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?

I would have gladly to u,if u haven't lied on the physics question, if u have done it,we can trade,as I have done this question

lol u think i care

answer ??

49 Its 61

To find the most number of runners that can be arranged in a circle, such that the product of the numbers on the bibs of any two neighboring runners is less than 1000, we can start by considering the prime factorization of 1000.

The prime factorization of 1000 is 2^3 * 5^3. This means that any number that can be expressed as a product of prime factors of 2 and/or 5 (with exponents less than or equal to 3) will have a product less than 1000.

Now let's analyze the possible arrangements:
1. If there are no prime factors of 2 or 5 in neighboring bib numbers, their product will not exceed 1. This means that any numbers that have prime factors other than 2 or 5 can be adjacent to each other.

2. If one neighbor has a prime factor of 2 or 5, then the other neighbor cannot have a prime factor of 2 or 5. Otherwise, their product would have a prime factor greater than 3 and exceed 1000.

Based on these observations, we can construct the following arrangement:

1 (no prime factors) - 2^3 (prime factor 2) - 3^3 (no prime factors) - 5^3 (prime factor 5) - 6^3 (prime factors 2 and 3) - 7^3 (no prime factors) - 10^3 (prime factors 2 and 5)

Each of these bib numbers can be adjacent to each other because they have the required properties. This arrangement consists of 7 runners.

Therefore, the most number of runners that can be arranged in a circle, such that the product of the numbers on the bibs of any two neighboring runners is less than 1000, is 7.