Five numbers are selected and deleted from a set of consecutive positive integers beginning with 1. The arithmetic mean of the remaining numbers in the set is 45.1.

Find the largest possible single number that could have been deleted from the original set.

The way to go about solving this is to find the sum of integers 1 to x whose average does not exceed 41.1.  In this case, x = 81, Σx = 3321, mean = 41.  In any drawing of 5 numbers, the way to ensure the lowest remaining average is to choose the five highest numbers.  Therefore, with integers 1 to 86, it is possible to draw 5 (82, 83, 84, 85, & 86) and leave a remaining mean of 41.

Since the remaining mean must equal 41.1, the sum must be a multiple of 411, and the number of remaining integers must be a multiple of 10.  So I find multiples of (41.1*10) that approach 3321:

{2466, 2877, 3288, 3699}

Beginning with an integer group of 86, I subtract the four highest integers from that group (83, 84, 85, 86) and sum the result:

Σ82 = 3403

I can reduce this sum by replacing 83 and removing a lower integer.  However, the most I can possibly reduce this total is by 82 which leaves 3321.  Since none of the multiples of 441 fall inside this range {3321 to 3403}, 86 cannot be the highest in the  group.

I now move down to 85.  I again subtract the four highest (82, 83, 84, 85) from the total.  This leaves:

Σ81 = 3321

The difference between this and the next lowest multiple of 441 is:

3321 - 3288 = 33

Therefore, choosing 33 as the final number to remove will yield a  sum of 3288 to be divided by 80 remaining integers, yielding an average of 41.1.

The largest possible single number that could have been deleted from the original set is 85.

Posted by hoodat on 2007-05-09 13:32:15