Difference between revisions of "1993 AJHSME Problems/Problem 9"

Revision as of 00:11, 5 July 2013

Consider the operation $*$ defined by the following table:

$\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}$

For example, $3*2=1$ . Then $(2*4)*(1*3)=$

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=\boxed{\text{(D)}\ 4}$ .

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

@@ Line 1: / Line 1: @@
-== Problem 9 ==
+== Problem ==
 Consider the operation <math>*</math> defined by the following table:
@@ Line 17: / Line 17: @@
 ==Solution==
 Using the chart, <math> (2*4)=3 </math> and <math> (1*3)=3 </math>. Therefore, <math> (2*4)*(1*3)=3*3=\boxed{\text{(D)}\ 4} </math>.
+==See Also==
+{{AJHSME box|year=1993|num-b=8|num-a=10}}
+{{MAA Notice}}