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> | ||
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~Interstigation | ~Interstigation | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let the common different 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 | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:15, 16 January 2023
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
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=409
~Interstigation
Solution 2
Let the common different 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 $15-9=\boxed{\textbf{(C) } 6$ (Error compiling LaTeX. Unknown error_msg).
~MathFun1000
Video Solution
~STEMbreezy
Video Solution
~savannahsolver
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.