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

Revision as of 12:54, 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

Video Solution 2 by SpreadTheMathLove

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

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.

@@ Line 22: / Line 22: @@
 ==Video Solution 1 (easy to digest) by Power Solve==
 https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239
+==Video Solution 2 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=L83DxusGkSY
 ==See Also==
 {{AMC8 box|year=2024|num-b=4|num-a=6}}
 {{MAA Notice}}

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