Art of Problem Solving

2022 AMC 8 Problems/Problem 6: Difference between revisions

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~Interstigation
~Interstigation
==Video Solution==
https://youtu.be/1xspUFoKDnU
~STEMbreezy


==See Also==  
==See Also==  
{{AMC8 box|year=2022|num-b=5|num-a=7}}
{{AMC8 box|year=2022|num-b=5|num-a=7}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 20:11, 25 December 2022

Problem

Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution

Let the smallest number be $x.$ It follows that the largest number is $4x.$

Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*} ~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=409

~Interstigation

Video Solution

https://youtu.be/1xspUFoKDnU

~STEMbreezy

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing