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

m (→‎Solution 2 (Brute force) VERY NOT RECOMMENDED)
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==Solution 2 (Brute force) UNRECOMMENDED==
 
==Solution 2 (Brute force) UNRECOMMENDED==
 
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>
 
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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−
~ xxsc
 
  
 
==See Also==
 
==See Also==

Revision as of 20:14, 4 January 2022

Problem

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

$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

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://youtu.be/rQUwNC0gqdg?t=300

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

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) UNRECOMMENDED

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

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
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

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