Problem: 21β23=\dfrac{2}{1-\dfrac{2}{3}}=1β32β2β=
Answer Choices:
A. β3-3β3 B. β43-\dfrac{4}{3}β34β C. 23\dfrac{2}{3}32β D. 222 E. 666
Solution:
21β23=213=2Γ3=6\dfrac{2}{1-\dfrac{2}{3}}=\dfrac{2}{\dfrac{1}{3}}=2 \times 3=61β32β2β=31β2β=2Γ3=6.
Answer: 6\boxed{6}6β.
The problems on this page are the property of the MAA's American Mathematics Competitions