Problem: If nβ‘m=n3m2n\heartsuit m=n^3m^2nβ‘m=n3m2, what is 2β‘44β‘2\dfrac{2\heartsuit 4}{4\heartsuit 2}4β‘22β‘4β?
Answer Choices:
A. 14\dfrac{1}{4}41β
B. 12\dfrac{1}{2}21β
C. 111
D. 222
E. 444
Solution:
2β‘44β‘2=23β 4243β 22=24=(B)12\dfrac{2 \heartsuit 4}{4 \heartsuit 2}=\dfrac{2^{3} \cdot 4^{2}}{4^{3} \cdot 2^{2}}=\dfrac{2}{4}= (\text{B})\boxed{\dfrac{1}{2}} 4β‘22β‘4β=43β 2223β 42β=42β=(B)21ββ
The problems on this page are the property of the MAA's American Mathematics Competitions