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> | ||
| − | ==Solution== | + | ==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> | ||
| Line 16: | Line 16: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | ~MathFun1000 | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | 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== | ==Video Solution== | ||
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=409 | https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=409 | ||
~Interstigation | ~Interstigation | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/fY87Z0753NI | ||
| + | |||
| + | ~harungurcan | ||
| + | |||
| + | ==Video Solution by Dr. David== | ||
| + | https://youtu.be/fIRZycIROwM | ||
==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 20:27, 25 November 2024
Contents
Problem
Three positive integers are equally spaced on a number line. The middle number is 
and the largest number is 
times the smallest number. What is the smallest of these three numbers?
Solution 1
Let the smallest number be 
It follows that the largest number is 
Since 
and 
are equally spaced on a number line, we have
~MRENTHUSIASM
Solution 2
Let the common difference of the arithmetic sequence be 
. Consequently, the smallest number is 
and the largest number is 
. As the largest number is times the smallest number, 
. Finally, we find that the smallest number is 
.
~MathFun1000
Solution 3
Let the smallest number be 
. Because and are equally spaced from 
, must be the average. By adding and and dividing by 
, we get that the mean is also 
. We get that 
, and solving gets 
.
~DrDominic
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=bwDG0eKuI9uNqoOW&t=677
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=409
~Interstigation
Video Solution
~harungurcan
Video Solution by Dr. David
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. 
