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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 difference 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

Solution 3

Let the smallest number be $x$. Since the integers are equally spaced, and there are three of them, the middle number ($15$) is the arithmetic mean of the other two numbers ($x$ and $4x$). Thus, we set up the equation $(4x + x)/3 = 15$, and, solving for $x$, get $x = 6$. Since $6$ is the smallest number out of the list $6, 15, 24$ ($24$ because it equals $4x$), the answer is $\boxed{\textbf{(C) } 6}$.

~scthecool

Solution 4

Let the smallest number be $x$. Because $x$ and $4x$ are equally spaced from $15$, $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$, we get that the mean is also $2.5x$. We get that $2.5x=15$, and solving gets $x=\boxed{\textbf{(C) } 6}$.

~DrDominic

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

https://youtu.be/oUEa7AjMF2A?si=bwDG0eKuI9uNqoOW&t=677

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/8PDJzeebmLw

~Education, the Study of Everything

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

https://youtu.be/fY87Z0753NI

~harungurcan

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
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

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