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)== |
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 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 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> | ||
Revision as of 13:13, 21 October 2020
Problem 25
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)
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 ending with 
. Now, by counting how many numbers are between these, we find the answer to be 
~ xxsc
Video Solution
https://www.youtube.com/watch?v=pbnddMinUF8 -Happytwin
https://youtu.be/ZZloby9pPJQ ~DSA_Catachu
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. 
