Difference between revisions of "2001 AMC 8 Problems/Problem 18"
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The only way to get a multiple of 5 is to have at least one 5. If the first dice rolls a 5, there are 6 ways to get a multiple of 5. If the second dice rolls a 5, there are also 6 ways. However, there is one case that is repeated: both dice roll a 5. Therefore, there are 6 + 6 - 1 = 11, and there is a total of 6 x 6 ways, so the probability is <math>\text{(D)}\ \dfrac{11}{36}</math> | The only way to get a multiple of 5 is to have at least one 5. If the first dice rolls a 5, there are 6 ways to get a multiple of 5. If the second dice rolls a 5, there are also 6 ways. However, there is one case that is repeated: both dice roll a 5. Therefore, there are 6 + 6 - 1 = 11, and there is a total of 6 x 6 ways, so the probability is <math>\text{(D)}\ \dfrac{11}{36}</math> | ||
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==Video Solution== | ==Video Solution== | ||
Revision as of 10:59, 5 January 2023
Problem
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
Solution 1
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 
Solution 2
The only way to get a multiple of 5 is to have at least one 5. If the first dice rolls a 5, there are 6 ways to get a multiple of 5. If the second dice rolls a 5, there are also 6 ways. However, there is one case that is repeated: both dice roll a 5. Therefore, there are 6 + 6 - 1 = 11, and there is a total of 6 x 6 ways, so the probability is 
-ILoveMath31415926535
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
