2008 AMC 8 Problems/Problem 24

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?

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

Video Solution

https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David

Video Solution by Soo

https://youtu.be/IUh28kMLLVE - Soo, DRMS, NM

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 $\boxed{\textbf{(C)}\ \frac{11}{60}}$ .

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2008 AMC 8 Problems/Problem 24

Contents

Problem

Video Solution

Video Solution by Soo

Solution

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