2008 AMC 8 Problems/Problem 24: Difference between revisions
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==Solution== | ==Solution== | ||
The numbers can at | The numbers can at mos 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:33, 6 October 2024
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?
Video Solution
https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David
Solution
The numbers can at mos 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
