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.

The number of integers that remain must be some multiple of 16. In order for the numerator of the reduced fraction to be odd, the multiple must be even, as removing only even numbers does not change the parity of the number.

The average of 825/16 is 51.5625, so there are about 100 integers altogether, which is about 6*16.

If six times 16 integers are to remain after removal of four, then the original must have been 100 integers exactly. The numerator after the removal must be 6*825=4950, while the total of the first 100 integers is 101*100/2 = 5050, requiring the removal of a total of 100. To be spread over 4 integers, those removed must be centered on 25, and so are 22, 24, 26 and 28.

Edited on August 5, 2009, 12:14 pm

Posted by Charlie on 2009-08-05 12:11:08