Difference between revisions of "2001 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | This is equivalent to asking for the probability that at least one of the numbers is a multiple of <math> 5 </math>, since if one of the numbers is a multiple of <math> 5 </math>, then the product with it and another integer is also a multiple of <math> 5 </math>, and if a number is a multiple of <math> 5 </math>, then since <math> 5 </math> is prime, one of the factors must also have a factor of <math> 5 </math>, and <math> 5 </math> is the only multiple of <math> 5 </math> on a die, so one of the numbers rolled must be a <math> 5 </math>. To find the probability of rolling at least one <math> 5 </math>, we can find the probability of not rolling a <math> 5 </math> and subtract that from <math> 1 </math>, since you either roll a <math> 5 </math> or not roll a <math> 5 </math>. The probability of not rolling a <math> 5 </math> on either dice is <math> (\frac{5}{6})(\frac{5}{6})=\frac{25}{36} </math>. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of <math> 5 </math>, is <math> 1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}} </math> | + | |
+ | This is equivalent to asking for the probability that at least one of the numbers is a multiple of <math> 5 </math>, since if one of the numbers is a multiple of <math> 5 </math>, then the product with it and another integer is also a multiple of <math> 5 </math>, and if a number is a multiple of <math> 5 </math>, then since <math> 5 </math> is prime, one of the factors must also have a factor of <math> 5 </math>, and <math> 5 </math> is the only multiple of <math> 5 </math> on a die, so one of the numbers rolled must be a <math> 5 </math>. To find the probability of rolling at least one <math> 5 </math>, we can find the probability of not rolling a <math> 5 </math> and subtract that from <math> 1 </math>, since you either roll a <math> 5 </math> or not roll a <math> 5 </math>. The probability of not rolling a <math> 5 </math> on either dice is <math> \left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36} </math>. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of <math> 5 </math>, is <math> 1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}} </math> | ||
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+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=17|num-a=19}} | {{AMC8 box|year=2001|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Revision as of 02:12, 3 March 2022
Contents
Problem
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
Solution
This is equivalent to asking for the probability that at least one of the numbers is a multiple of , since if one of the numbers is a multiple of , then the product with it and another integer is also a multiple of , and if a number is a multiple of , then since is prime, one of the factors must also have a factor of , and is the only multiple of on a die, so one of the numbers rolled must be a . To find the probability of rolling at least one , we can find the probability of not rolling a and subtract that from , since you either roll a or not roll a . The probability of not rolling a on either dice is . Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of , is
Video Solution
https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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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