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

(Video Solution 1(easy to digest) by Power Solve)
(Video Solution by Central Valley Math Circle (Goes through full thought process))
 
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==Solution 1==
 
==Solution 1==
  
Using the process of elimination, we can find the following:
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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:  
A is possible: <math>2*3</math>
 
C is possible: <math>1*6</math>
 
D is possible: <math>2*6</math>
 
E is possible: <math>3*6</math>
 
So therefore, the only integer that cannot be the sum is <math>(B) \boxed{6}</math>.
 
-ILoveMath31415926535
 
  
==Video Solution 1 (easy to digest) by Power Solve==
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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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The only integer that cannot be the sum is <math>\boxed{\textbf{(B) } 6}</math>.
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==Solution 2==
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First, see that in order for the two numbers to have a product of 6, it must either have 6 as a factor or have factors that have a product of 6 (Notably <math>2\times 3</math>).
 +
 
 +
When we look at the answer choices, we see that answers <math>C, D, E</math> all are more than 6, so they can all be added as a sum of 6. Answer <math>A</math> works because <math>2+3=5</math>, so 6 is the only answer choice that doesn't work. Thus, the answer is <math>\boxed{\textbf{(B) } 6}</math>.
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== Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
 +
Youtube Link ⬇️
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https://youtu.be/SreOL9LLmiI
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~ ChillGuyDoesMath :)
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==Video Solution by Central Valley Math Circle (Goes through full thought process)==
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https://youtu.be/SjMK0jrVgj0
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~mr_mathman
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== Video by MathTalks_Now ==
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https://www.youtube.com/watch?v=crn37TRMLv4
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-rc1219
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==Video Solution by Math-X (MATH-X)==
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https://youtu.be/BaE00H2SHQM?si=W8N4vwOx84KauOA6&t=1108
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~Math-X
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==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=eqTGEN2doXnn-oZE&t=443
 +
 
 +
~hsnacademy
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 +
==Video Solution  (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
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 +
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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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==Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)==
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https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR
 +
 +
~Thesmartgreekmathdude
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==Video Solution by WhyMath==
 +
https://youtu.be/QD016SsgJBQ
 +
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Latest revision as of 18:37, 20 January 2025

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: $2\times 6$

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

Solution 2

First, see that in order for the two numbers to have a product of 6, it must either have 6 as a factor or have factors that have a product of 6 (Notably $2\times 3$).

When we look at the answer choices, we see that answers $C, D, E$ all are more than 6, so they can all be added as a sum of 6. Answer $A$ works because $2+3=5$, so 6 is the only answer choice that doesn't work. Thus, the answer is $\boxed{\textbf{(B) } 6}$.


Video Solution 1 (Detailed Explanation) 🚀⚡📊

Youtube Link ⬇️

https://youtu.be/SreOL9LLmiI

~ ChillGuyDoesMath :)


Video Solution by Central Valley Math Circle (Goes through full thought process)

https://youtu.be/SjMK0jrVgj0

~mr_mathman

Video by MathTalks_Now

https://www.youtube.com/watch?v=crn37TRMLv4

-rc1219

Video Solution by Math-X (MATH-X)

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

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=eqTGEN2doXnn-oZE&t=443

~hsnacademy

Video Solution (easy to digest) by Power Solve

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


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

Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)

https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR

~Thesmartgreekmathdude

Video Solution by WhyMath

https://youtu.be/QD016SsgJBQ

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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