1993 AJHSME Problems/Problem 9

Problem

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 $(2*4)*(1*3)=$

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

Solution 1

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

Solution 2

By the chart, we can see that the " $*$ " operation is actually multiplication modulo $5$ . Thus, we can do $(2*4)*(1*3)\rightarrow(2\cdot4)\cdot(1\cdot3)=8\cdot3=24\rightarrow\boxed{\text{(D)}\ 4}$

-JeffersonJ

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1993 AJHSME Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See Also