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Obtain the sum of all positive integers up to 1000, which [#permalink]
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08 Jun 2008, 11:05
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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
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Re: sum of positive integers [#permalink]
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08 Jun 2008, 11:32
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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



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Re: sum of positive integers [#permalink]
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10 Jun 2008, 12:02
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



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Re: sum of positive integers [#permalink]
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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.
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Re: sum of positive integers [#permalink]
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10 Jun 2008, 12:57
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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.



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Re: sum of positive integers [#permalink]
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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.
Ans = 5 (100^2)=50000 .D



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Re: sum of positive integers [#permalink]
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02 Jul 2008, 04:04
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 \(\sum_{k=1 and k is odd}^{200} 5*k = \sum_{i=0}^{99} 5(2i+1) = 5 \left( \sum_{i=0}^{99} 2i + \sum_{i=0}^{99} 1 \right) = 5 \left( 2 \sum_{i=0}^{99} i + 100 \right) = 5 \left( 2 \frac{99*100}{2} + 100 \right) = 5 \left( 9,900 + 100 \right) = 50,000\)



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Re: sum of positive integers [#permalink]
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02 Jul 2008, 08:36
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.



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Re: sum of positive integers [#permalink]
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03 Feb 2011, 05:47
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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
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Re: sum of positive integers [#permalink]
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03 Feb 2011, 07:41
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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.
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Re: sum of positive integers [#permalink]
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03 Feb 2011, 07:58
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
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Re: sum of positive integers [#permalink]
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04 Feb 2011, 07:50
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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"
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Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
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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.



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Re: sum of positive integers [#permalink]
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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= \(\frac{(10000)}{5} + 1= 201\). 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



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Re: sum of positive integers [#permalink]
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05 Feb 2014, 22:53
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.
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Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
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05 Feb 2016, 06:37
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
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Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
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13 Mar 2016, 23:59
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
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Obtain the sum of all positive integers up to 1000, which [#permalink]
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06 Dec 2017, 21:59
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 2So, 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+(n1)d} where s=sum of numbers, n=number of numbers a=first term and d=common difference So, s=100/2*{2*5+(1001)*10}=50,000 ...OPTION D *number of numbers is 100 because between 110, we have only one number, i.e. 5, now multiply it to our total 11000, i.e. 100.
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Obtain the sum of all positive integers up to 1000, which
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