Difference between revisions of "2018 AMC 8 Problems/Problem 25"
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<math>\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58</math> | <math>\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58</math> | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=300 | ||
==Solution 1== | ==Solution 1== |
Revision as of 19:30, 27 October 2020
Contents
Problem 25
How many perfect cubes lie between and , inclusive?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=300
Solution 1
We compute . We're all familiar with what is, namely , which is too small. The smallest cube greater than it is . is too large to calculate, but we notice that , which therefore clearly will be the largest cube less than . So, the required number of cubes is
Solution 2 (Brute force)
First, . Then, . Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from and ending with . Now, by counting how many numbers are between these, we find the answer to be
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Video Solution
https://www.youtube.com/watch?v=pbnddMinUF8 -Happytwin
https://youtu.be/ZZloby9pPJQ ~DSA_Catachu
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.