Difference between revisions of "2022 AMC 8 Problems/Problem 6"
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==Solution 3== | ==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>. |
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+ | ~DrDominic | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
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~harungurcan | ~harungurcan | ||
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==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 16:11, 21 April 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
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.