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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.

See The Solution Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsThanks JustinSteve Herman2011-01-08 14:52:07
re(2): Eureka! (spoiler)Justin2009-12-07 04:33:40
re(2): Eureka! (spoiler)Steve Herman2009-10-22 16:44:45
re: Eureka! (spoiler)Jer2009-10-22 10:42:22
SolutionEureka! (spoiler)Steve Herman2009-10-21 18:38:26
Some ThoughtsSome thoughtsSteve Herman2009-10-21 18:10:08
Some ThoughtsStumpedJer2009-10-21 15:35:24
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