Problem:
1000Γ1993Γ0.1993Γ10=
Answer Choices:
A. 1.993Γ103
B. 1993.1993
C. (199.3)2
D. 1,993,001.993
E. (1993)2
Solution:
1000Γ1993Γ0.1993Γ10β=((1000Γ10)Γ0.1993)Γ1993=(10,000Γ0.1993)Γ1993=1993Γ1993=(1993)2.β
Answer: Eβ.
The problems on this page are the property of the MAA's American Mathematics Competitions