2018 AMC 8 Problems/Problem 25

Revision as of 10:29, 23 November 2018 by Frostfox (talk | contribs) (Solution)

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

We compute $2^8+1=257$. We're all familiar with what $6^3$ is, it's $256$, barely smaller. 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 is the largest cube less than $2^{18}+1$. Therefore, the amount of cubes is $64-7+1= \boxed{\textbf{(E) }58}$

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