Problem:
For each pair of real numbers aξ =b, define the operation β as
(aβb)=aβba+bβ
What is the value of ((1β2)β3)?
Answer Choices:
A. β32β
B. β51β
C. 0
D. 21β
E. This value is not defined.
Solution:
First we have
(1β2)=1β21+2β=β3
Then
((1β2)β3)=(β3β3)=β3β3β3+3β=β60β=0
Answer: Cβ.
The problems on this page are the property of the MAA's American Mathematics Competitions