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Difference between revisions of "2024 AMC 8 Problems/Problem 5"
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=296
  
 
==See Also==
 
==See Also==
 
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{{AMC8 box|year=2024|num-b=4|num-a=6}}
 
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Revision as of 02:28, 6 February 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

First, figure out all pairs of numbers whose product is 6. Then, 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 & Nivaar

Video Solution 1 (easy to digest) by Power Solve

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

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=W8N4vwOx84KauOA6&t=1108

~Math-X

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=Ylw-kJkSpq8

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=51pqs80PUnY

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=296

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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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