Difference between revisions of "2018 AMC 8 Problems/Problem 25"
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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 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> | ||
| − | ==Solution 2 (Brute force) | + | ==Solution 2 (Brute force) (works only if you have enough time for calculations)== |
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> | ||
Revision as of 17:29, 1 January 2023
Contents
Problem
How many perfect cubes lie between 
and 
, inclusive?
Solution 1
We compute 
. We're all familiar with what 
is, namely 
, which is too small. The smallest cube greater than it is 
. is too large to calculate, but we notice that 
, which therefore clearly will be the largest cube less than . So, the required number of cubes is 
Solution 2 (Brute force) (works only if you have enough time for calculations)
First, . Then, 
. Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from 
and ends with 
. Now, by counting how many numbers are between these, we find the answer to be 
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 
. People will forget the 
so the only possibilities are C and E. We can clearly see that C is too small so our answer is .
~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
~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) | ||
| 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. 
