Difference between revisions of "2008 AMC 8 Problems/Problem 24"

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==Problem 24==
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==Problem==
Ten tiles numbered <math>1</math> through <math>10</math> 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?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Ten tiles numbered <math>1</math> through <math>10</math> 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?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
<math>\textbf{(A)}\ \frac{1}{10}\qquad
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<math>\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}</math>
\textbf{(B)}\ \frac{1}{6}\qquad
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\textbf{(C)}\ \frac{11}{60}\qquad
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==Solution==
\textbf{(D)}\ \frac{1}{5}\qquad
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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>.
\textbf{(E)}\ \frac{7}{30}</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}}
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{{MAA Notice}}

Revision as of 22:20, 11 February 2020

Problem

Ten tiles numbered $1$ through $10$ 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?

$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}$

Solution

The numbers can at most multiply to be $60$. The squares less than $60$ are $1,4,9,16,25,36,$ and $49$. The possible pairs are $(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),$ and $(9,4)$. There are $11$ choices and $60$ possibilities giving a probability of $\boxed{\textbf{(C)}\ \frac{11}{60}}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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

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