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

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==Solution 2 (Brute force) (works only if you have enough time for calculations)==
 
==Solution 2 (Brute force) (works only if you have enough time for calculations)==
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First, <math>2^8+1=257</math>. Then, <math>2^{18}+1=262145</math>. Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from <math>7</math> and ends with <math>64</math>. Now, by counting how many numbers are between these, we find the answer to be <math>\boxed{\textbf{(E) }58}</math>  
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First, <math>2^8+1=257</math>. Then, <math>2^{18}+1=262145</math>. Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from <math>7</math> and ends with <math>64</math>. Now, by counting how many numbers are between these, we find the answer to be <math>\boxed{\textbf{(E) }58}</math>.
  
 
~ edited by gbatkhuu1
 
~ edited by gbatkhuu1

Revision as of 11:31, 7 January 2023

Problem

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) (works only if you have enough time for calculations)

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 $\boxed{\textbf{(E) }58}$.

~ edited by gbatkhuu1

Solution 3 (Guessing)

First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number $-$ the lowest number $+ 1$. People will forget the $+1$ so the only possibilities are C and E. We can clearly see that C is too small so our answer is $\boxed{\textbf{(E) }58}$.

~MathFun1000

Video Solutions

https://www.youtube.com/watch?v=pbnddMinUF8 -Happytwin

https://youtu.be/ZZloby9pPJQ ~DSA_Catachu

https://www.youtube.com/watch?v=2e7gyBYEDOA - MathEx

https://euclideanmathcircle.wixsite.com/emc1/videos?wix-vod-video-id=5f1ae882ac754e54906db7cfb62c61f6&wix-vod-comp-id=comp-kn8844mv

https://youtu.be/geZupO75zUw

~savannahsolver

https://www.youtube.com/watch?v=r0e_6dXViRI

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=297

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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