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2/5 ≤ m/n ≤ 3/5 (Posted on 2009-10-20) Difficulty: 3 of 5
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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re(2): Eureka! (spoiler) | Comment 5 of 7 |
(In reply to re: Eureka! (spoiler) by Jer)


Excellent!  Thanks for the existence example.


  Posted by Steve Herman on 2009-10-22 16:44:45
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