AMC8 2018 8, Problem 25

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

Solution: 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 $64-7+1= \fbox{58}$.

Answer: 58