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.

I was interested in your examples.

We had already proved that m and n must either be both odd or both even, and that 2/5 <= m/n <= 3/5, so that means that m must be the following (as a function of n)

n m

-- -------

3 none

4 2 (but impossible for other reasons)

5 3

6 none

7 3

8 4

9 5

10 4 or 6

11 5

12 6

13 5 or 7

14 6 or 8

15 7 or 9 or 11

etc.