Five distinct ndigit 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
re: Eureka! (spoiler) by Jer)
I know the problem didn't ask for values, only proof, but I was bored and decided to make a program just to see how many values there are for different cases.
There are a total of 32 solutions for n=5, m=3 with the last one being:
01111
10111
11011
11101
11110
No solutions exist for n=6
There are a total of 26,880 solutions for n=7, m=3 with the last one being:
0111111
1010110
1011001
1100101
1101010
There are a total of 430,080 solutions for n=8, m=4 with the last one being:
01111111
10111100
11011010
11100110
11110001
That's as far as the program has gotten so far, as Python isn't as fast as I would like.

Posted by Justin
on 20091207 04:33:40 