All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Even Erasure (Posted on 2009-08-05) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 2 of 3 |

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

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information