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.

Submitted by K Sengupta

Rating: 4.5000 (2 votes)

Solution: (Hide)

Refer to the solution submitted by Steve Herman in this location.
Jer has proved in this location that there actually exist 5 distinct n-digit binary numbers such that for any two numbers chosen from them the digits will coincide in precisely m place.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Thanks Justin	Steve Herman	2011-01-08 14:52:07
re(2): Eureka! (spoiler)	Justin	2009-12-07 04:33:40
re(2): Eureka! (spoiler)	Steve Herman	2009-10-22 16:44:45
re: Eureka! (spoiler)	Jer	2009-10-22 10:42:22
Eureka! (spoiler)	Steve Herman	2009-10-21 18:38:26
Some thoughts	Steve Herman	2009-10-21 18:10:08
Stumped	Jer	2009-10-21 15:35:24

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