101)\) = \(500*100\) = \(50,000\). Answer. Thus, the answer is D. 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15994

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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



SVP
Joined: 08 Jul 2010
Posts: 1737
Location: India
GMAT: INSIGHT
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 Answer: Option D
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2185

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+(n1)D So N=100 Sum =100/2 * [5+995] = 50,000 hence D
_________________
Give me a hell yeah ...!!!!!




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







