Difference between revisions of "2018 AMC 8 Problems/Problem 25"
(→Solution) |
m (→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | We compute <math>2^8+1=257</math>. We're all familiar with what <math>6^3</math> is, it's <math> | + | We compute <math>2^8+1=257</math>. We're all familiar with what <math>6^3</math> is, it's <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 is the largest cube less than <math>2^{18}+1</math>. Therefore, the amount of cubes is <math>64-7+1= \boxed{\textbf{(E) }58}</math> |
==See Also== | ==See Also== | ||
Revision as of 13:03, 23 November 2018
Problem 25
How many perfect cubes lie between 
and 
, inclusive ?
Solution
We compute 
. We're all familiar with what 
is, it's 
, which is too small. The smallest cube greater than it is 
. is too large to calculate, but we notice that 
which is the largest cube less than . Therefore, the amount of cubes is 
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. 
