Posted by Mathsfreak on .
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,
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,
I tried using circular permutations......but the answer didn't tally.Also..............31 is a wrong answer

Maths 
Steve,
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.

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Mathsfreak,
In any order ,I guess, as it hasn't been particularly specified!The question is an exact copy from the springer book on 'Combinatorics'.

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BrilliantLover,
The question is an exact copy from the springer book on 'Combinatorics'.

Maths 
Mathsfreak,
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