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

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==Problem==
 
==Problem==
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Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of <math>6</math>. Which of the following integers cannot be the sum of the two numbers?
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<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math>
  
 
==Solution 1==
 
==Solution 1==
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First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:
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<math>\textbf{(A)}</math> is possible: <math>2\times 3</math>
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<math>\textbf{(C)}</math> is possible: <math>1\times 6</math>
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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 & Nivaar
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==Video Solution 1 (easy to digest) by Power Solve==
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https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239
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==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=W8N4vwOx84KauOA6&t=1108
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~Math-X
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=Ylw-kJkSpq8
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~NiuniuMaths
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=L83DxusGkSY
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=51pqs80PUnY
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=296
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==See Also==
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{{AMC8 box|year=2024|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 09:35, 8 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 (ProblemsAnswer KeyResources)
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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