Problem:
The operation β is defined for all nonzero numbers by aβb=ba2β. Determine [(1β2)β3]β[1β(2β3)].
Answer Choices:
A. β32β
B. β41β
C. 0
D. 41β
E. 32β
Solution:
We have
(1β2)β3=212ββ3=21ββ3=3(21β)2β=341ββ=121β
and
1β(2β3)=1β(322β)=1β34β=34β12β=43β
Therefore,
(1β2)β3β1β(2β3)=121ββ43β=121ββ129β=β128β=β32β.
Answer: Aβ.
The problems on this page are the property of the MAA's American Mathematics Competitions