Difference between revisions of "2018 AMC 8 Problems/Problem 25"
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~MathFun1000 | ~MathFun1000 | ||
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+ | ==Solution 4 (EVEN MORE BRUTE FORCE)== | ||
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+ | Just Throw this: `int(((2**18+1)**(1/3)//1)-((2**8+1)**(1/3)//1+1)+1)` into Python. | ||
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+ | Or, calculate it by yourself. | ||
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+ | ~hefei417 | ||
==Video Solutions== | ==Video Solutions== |
Revision as of 03:21, 6 May 2022
Contents
Problem
How many perfect cubes lie between and
, inclusive?
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) UNRECOMMENDED
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 ends with
. Now, by counting how many numbers are between these, we find the answer to be
Solution 3 (Guessing)
First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number the lowest number
. People will forget the
so the only possibilities are C and E. We can clearly see that C is too small so our answer is
.
~MathFun1000
Solution 4 (EVEN MORE BRUTE FORCE)
Just Throw this: `int(((2**18+1)**(1/3)//1)-((2**8+1)**(1/3)//1+1)+1)` into Python.
Or, calculate it by yourself.
~hefei417
Video Solutions
https://www.youtube.com/watch?v=pbnddMinUF8 -Happytwin
https://youtu.be/ZZloby9pPJQ ~DSA_Catachu
https://www.youtube.com/watch?v=2e7gyBYEDOA - MathEx
https://youtu.be/rQUwNC0gqdg?t=300
~savannahsolver
https://www.youtube.com/watch?v=r0e_6dXViRI
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.