Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
 
(54 intermediate revisions by 34 users not shown)
Line 1: Line 1:
−
==Problem 25==
+
==Problem==
 
How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive?
 
How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive?
  
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
−
We compute <math>2^8+1=257</math>. We're all familiar with what <math>6^3</math> is, namely <math>216</math>, which is too small. The smallest cube greater than it is <math>7^3=343</math>. <math>2^{18}+1</math> is too large to calculate, but we notice that <math>2^{18}=(2^6)^3=64^3</math>, which therefore clearly will be the largest cube less than <math>2^{18}+1</math>. So, the required number of cubes is <math>64-7+1= \boxed{\textbf{(E) }58}</math>
+
We compute <math>2^8+1=257</math>. We're all familiar with what <math>6^3</math> is, namely <math>216</math>, which is too small. The smallest cube greater than it is <math>7^3=343</math>. <math>2^{18}+1</math> is too large to calculate, but we notice that <math>2^{18}=(2^6)^3=64^3</math>, which therefore will clearly be the largest cube less than <math>2^{18}+1</math>. So, the required number of cubes is <math>64-7+1= \boxed{\textbf{(E) }58}</math>.
  
−
==Solution 2==
+
==Solution 2 (Brute force)==
−
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 because we have enough time to do this, we find that it starts from <math>7</math> and ending 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>.
  
−
Fun Fact: <math>2^{18}+1=262145</math> ~ xxsc
+
==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 <math>-</math> the lowest number <math>+ 1</math>. People will forget the <math>+1</math> so the only possibilities are C and E. We can clearly see that C is too small so our answer is <math>\boxed{\textbf{(E) }58}</math>.
 +
 
 +
==Solution 4 (If you do not notice that <math>2^{18}=(2^6)^3=64^3</math>)==
 +
 
 +
There is not so much guessing and checking after we find that it starts from <math>7</math> because <math>7^3=343</math>, which is over <math>2^8+1=257</math>. We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and <math>17^3 = 4,913</math>, which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gives us 64 and <math>64^3 = 262,144</math>, which is perfect for the <math>2^{18}+1=262,145</math>. But we are not done. Since it is inclusive, we must add 1 to this solution, which give us <math>\boxed{\textbf{(E) }58}</math>.
 +
 
 +
== Video Solution by Pi Academy ==
 +
 
 +
https://youtu.be/oSjDIojucGo?si=b_FsVd470FLZYSjx
 +
 
 +
~ jj_empire10
 +
 
 +
==Video Solutions==
 +
 
 +
https://www.youtube.com/watch?v=pbnddMinUF8
 +
 
 +
-Happytwin
 +
 
 +
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 
 +
 
 +
~David
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/rQUwNC0gqdg?t=297
  
 
==See Also==
 
==See Also==

Latest revision as of 19:23, 20 October 2024

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

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

Solution 4 (If you do not notice that $2^{18}=(2^6)^3=64^3$)

There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$, which is over $2^8+1=257$. We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$, which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gives us 64 and $64^3 = 262,144$, which is perfect for the $2^{18}+1=262,145$. But we are not done. Since it is inclusive, we must add 1 to this solution, which give us $\boxed{\textbf{(E) }58}$.

Video Solution by Pi Academy

https://youtu.be/oSjDIojucGo?si=b_FsVd470FLZYSjx

~ jj_empire10

Video Solutions

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

-Happytwin

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

~David

Video Solution by OmegaLearn

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_25&oldid=230206"