Problem:
How many distinct four-tuples (a,b,c,d) of rational numbers are there with
alog102+blog103+clog105+dlog107=2005?
Answer Choices:
A. 0
B. 1
C. 17
D. 2004
E. infinitely many
Solution:
The given equation is equivalent to
log10(2a⋅3b⋅5c⋅7d)=2005, so 2a⋅3b⋅5c⋅7d=102005=22005⋅52005
Let M be the least common denominator of a,b,c and d. It follows that
2Ma⋅3Mb⋅5Mc⋅7Md=22005M⋅52005M
Since the exponents are all integers, the Fundamental Theorem of Arithmetic implies that
Ma=2005M,Mb=0,Mc=2005M, and Md=0
Hence the only solution is (a,b,c,d)=(2005,0,2005,0).
The problems on this page are the property of the MAA's American Mathematics Competitions