wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> An Average Problem
        
(Message started by: ThudanBlunder on Mar 17^th, 2009, 9:01pm)

Title: An Average Problem
Post by ThudanBlunder on Mar 17^th, 2009, 9:01pm

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?

Title: Re: An Average Problem
Post by Leo Broukhis on Mar 17^th, 2009, 11:01pm

[hide]22, 24, 26, 28[/hide]

Title: Re: An Average Problem
Post by ThudanBlunder on Mar 18^th, 2009, 2:01pm

That's that settled then. Perhaps I should move this to CS.

Title: Re: An Average Problem
Post by towr on Mar 18^th, 2009, 2:56pm

Oh poor you ::)

[hide]
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
[/hide]

Title: Re: An Average Problem
Post by Immanuel_Bonfils on Mar 22^nd, 2009, 9:51am

[hide]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".[/hide]

Title: Re: An Average Problem
Post by towr on Mar 22^nd, 2009, 10:05am

on 03/22/09 at 09:51:55, Immanuel_Bonfils wrote:

[hide]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".[/hide]

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 ;)

Title: Re: An Average Problem
Post by Immanuel_Bonfils on Mar 22^nd, 2009, 9:12pm

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.