Posted by Mathsfreak on Thursday, April 18, 2013 at 10:38am.
There are 100 runners, each given a distinct bib labeled 1 to 100. What is the most number of runners that we could arrange in a circle, such that the product of the numbers on the bibs of any 2 neighboring runners, is less than 1000?

Maths  Steve, Thursday, April 18, 2013 at 10:45am
Are the runners arranged in any particular order? If not, then any two numbers might be next to each other. So only 1 runner can be used.
If we pick the lowestnumbered n runners, then since √1000 = 31.6, we need to make sure that all numbers are less than that. So, if there are only 31 runners, then no two numbers can multiply to be greater than 1000.

Maths  Mathsfreak, Thursday, April 18, 2013 at 10:53am
I tried using circular permutations......but the answer didn't tally.Also..............31 is a wrong answer

Maths  Steve, Thursday, April 18, 2013 at 10:57am
well, there must be some other restriction on which numbers may be chosen, or how they may be arranged. I may have to get back to you on that.

Maths  Mathsfreak, Thursday, April 18, 2013 at 11:04am
In any order ,I guess, as it hasn't been particularly specified!The question is an exact copy from the springer book on 'Combinatorics'.

Maths  BrilliantLover, Friday, April 19, 2013 at 1:55am
The question is an exact copy from the springer book on 'Combinatorics'.

Maths  Mathsfreak, Friday, April 19, 2013 at 4:15am
Mind ur language,geezer.What on this earth is Brilliant anyways.Please check out 'The 1001 combinatorics problems' by 'Springer' problem #79 before blaming me..........:p
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