<pre>
Let a be the number of students, b be the number of buses, and c the
number of students per bus for the second try.

For more generality replace 33 with n.  We then have that (a-1)/n = b (first
trip) and a/(b-1) <code> c (second trip) </code><code>> a </code> n*b + 1 <code></code>> c <code> a/(b-1) </code>
(n'''b+1)/(b-1) <code> (n'''b-n)/(b-1) + (n+1)/(b-1) </code> n + (n+1)/(b-1).  From
this all solutions may be derived, and they are

     (a,b,c) = (n*(n+1)/d + n + 1, (n+1)/d + 1, n + d)

where d || (n+1) (i.e. d divides n+1).  The answer with smallest a
results from letting d<code>n+1 and it is (a,b,c)</code>(2n+1, 2, 2n+1);
likewise the answer with largest a is ((n+1)^2,n+2,n+1).

In the specific case we have n<code>33, so n+1</code>34=2*17 and we have
4 answers.  They are:

 (a,b,c) = (67,2,67), (100,3,50), (595,18,35), (1156,35,34)

--Chris Long (clong@cnj.digex.com)
</pre>
