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Difference between revisions of "2022 AMC 8 Problems/Problem 6"
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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. 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. These numbers are equivalent, so we can see that 9 indeed does work. Therefore, the answer is \boxed{\textbf{(C) } 6}.
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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>
  

Revision as of 13:57, 26 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. 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. These numbers are equivalent, so we can see that 9 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 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

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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