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OG 12 PS #157

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OG 12 PS #157 [#permalink] New post 26 Jul 2010, 16:56
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Can somebody please elaborate further on the solution for problem 157 on OG 12 Problem Solving? I don't see where the 50 and the 150 came from. Why divide the original numbers by 2?

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Re: OG 12 PS #157 [#permalink] New post 26 Jul 2010, 20:08
Can you post the exact problem here
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Re: OG 12 PS #157 [#permalink] New post 26 Jul 2010, 20:22
Hi,

I remember this problem and yeah the explanation in the OG sucks :)

Hope what follows will elaborate a bit more.

The sum of the first 300 integers is 300*301/2. Since you are considering only even integers, then you will divide it once again by 2 since the number of even integers = number of odd integers.
Since you are considering only even integers you will go with 1/2(300*302/2).
Similarly the sum of even integers up to 99 will be 1/2(98*100/2).

So the sum of even integers from 99 to 300 will be their difference which is 20200.

I know this is still confusing but I guess the explanation is a bit better than the OG's one.

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Re: OG 12 PS #157 [#permalink] New post 27 Jul 2010, 20:04
Hi again,

Solving this problem by arithmetic progression is much simpler and straight forward.

Tn = a + (n-1)*d
Sn = n/2 * (2a + (n-1)*d)

Where:
Tn is the nth term
a is the first term
d is difference between the terms
n is number of terms

So a=100 (not 99 since 99 is odd)
d=2
Tn=300

300 = 98 + (n-1)*2, n = 101
Sn = 101/2 * (2*98 + (101-1)*2) = 20200


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Re: OG 12 PS #157 [#permalink] New post 27 Jul 2010, 20:07
crack700 wrote:
Can you post the exact problem here



Crack 700, below is question 157 from OG12

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
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Re: OG 12 PS #157 [#permalink] New post 27 Jul 2010, 23:28
jakolik wrote:
crack700 wrote:
Can you post the exact problem here



Crack 700, below is question 157 from OG12

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150



Easy...

question asks what is...

100+102+104.......+300

We can solve it in 2 ways...

First Method

Average of 100 and 300 is 200
Total even number between 100 and 300 (inclusive) = [(300-100)/2] + 1 = 101
therefore 101*200 = 20200 (ans B)


Second Method

100+102+104.......+300
=2(50+51+52+....+150)

Sum of (50+51+52+....+150) equals

Sum of (1+2+3.....+150) - (1+2+3+...50)
= 11325-1275
= 10050

Now we have the answer as 2*10050 = 20200


Hope this helps.....
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Re: OG 12 PS #157   [#permalink] 27 Jul 2010, 23:28
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