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.
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
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Posted by armando
on 2016-02-26 03:18:29 |
quite general
