Difference between revisions of "2018 AMC 8 Problems/Problem 25"

Revision as of 23:08, 27 November 2020

Problem 25

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

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

Video Solutions

https://youtu.be/rQUwNC0gqdg?t=300

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

https://youtu.be/ZZloby9pPJQ ~DSA_Catachu

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

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)

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

~ xxsc

2018 AMC 8 (Problems • Answer Key • Resources)
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