This problem gave me a great deal of trouble when I first encountered it! I was able to figure out the maximum case (see Tanisha Gupta's answer if you're having trouble) quickly. I also figured out that for the minimum case the solution was to put students in as "uniform" of teams as possible. However, it was figuring out how to put participants in these uniform teams which was the big challenge!

Note: as a matter of notation, the quantity n / m denotes integer truncating division throughout.

In hindsight, the reason this gave me so much trouble was because I was lacking the interpretation of n % m necessary for this problem. My tried and true understand of the quantity n % m has always been the following:

n = (n/m)*m + n%m

Intuitively, this interpretation of n % m can be understood by its computation; by repeatedly summing m (i.e. n/m times) until you get "close" to n and then calling the remainder n % m. Another way to view this procedure is that we start off with n pieces and split them into groups of size m (there are n/m of these groups) and n % m is the number of leftover pieces which don't fit in any of the groups.

Crucially for this problem, it turns out there is actually another interpretation of the quantity n % m, which can be expressed as follows:

n = m*(n/m) + n%m

Equation 2 looks almost identical to Equation 1, but it's computed in a very different way! n % m still represents the number of leftover pieces after forming groups, but the nature of the groups differ here. In Equation 1 we had (n/m) groups (each of size m), whereas here we have m groups (each of size n/m). If this is enough information for you, then go now and solve the problem! If not, then read on for a more detailed explanation.

In order to explain Equation 2, I am first going to illustrate Equation 1 with an example. To this end, consider the quantity 8 % 3. We can write it down as follows:

8 = (3 + 3) + 2

That is, from 8 pieces we are able to make 2 groups (each of size 3), but we cannot make a third, so the number of leftover pieces (i.e. 8 % 3) is 2.

With this understanding, consider the following illustration of 8 / 3:

8 = 2.67 + 2.67 + 2.67

Here we start with 8 pieces and split them into 3 groups. The quantity 8 / 3 represents the number of pieces in each group (which is 2.67). Now we can manipulate the expression in Equation 3 in the following way:

8 = 2.67 2.67 2.67 = (2.0 + 2.0 + 2.0) + (.67 + .67 + .67) = (2 + 2 + 2) + 2 = 3*2 + 2.

All we did was strip off the decimal components (i.e. leftover portions) of each group and moved them off to the right. Notice that the rightmost expression now matches Equation 2 very closely! Concretely, relating this back to Equation 2 we have

8 = 3*2 + 2 = 3*(8/3) + 2.

Even though this equation looks almost identical to Equation 1, we now have exactly m groups (with some leftover), which is exactly what Problem 478B calls for! All that remains now is to distribute the leftover 2 (i.e. n % m) participants as evenly as possible into the m groups we've computed. It's not hard to see the way to do this is to add one participant to each group until we run out of leftover participants. Concretely, in our running example this would be the following:

8 = (2 + 2 + 2) + 2 = (2+1) + (2+1) + 2.

Hence to distribute n participants as evenly as possible into m teams, we will wind up with n % m teams (each of size n/m + 1) and m — (n % m) teams (each of size n / m).