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# The sum of the even numbers between 1 and n is 79*80, where

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The sum of the even numbers between 1 and n is 79*80, where [#permalink]

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18 Aug 2006, 01:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=?

HI guys I'm looking for a faster way to solve this problem.

I just got the range from 2 to n-1. Obtained the average (n-1+2)2
multiplied it by the number of terms (n-1)/2, and equated it to 79 and 80.
Can anyone suggest a faster way of arriving at the answer? I think the method that I used is time consuming.
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Re: question bank problem solving [#permalink]

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18 Aug 2006, 02:25
apollo168 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=?

HI guys I'm looking for a faster way to solve this problem.

I just got the range from 2 to n-1. Obtained the average (n-1+2)2
multiplied it by the number of terms (n-1)/2, and equated it to 79 and 80.
Can anyone suggest a faster way of arriving at the answer? I think the method that I used is time consuming.

I know know whether this is any better, but here it goes:

If n is an odd number (obviously positive) we can write n=2k+1 where k is a positive integer. The greatest even number less than n is 2k.

So the sum of the positive even numbers less than n= 2+4+6+...+2k =2(1+2+3....+k)=2(k+1)k/2 =(k)(k+1)=79(80)

Thus k=79 and n=2(79)+1=159 This took about 1 minute
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18 Aug 2006, 02:39
Hi kevin,

Certainly looks more efficient. But can't seem to decipher this part of your solution

2(k+1)k/2 --> I know that the average of the evenly spaced even numbers is 2(k+1)/2 ...but what I don't know is why you just multiplied the average by k? Won't multiplying the average by k also include the odd numbers? Sorry to inconvenience you again.
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18 Aug 2006, 02:43
Sorry tried it out again. I got it!

Thanks
18 Aug 2006, 02:43
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