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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)
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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.
Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
05 Feb 2016, 03:19
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Re: Obtain the sum of all positive integers up to 1000, which [#permalink]
05 Feb 2016, 06:37
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
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Re: Obtain the sum of all positive integers up to 1000, which
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05 Feb 2016, 06:37
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