2022 AMC 8 Problems/Problem 6

Revision as of 20:07, 10 January 2024 by Scthecool (talk | contribs)

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

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 (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

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