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

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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>
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<math>\textbf{(A)}</math> is possible: <math>2\times 3</math>  
−
D is possible: <math>2*6</math>
+
 
−
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>.
+
 
−
-ILoveMath31415926535
+
<math>\textbf{(D)}</math> is possible: <math>2\times 6</math>  
 +
 
 +
<math>\textbf{(E)}</math> is possible: <math>3\times 6</math>
 +
 
 +
The only integer that cannot be the sum is <math>\boxed{\textbf{(B)\ 6}}.</math>
 +
 
 +
-ILoveMath31415926535 & countmath1 & Nivaar
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=eqTGEN2doXnn-oZE&t=443
 +
 
 +
~hsnacademy
 +
 
 +
==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
 +
 
 +
==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==
 +
{{AMC8 box|year=2024|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Latest revision as of 14:57, 9 November 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 (A Clever Explanation You’ll Get Instantly)

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

~hsnacademy

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

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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