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

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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 1==
 
==Solution 1==
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Since the problem is giving us the options, we can try all of them. The differences MUST be equal because the numbers have to be evenly spaced on the number line. We see that 4 cannot work because 15 - 4 = 11, and 4 * 4 = 16. 16 - 15 is not 11, so this option does not work. Next, we can try 5. We see that this also does not work because 15 - 5 = 10, and 5 * 4 = 20. 20 - 15 = 5, and this is not 10. We can then try 6 to see that it does work. 15 - 6 = 9, and 6 * 4 = 24. 24 - 15 = 9. The differences are equivalent, so we can see that 6 indeed does work. Therefore, the answer is <math>\boxed{\textbf{(C)} ~6}</math>.
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~~Brainiacs77~~
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 +
==Solution 2==
  
 
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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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2==
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==Solution 3==
  
 
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>.
 
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
 
~MathFun1000
  
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==Solution 3==
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==Solution 4==
  
 
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>.
 
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>.

Latest revision as of 12:45, 27 December 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

Since the problem is giving us the options, we can try all of them. The differences MUST be equal because the numbers have to be evenly spaced on the number line. We see that 4 cannot work because 15 - 4 = 11, and 4 * 4 = 16. 16 - 15 is not 11, so this option does not work. Next, we can try 5. We see that this also does not work because 15 - 5 = 10, and 5 * 4 = 20. 20 - 15 = 5, and this is not 10. We can then try 6 to see that it does work. 15 - 6 = 9, and 6 * 4 = 24. 24 - 15 = 9. The differences are equivalent, so we can see that 6 indeed does work. Therefore, the answer is $\boxed{\textbf{(C)} ~6}$.

~~Brainiacs77~~


Solution 2

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 3

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 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

Video Solution by Dr. David

https://youtu.be/fIRZycIROwM

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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