Date: Tue, 3 Sep 1996 21:01:35 PST 
Subject: math problem
Can you prove that "2n choose n" is not divisible by 3, 5, and 7 for infinitely many n?

Thanks.
Kathy

Your question is an unsolved problem.

Erdos, Graham, Rusza and Straus (Math. of Comp., 29(1975),pp 83-92) show that for any two primes p and q there exist infinitely many integers n for which (C(2n,n),pq) = 1. They remark that nothing is known for three primes and, in particular, they ask whether there are infinitely many n for which (C(2n,n),105) = 1 and this is your problem. As far as I know this is still unsettled.

I have put a note on the divisibility of 2n choose n by individual primes in the Resource Room.

Penny

Go to Math Central

To return to the previous page use your browser's back button.