Problem:
How many ordered triples of integers (a,b,c), with a≥2,b≥1, and c≥0, satisfy both logab=c2005 and a+b+c=2005?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
The two equations are equivalent to b=a(c2005) and c=2005−b−a, so
c=2005−a(c2005)−a
If c>1, then
b≥2(22005)>2005>2005−a−c=b
which is a contradiction. For c=0 and for c=1, the only solutions are the ordered triples (2004,1,0) and (1002,1002,1), respectively. Thus the number of solutions is 2.
The problems on this page are the property of the MAA's American Mathematics Competitions