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

Home > Just Math
Constant and Ratio Puzzle (Posted on 2016-01-30) Difficulty: 3 of 5
Determine all possible values of a positive integer constant C for which there exists an infinite number of pairs (A, B) of positive integers such that:
(A+B-C)!/(A!*B!) is an integer.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
quite general Comment 1 of 1
As posted the puzzle seems quite general. 

With B>C it's always posible to adjust A to have an integer for every B value. If this is true the answer would be that for every value of C there exists an infinite number of pairs....

For example: B>C, B-C=D the expression is (A+D!)/A!*B! which result in: ((A+D)(A+D-1)...(A+1))/B!
But now imposing A=(B!)-D the expression is
((B!)(B!-1)...(B!-D+1))/(B!) which is obviously an integer. 

Ex: for C=4
We pick ad es. B=7 and so B-C=3 and A=7!-3=5037
(A+B-C)!=5040!
and expression is =5040!/(5037!*7!)=5040*5039*5038/5040=5039*5038.

Edited on February 26, 2016, 5:08 am
  Posted by armando on 2016-02-26 03:18:29

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


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information