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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)

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$$.

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

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.