Small Bib Products

Since $32\times 33 > 1000$ , no 2 runners with bibs above $32$ , could stand next to each other.
This shows that there are at most $31 \times 2 = 62$ runners placed in the circle.

If there are 62 runners, we know that those with numbers 32 to 62 must be placed in every alternate position.
Now consider where runner $31$ goes. Since $31 \times 33 = 1023$ , it follows that we would be unable to place runner 31 in this circle.

Thus, there are at most 61 runners.
This can be done by having the runners arranged as $1, 61, 2, 60, 3, 59, \ldots, 29, 33, 30, 32, 31$ .

First if $a*b\le 1000$ and $a\le b$ we have $a\le 31$ .

Thus if we pick up any two block of people, one of their numbers is less than or equal to $31$ .

Let us first arrange #1 to 31 in the circle. Now observe that if we insert two number greater than $31$ as a neighbor of each of these numbers then we will be able to maximize the number of people.

But the only number $>31$ which when multiplied by $31$ gives a product less than $1000$ is $32$ . So $31$ cannot have two neighbors.

Now to the counting, first we choose $\{1,2,3,...,31\}$ which gives us $31$ runners.

Each one of them has $2$ neighbors, but we'll count only the 'left' neighbors to avoid overcounting which gives us $31$ more runners.

But as discussed above, one of them cannot have two neighbors, so minus $1$ . So $2*31-1=\boxed{61}$

Now if we insert any other number then it will make two numbers greater than $31$ adjacent. So this indeed the maximum!

first,what you can do is find consecutive numbers where their product is less than one thousand,that is 1-31.then order them in a circlle so that you can fill them with another number.By the circular permuation rule if there are 31 number you can fill in (n-1) numbers between them.so that is 31 + 30 makes 61 numbers. To make it more clear match the lowest number with the highest number,then the second highest with the second lowest,and so forth... 1 - 61 2 - 60 3 - 59 till 30 - 31 note:- it necessarily doesn't have to be the lowest number with highest number.It is used for illustation purpose

First, we will put the runners with numbers from $1$ to $32$ in the circle. This is because $32 \times 33 = 1056$ , while $31 \times 32 = 992$ .

Next, we will insert the runners with numbers from $33$ and onwards in between the already positioned runners, until no more is possible.

Since, the numbers of the runners will be increasing, we will put the runner with the bib number $33$ between two runners with the largest numbers on the bib possible. This means the runner with the number $33$ is put between the runners with the numbers $29$ and $30$ . We next put the runner with the number $34$ between the runners numbered $28$ and $29$ , and so on, until we put the runner numbered $61$ in between $1$ and $2$ .

This is the largest number of runners possible, since we used the runners with the smallest numbers possible.

|8|100|9|99|10|98|7|97|6|96|5|95|4|94| 3|92|2|91|1|90|11|83|12|76|13|71|14| 66|16|62|16|58|17|55|18|52|19|49|20| 47|21|45|22|43|23|41|24|39|25|38|26| 37|27|35|28|34|29|33|30|32|31

with 8 being next to 31 to make the circle.

31x32 = 992 We cannot have a larger pair of numbers than this. But we can pair larger numbers with smaller numbers to get 31,32, 30,33, 29,34 etc. By pairing numbers off like this we can loop right around the circle and find the number 61 at the end which is the maximum possible number of runners. Although we could use 100x1, there are insufficient smaller numbers to deal with the 90s and 80s.

We want to optimize the product of both neighbors as much as possible. We know that $\sqrt{1000} = 31.6$ approx
. Thus, let start with runners with bib labeled 31 and 32, and x : 31, 32, x We want 32.x to be less than 1000, so maximum x is 31. But 31 is already his neighbor, thus the next number must be 30. So, 31, 32, 30, y. Using the same method, we will get the arrangement is : 31, 32, 30, 33, 29, 34, 28, 35, 27, 37, 26, 38, 25, 39, 24, 41, 23, 42, 22, 45, 21, 47, 20, 49, 19, 52, 18, 55, 17, 58, 16, 62, 15, 66, 14, 71, 13, 76, 12, 83, 11, 90, 10, 91, 9, 92, 8, 93, 7, 94, 6, 95, 5, 96, 4, 97, 3, 98, 2, 99, 1 with 1 is next to 31. So there are total of 61 runners can be arrange in a circle.

We optimize the product of both neighbours as much as possible. We know that √1000 = 31.623. So, let start with runners with bib labeled 31 and 32, and x : 31, 32, x We want 32x to be less than 1000, so maximum x is 31. But 31 is already his neighbour, thus the next number must be 30. So, 31, 32, 30, y.

Using similar method, we'll get the arrangement is:

31, 32, 30, 33, 29, 34, 28, 35, 27, 37, 26, 38, 25, 39, 24, 41, 23, 42, 22, 45, 21, 47, 20, 49, 19, 52, 18, 55, 17, 58, 16, 62, 15, 66, 14, 71, 13, 76, 12, 83, 11, 90, 10, 91, 9, 92, 8, 93, 7, 94, 6, 95, 5, 96, 4, 97, 3, 98, 2, 99, 1

From the list there are total of 61 runners can be arrange in a circle.

We know that √1000 = 31.6 approx. Thus, we start with runners with bib labeled 31,32 and x. Now according to question, We want 32x to be less than 1000, so maximum x we get is 31. But 31 being his neighbor, next number we get must be 30. Similarly we check for other numbers, as follows Now take 31, 32, 30, y. Using the same method, we will get the arrangement as : 31, 32, 30, 33, 29, 34, 28, 35, 27, 37, 26, 38, 25, 39, 24, 41, 23, 42, 22, 45, 21, 47, 20, 49, 19, 52, 18, 55, 17, 58, 16, 62, 15, 66, 14, 71, 13, 76, 12, 83, 11, 90, 10, 91, 9, 92, 8, 93, 7, 94, 6, 95, 5, 96, 4, 97, 3, 98, 2, 99, 1 with 1 is next to 31. So there are total of 61 runners that can be arranged in a circle.

The best way to start such a pattern would be 1, 100, 2, 99, 3, 98 ... until the adjacent numbers have product above 1000. In this case, this happens when ... 93, 9, 92, 10, 90*,11 ...

* Here, 91 $\times$ 11 > 1000. Investigating the pattern, and how subsequent numbers are come up with, I got that 1000/n, where n is a positive integer (starting from 1), rounded down, would give the number before n. i.e. 1000/10 = 100. However, 100 has already been used, with the next-largest unused integer being 92. This comes into play at 11, where the number before 11 is gotten by 1000/11 rounded down. (note in cases such as 1000/20=50, 49 has to be used, but this is not needed in the question.)

This pattern continues, such that as long as 1000/n > n, there are (2n-1) runners that can be in the circle. (1 to n, with an extra (large-numbered) runner in the middle of each runner with a small integer)

1000/n has to be larger than n, as all integers from 1 to n would have been used before n.

This continues all the way until 1000/31(rounded down) = 32. Here, 32 goes in front of 31, as per the pattern. However, after 31, no other integer can fit. All integers below 32 have been used.

Hence, there are 31 $\times$ 2 - 1 runners in the circle (keeping in mind 1 is the first integer, the fact that this is a circle does not matter).

31 $\times$ 2 - 1 = 61.

And the number pattern goes as such: 1, 100, 2, 99, 3, 98, 4, 97, 5, 96, 6, 95, 7, 94, 8, 93, 9, 92, 10, 90, 11, 83, 12, 76, 13, 71, 14, 66, 15, ... , 28 , 34, 29, 33, 30, 32, 31.

For a total of 61 runners.