Five distinct n-digit binary numbers are such that for any two numbers chosen from them the digits will coincide in precisely m places. There is no place with the common digit for all the five numbers. At least one of the binary numbers contains leading zero.

Prove that 2/5 ≤ m/n ≤ 3/5.

(In reply to Eureka! (spoiler) by Steve Herman)

Nice proof.

In response to your part D)
n=5, m=2

00000
00011
00101
01001
10001

which can be extended to n=5a, m=2a

0000000000
0000001111
0000110011
0011000011
1100000011

but have not found others

Posted by Jer on 2009-10-22 10:42:22