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

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Obtain the sum of all positive integers up to 1000, which [#permalink] New post 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] New post 08 Jun 2008, 11:32
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
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Re: sum of positive integers [#permalink] New post 10 Jun 2008, 11:55
OA is D

Could you please explain the following calculation


sum= 100/2(10+(100-1)10)
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Re: sum of positive integers [#permalink] New post 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 100,500-50,500 = 50,000 -> D
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Re: sum of positive integers [#permalink] New post 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] New post 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+(n-1)*d – just put in numbers and solve for n.
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Re: sum of positive integers [#permalink] New post 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] New post 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] New post 02 Jul 2008, 08:36
5,15,25,.....,995
total terms = 100

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

n = 100
a = 5
d = 10

Sum = 50000

D is the answer.
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Re: sum of positive integers [#permalink] New post 03 Feb 2011, 05:47
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
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Re: sum of positive integers [#permalink] New post 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

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

Thus, the answer is D.
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Re: sum of positive integers [#permalink] New post 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+(n-1)*d – just put in numbers and solve for n.


This is the quickest approach IMO. you really dont need anything else!
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Re: sum of positive integers [#permalink] New post 04 Feb 2011, 07:50
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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"
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Re: Obtain the sum of all positive integers up to 1000, which [#permalink] New post 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] New post 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

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

Thus, the answer is D.


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

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Re: sum of positive integers [#permalink] New post 05 Feb 2014, 22:53
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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

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

Thus, the answer is D.


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

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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: sum of positive integers   [#permalink] 05 Feb 2014, 22:53
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