2018 AMC 8 Problems/Problem 25

Problem

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?

$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution 1

We compute $2^8+1=257$. We're all familiar with what $6^3$ is, namely $216$, which is too small. The smallest cube greater than it is $7^3=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$, which therefore clearly will be the largest cube less than $2^{18}+1$. So, the required number of cubes is $64-7+1= \boxed{\textbf{(E) }58}$

Solution 2 (Brute force) UNRECOMMENDED

First, $2^8+1=257$. Then, $2^{18}+1=262145$. Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ends with $64$. Now, by counting how many numbers are between these, we find the answer to be $\boxed{\textbf{(E) }58}$

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 $+ 1$. People will forget the $+1$ so the only possibilities are C and E. We can clearly see that C is too small so our answer is $\boxed{\textbf{(E) }58}$.

~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

https://euclideanmathcircle.wixsite.com/emc1/videos?wix-vod-video-id=5f1ae882ac754e54906db7cfb62c61f6&wix-vod-comp-id=comp-kn8844mv

https://youtu.be/geZupO75zUw

~savannahsolver

https://www.youtube.com/watch?v=r0e_6dXViRI

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS