perplexus.info :: Just Math : 2/5 ≤ m/n ≤ 3/5

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.

See The Solution Submitted by K Sengupta

Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

re(2): Eureka! (spoiler)

| Comment 5 of 7 |

(In reply to re: Eureka! (spoiler) by Jer)

Jer:

Excellent! Thanks for the existence example.

Steve

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

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