# 2018 AMC 8 Problems/Problem 25

## Problem 25

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? $\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

## 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 ending with $64$. Now, by counting how many numbers are between these, we find the answer to be $\boxed{\textbf{(E) }58}$

~ xxsc

## Video Solution

https://youtu.be/ZZloby9pPJQ ~DSA_Catachu

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 