2022 AMC 8 Problems/Problem 6

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 1

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

Solution 2

Let the common different of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\implies d=9$ . Finally, we find that the smallest number is $15-9=\boxed{\textbf{(C) } 6}$ .

~MathFun1000

Video Solution

https://youtu.be/1xspUFoKDnU

~STEMbreezy

Video Solution

https://youtu.be/evYD-UMJotA

~savannahsolver

Video Solution

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

~Interstigation

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/8PDJzeebmLw

~Education, the Study of Everything

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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

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2022 AMC 8 Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

Video Solution

Video Solution

Video Solution

Video Solution (CREATIVE THINKING!!!)

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