Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 8 Problems/Problem 6"

(→‎Solution 4)
(→‎Solution 3)
(5 intermediate revisions by 3 users not shown)
Line 24: Line 24:
  
 
==Solution 3==
 
==Solution 3==
−
Let the smallest number be <math>x</math>. Since the integers are equally spaced, and there are three of them, the middle number (<math>15</math>) is the arithmetic mean of the other two numbers (<math>x</math> and <math>4x</math>). Thus, we set up the equation <math>(4x + x)/3 = 15</math>, and, solving for <math>x</math>, get <math>x = 6</math>. Since <math>6</math> is the smallest number out of the list <math>6, 15, 24</math> (<math>24</math> because it equals <math>4x</math>), the answer is <math>\boxed{\textbf{(C) }6}</math>.
 
  
−
~scthecool
+
Let the smallest number be <math>x</math>. Because <math>x</math> and <math>4x</math> are equally spaced from <math>15</math>, <math>15</math> must be the average. By adding <math>x</math> and <math>4x</math> and dividing by <math>2</math>, we get that the mean is also <math>2.5x</math>. We get that <math>2.5x=15</math>, and solving gets <math>x=\boxed{\textbf{(C) } 6}</math>.
  
−
==Solution 4==
+
~DrDominic
−
 
 
−
Let the smallest number be <math>x</math>. Because <math>x</math> and <math>4x</math> are equally spaced from 15, <math>15</math> must be the average. By adding <math>x</math> and <math>4x</math> and dividing by <math>2</math>, we get that the mean is also <math>2.5x</math>. We get that <math>2.5x</math>=<math>15</math>, and solving gets that <math>x</math>=<math>\boxed{\textbf{(C) }6}</math>.
 
  
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==

Revision as of 15:11, 21 April 2024

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 $x$. 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_6&oldid=218419"