2001 AMC 8 Problems/Problem 18: Difference between revisions

Latest revision as of 23:44, 2 November 2025

Problem

Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?

$\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$

Solution 1

This is equivalent to asking for the probability that at least one of the numbers is a multiple of $5$ , since if one of the numbers is a multiple of $5$ , then the product with it and another integer is also a multiple of $5$ , and if a number is a multiple of $5$ , then since $5$ is prime, one of the factors must also have a factor of $5$ , and $5$ is the only multiple of $5$ on a die, so one of the numbers rolled must be a $5$ . To find the probability of rolling at least one $5$ , we can find the probability of not rolling a $5$ and subtract that from $1$ , since you either roll a $5$ or not roll a $5$ . The probability of not rolling a $5$ on either dice is $\left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36}$ . Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $5$ , is $1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}}$

Solution 2 (quick & easy)

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, one case is repeated: both dice roll a $5$ . Therefore, there are $6 + 6 - 1 = 11$ , and there is a total of $6 \times 6$ ways, so the probability is $\text{(D)}\ \dfrac{11}{36}$

Solution by IHateMath314265169

Video Solution

https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM

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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@@ Line 5: / Line 5: @@
 <math>\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}</math>
-==Solution==
+==Solution 1==
-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})(\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>
+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>
+==Solution 2 (quick & easy)==
+The only way to get a multiple of <math> 5 </math> is to have at least one <math> 5 </math>. If the first dice rolls a <math> 5 </math>,  there are <math> 6 </math> ways to get a multiple of <math> 5 </math>. If the second dice rolls a <math> 5 </math>, there are also <math> 6 </math> ways. However, one case is repeated: both dice roll a <math> 5 </math>. Therefore, there are <math> 6 + 6 - 1 = 11 </math>, and there is a total of <math> 6 \times 6 </math> ways, so the probability is <math>\text{(D)}\ \dfrac{11}{36}</math>
+Solution by [[User:IHateMath314265169|IHateMath314265169]]
+==Video Solution==
+https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM
 ==See Also==
 {{AMC8 box|year=2001|num-b=17|num-a=19}}
 {{MAA Notice}}

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