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

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

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25 May 2005, 13:47
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The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
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If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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Director
Joined: 18 Apr 2005
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Location: Canuckland

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25 May 2005, 23:32
HIMALAYA wrote:
79. explain later

HImalaya, could you please explain, and if there is a shortcut.

My long solution

If you divide the sequence by 2, you get a sequence of consequtive integers from 1 to (n-1)/2, its sum equals 40*79

The sum of consequtive integers is, according to the formula, ((n-1)/4)(1+ (n-1)/2) = 40*79

8*40*79 = n^2 - 1

so, n = sqrt(8*40*79 + 1) = 159, but the pain of this solution is sqrt(8*40*79 + 1), can't figure it fast if at all.

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SVP
Joined: 05 Apr 2005
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26 May 2005, 07:13
sparky wrote:
HIMALAYA wrote:
79. explain later

HImalaya, could you please explain, and if there is a shortcut.
My long solution
If you divide the sequence by 2, you get a sequence of consequtive integers from 1 to (n-1)/2, its sum equals 40*79
The sum of consequtive integers is, according to the formula, ((n-1)/4)(1+ (n-1)/2) = 40*79
8*40*79 = n^2 - 1
so, n = sqrt(8*40*79 + 1) = 159, but the pain of this solution is sqrt(8*40*79 + 1), can't figure it fast if at all.

the clues are
first, "The sum of the even numbers between 1 and n".
second, sum = 79*80,
third, n is an odd number.

thats enough. sum of the even numbers between 1 and n, where n is an odd number, is always consequtive numbers and n can not be the larger number... if sum of positive even numbers 1 to n is 35*36, n cannot be 36 and n ust be 35. this is a fact. try some.

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VP
Joined: 30 Sep 2004
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27 May 2005, 00:30
OA is 159

1+2+3+..+n = n*(n+1)/2
2(1+2+3+...+n) = n*(n+1)
2+4+...+2n = n*(n+1)
the sum of even numbers till 2n = n*(n+1)
We need to know 2n given that sum till 2n = 79*80
n = 79 and 2n = 158
odd term => 2n+1 = 159
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Director
Joined: 27 Dec 2004
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27 May 2005, 06:49
Hi Christoph, could you please tell me the source of this question?
Thanks

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VP
Joined: 18 Nov 2004
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27 May 2005, 07:04
First even number 2 and last is n-1......number of even numbers =
(n-3)/2 + 1 = (n-1)/2

Avg = Sum / (n-1)/2 ===>

Avg of equally spaced numbers: (2+n-1) / 2 = (n+1)/2

Sum = 79*80*2*2 = (n-1)(n+1) ....n = 159

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SVP
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27 May 2005, 18:51
christoph wrote:
OA is 159

1+2+3+..+n = n*(n+1)/2
2(1+2+3+...+n) = n*(n+1)
2+4+...+2n = n*(n+1)
the sum of even numbers till 2n = n*(n+1)
We need to know 2n given that sum till 2n = 79*80
n = 79 and 2n = 158
odd term => 2n+1 = 159

ok, i calculated only number of even numbers. i thought the question s about how many even numbers are there whose sum is 79*80. thats ok. agreed.

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Senior Manager
Joined: 21 Mar 2004
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29 May 2005, 11:45
Some formulae can be used to get the answers neatly.

Suppose you have a series of numbers.

Sum of 1st n odd nos = n^2
Sum of 1st n even nos = n(n+1)

Given n is odd here. There are (n-1)/2 even integers here

Sum of 1st (n-1)/2 even integers = (n-1)/2 * ((n-1)/2 +1 ) =79*80

DO NOT Multiply yet. Simplify

(n-1)(n+1) = 79*80*4 = 158 * 160

n =159
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ash
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29 May 2005, 11:45
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