Putting 33 students in each bus, the teacher has one student left over.

Let a be the number of people, b be the number of cars, and c the
number of people per car for the second try.  Furthermore, for more
generality replace 33 with n.  We then have that (a-1)/n = b (first
trip) and a/(b-1) = c (second trip) ==> a = n*b + 1 ==> c = a/(b-1) =
(n*b+1)/(b-1) = (n*b-n)/(b-1) + (n+1)/(b-1) = 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=n+1 and it is (a,b,c)=(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=33, so n+1=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)