Difference between revisions of "2008 AMC 8 Problems/Problem 24"
| Line 7: | Line 7: | ||
\textbf{(D)}\ \frac{1}{5}\qquad | \textbf{(D)}\ \frac{1}{5}\qquad | ||
\textbf{(E)}\ \frac{7}{30}</math> | \textbf{(E)}\ \frac{7}{30}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | The numbers can at most multiply to be <math>60</math>. The squares less than <math>60</math> are <math>1,4,9,16,25,36,</math> and <math>49</math>. The possible pairs are <math>(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),</math> and <math>(9,4)</math>. There are <math>11</math> choices and <math>60</math> possibilities giving a probability of <math>\boxed{\textbf{(C)}\ \frac{11}{60}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=23|num-a=25}} | {{AMC8 box|year=2008|num-b=23|num-a=25}} | ||
Revision as of 03:14, 25 December 2012
Problem
Ten tiles numbered 
through 
are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
Solution
The numbers can at most multiply to be 
. The squares less than are 
and 
. The possible pairs are 
and 
. There are 
choices and possibilities giving a probability of 
.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
