2022 AMC 8 Problems/Problem 6
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
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 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.