Difference between revisions of "2024 AMC 8 Problems/Problem 5"

(Video Solution 1(easy to digest) by Power Solve)
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==Solution 1==
 
==Solution 1==
  
Using the process of elimination, we can find the following:
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Using the process of elimination, we can find the following:  
A is possible: <math>2*3</math>
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C is possible: <math>1*6</math>
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<math>\textbf{(A)}</math> is possible: <math>2\times 3</math>  
D is possible: <math>2*6</math>
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E is possible: <math>3*6</math>
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<math>\textbf{(C)}</math> is possible: <math>1\times 6</math>  
So therefore, the only integer that cannot be the sum is <math>(B) \boxed{6}</math>.
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-ILoveMath31415926535
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<math>\textbf{(D)}</math> is possible: <math>2\times 6</math>  
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<math>\textbf{(E)}</math> is possible: <math>3\times 6</math>
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The only integer that cannot be the sum is <math>\boxed{\textbf{(B)\ 6}}.</math>
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-ILoveMath31415926535 & countmath1
  
 
==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==
 
https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239
 
https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239

Revision as of 13:37, 26 January 2024

Problem

Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$. Which of the following integers cannot be the sum of the two numbers?

$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$

Solution 1

Using the process of elimination, we can find the following:

$\textbf{(A)}$ is possible: $2\times 3$

$\textbf{(C)}$ is possible: $1\times 6$

$\textbf{(D)}$ is possible: $2\times 6$

$\textbf{(E)}$ is possible: $3\times 6$

The only integer that cannot be the sum is $\boxed{\textbf{(B)\ 6}}.$

-ILoveMath31415926535 & countmath1

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239