Four consecutive even positive integers are removed from a set containing precisely N consecutive positive integers starting with 1, such that the arithmetic mean of the remaining numbers is equal to 825/16.

Determine the four integers that were removed from the set.

Let the number of consecutive integers written on the board be n.
Let the integers removed be 2a,2(a+1),2(a+2),2(a+3)
then the average of the remaining would be
(n(n+1)-4(4a+6))/(2(n-4))
now this needs to simplify to 825/16 so we have
2(n-4)=16k for some integer k
2n-8=16k
2n=16k+8
n=8k+4

and the we also have

n(n+1)-4(4a+6)=825k
(8k+4)(8k+5)-16a-24=825k
16a=64k^2-753k-4
a=(64k^2-753k-4)/16

we have the restrictions
a>=1 and 2*(a+3)<=n
and this narrows down k to the sole value of 12

thus n=8*12+4=100 and a=11
thus the 4 number erased were
22,24,26,28

Posted by Daniel on 2009-08-05 12:01:19