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

(Solution 1)
(Problem)
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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>
 
<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math>
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==Solution 1==
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Using the process of elimination, we can find the following:
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A is possible: <math>2*3</math>
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C is possible: <math>1*6</math>
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D is possible: <math>2*6</math>
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E is possible: <math>3*6</math>
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So therefore, the only integer that cannot be the sum is <math>B\boxed{6}</math>.
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-ILoveMath31415926535

Revision as of 15:05, 25 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: A is possible: $2*3$ C is possible: $1*6$ D is possible: $2*6$ E is possible: $3*6$ So therefore, the only integer that cannot be the sum is $B\boxed{6}$. -ILoveMath31415926535