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Difference between revisions of "2022 AMC 8 Problems/Problem 6"
(→‎Problem 6: Typically we just write PROBLEM without the number in this page.)
 
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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math>
 
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math>
  
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==Solution==
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==Solution 1==
  
 
Let the smallest number be <math>x.</math> It follows that the largest number is <math>4x.</math>
 
Let the smallest number be <math>x.</math> It follows that the largest number is <math>4x.</math>
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x &= \boxed{\textbf{(C) } 6}.
 
x &= \boxed{\textbf{(C) } 6}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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~MRENTHUSIASM
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 +
==Solution 2==
 +
 +
Let the common difference of the arithmetic sequence be <math>d</math>. Consequently, the smallest number is <math>15-d</math> and the largest number is <math>15+d</math>. As the largest number is <math>4</math> times the smallest number, <math>15+d=60-4d\implies d=9</math>. Finally, we find that the smallest number is <math>15-9=\boxed{\textbf{(C) } 6}</math>.
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 +
~MathFun1000
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 +
==Solution 3==
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 +
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>.
 +
 +
~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==  
 
==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}}

Latest revision as of 19:33, 2 September 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

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