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+BC)!/(A!*B!) is an integer.
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, BC=D the expression is (A+D!)/A!*B! which result in: ((A+D)(A+D1)...(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 BC=3 and A=7!3=5037
(A+BC)!=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 20160226 03:18:29 