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 will clearly 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)

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 $\boxed{\textbf{(E) }58}$ .

Solution 4 (If you do not notice that $2^{18}=(2^6)^3=64^3$ )

There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$ , which is over $2^8+1=257$ . We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$ , which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gives us 64 and $64^3 = 262,144$ , which is perfect for the $2^{18}+1=262,145$ . But we are not done. Since it is inclusive, we must add 1 to this solution, which give us $\boxed{\textbf{(E) }58}$ .

Video Solutions

https://www.youtube.com/watch?v=pbnddMinUF8

-Happytwin

https://www.youtube.com/watch?v=2e7gyBYEDOA

- MathEx

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

~David

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=297

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

Page

Toolbox

Search

2018 AMC 8 Problems/Problem 25

Contents

Problem

Solution 1

Solution 2 (Brute force)

Solution 3 (Guessing)

Solution 4 (If you do not notice that $2^{18}=(2^6)^3=64^3$ )

Video Solutions

Video Solution by OmegaLearn

See Also

Page

Toolbox

Search

2018 AMC 8 Problems/Problem 25

Contents

Problem

Solution 1

Solution 2 (Brute force)

Solution 3 (Guessing)

Solution 4 (If you do not notice that )

Video Solutions

Video Solution by OmegaLearn

See Also

Solution 4 (If you do not notice that $2^{18}=(2^6)^3=64^3$ )