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 2/5 ≤ m/n ≤ 3/5 (Posted on 2009-10-20)
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.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 Thanks Justin Comment 7 of 7 |
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.

 Posted by Steve Herman on 2011-01-08 14:52:07

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