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?

Answer 1

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.

Answer 2

I tried using circular permutations......but the answer didn't tally.Also..............31 is a wrong answer

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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