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?

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**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 lowest-numbered 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.

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**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

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**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.

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**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'.

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

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**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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