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

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==Problem 25==
 
==Problem 25==
How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive ?
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How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive?
  
 
<math>\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58</math>
 
<math>\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58</math>

Revision as of 02:14, 2 December 2018

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, 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 $2^{18}+1$. Therefore, the required number 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
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All AJHSME/AMC 8 Problems and Solutions

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