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Topic: An Average Problem (Read 1244 times) |
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ThudnBlunder
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An Average Problem
« on: Mar 17th, 2009, 9:01pm » |
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Four consecutive even numbers are removed from the set S = {1,2,3,4,5,.....n}.
If the average of the remaining numbers is 51.5625 which four numbers were removed?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Leo Broukhis
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Re: An Average Problem
« Reply #1 on: Mar 17th, 2009, 11:01pm » |
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22, 24, 26, 28
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ThudnBlunder
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Re: An Average Problem
« Reply #2 on: Mar 18th, 2009, 2:01pm » |
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That's that settled then. Perhaps I should move this to CS.
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towr
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Re: An Average Problem
« Reply #3 on: Mar 18th, 2009, 2:56pm » |
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Oh poor you 
Take as the missing numbers x-3, x-1, x+1, x+3
Then we have n(n+1)/2 - 4 x = 51.5625 (n-4)
multiply both sides by 2, n(n+1) - 8 x = 103.125 (n-4)
The left side is an integer, so the right side should also be an integer
Therefore 8 | n-4 => n = 8k+4
Make the substitution, (8k+4)(8k+5) - 825 k = 8 x
8 should divide the left hand side, since it's on the right: 8 | 7 k + 4
So k = 4 + 8 l
n = 8 (4 + 8 l) + 4 = 64 l + 36
Find the minimal and maximal values of n
n(n+1)/2 - 10 = 51.5625 (n-4)
[edit](n-4)(n-3)/2 - 10 = 51.5625 (n-4)
(n-4)(n-3)/2 = 51.5625 (n-4)[/edit]
Which gives 99 <= n <= 106
So therefore l = 1, n = 100, x = 25
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| « Last Edit: Mar 22nd, 2009, 10:01am by towr » |
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Immanuel_Bonfils
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Re: An Average Problem
« Reply #4 on: Mar 22nd, 2009, 9:51am » |
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Shouldn't the equations for minimal and maximal be n(n+1)/2 - 16 = 51.5625(n-4) and (n-4)(n-3)/2 =51.5625(n-4) ? The results are OK, but just to make it more "visible".
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towr
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Re: An Average Problem
« Reply #5 on: Mar 22nd, 2009, 10:05am » |
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on Mar 22nd, 2009, 9:51am, Immanuel_Bonfils wrote:| Shouldn't the equations for minimal and maximal be n(n+1)/2 - 16 = 51.5625(n-4) and (n-4)(n-3)/2 =51.5625(n-4) ? The results are OK, but just to make it more "visible". |
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Well, the second one I certainly made a mistake. For the first one I subtracted the first 4 numbers, rather than the first 4 even ones. But 2+4+6+8 = 20, not 16 
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Immanuel_Bonfils
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Re: An Average Problem
« Reply #6 on: Mar 22nd, 2009, 9:12pm » |
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Thanks, may be because I find it odd... Also I should say "invisible" as long as it's hiden...
But, anyhow, as always a nice solution.
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| « Last Edit: Mar 22nd, 2009, 9:13pm by Immanuel_Bonfils » |
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