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

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<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>
  
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==Video Solution==
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==Video Solutions==
 
https://youtu.be/rQUwNC0gqdg?t=300
 
https://youtu.be/rQUwNC0gqdg?t=300
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https://www.youtube.com/watch?v=pbnddMinUF8
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-Happytwin
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https://youtu.be/ZZloby9pPJQ ~DSA_Catachu
  
 
==Solution 1==
 
==Solution 1==
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==Video Solution==
 
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https://www.youtube.com/watch?v=pbnddMinUF8
 
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-Happytwin
 
  
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https://youtu.be/ZZloby9pPJQ ~DSA_Catachu
 
  
 
==See Also==
 
==See Also==

Revision as of 14:00, 6 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

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


See Also

2018 AMC 8 (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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