Problem:
How many ordered triples of integers (a,b,c), with a≥2,b≥1, and c≥0, satisfy both loga​b=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