Difference between revisions of "1993 AJHSME Problems/Problem 9"
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5</math> | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | By the chart, we can see that the "<math>*</math>" operation is actually multiplication modulo <math>5</math>. Thus, we can do <math>(2*4)*(1*3)\rightarrow(2\cdot4)\cdot(1\cdot3)=8\cdot3=24\rightarrow\boxed{\text{(D)}\ 4}</math> | ||
| + | |||
| + | -JeffersonJ | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:44, 16 June 2022
Contents
Problem
Consider the operation 
defined by the following table:
![\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]](http://latex.artofproblemsolving.com/8/4/f/84f712e7e4892f74560db780058a7bd3b1c60a62.png)
For example, 
. Then 
Solution 1
Using the chart, 
and 
. Therefore, 
.
Solution 2
By the chart, we can see that the "" operation is actually multiplication modulo 
. Thus, we can do 
-JeffersonJ
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
