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Difference between revisions of "2022 AMC 8 Problems/Problem 13"

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/oUEa7AjMF2A?si=q1GMT9DB8-AY0Ld5&t=1882
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~Math-X
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/YnRJ1Q0GxgU
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/Ij9pAy6tQSg?t=1110
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~Interstigation
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==Video Solution==
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https://youtu.be/p29Fe2dLGs8?t=139
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~STEMbreezy
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==Video Solution==
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https://youtu.be/P3IgH0ns-7M
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~savannahsolver
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==Video Solution==
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https://youtu.be/yT-9qbMw8y8
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~harungurcan
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=12|num-a=14}}
 
{{AMC8 box|year=2022|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:34, 2 September 2024

Problem

How many positive integers can fill the blank in the sentence below?

“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$.”

$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Solution

Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$

By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there are $\boxed{\textbf{(D) } 9}$ possible values for $d.$

~MRENTHUSIASM

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=q1GMT9DB8-AY0Ld5&t=1882

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/YnRJ1Q0GxgU

~Education, the Study of Everything

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1110

~Interstigation

Video Solution

https://youtu.be/p29Fe2dLGs8?t=139

~STEMbreezy

Video Solution

https://youtu.be/P3IgH0ns-7M

~savannahsolver

Video Solution

https://youtu.be/yT-9qbMw8y8

~harungurcan

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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