Difference between revisions of "2018 AMC 8 Problems/Problem 25"
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==Solution== | ==Solution== | ||
| − | We compute <math>2^8+1=257</math>. 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>, | + | We compute <math>2^8+1=257</math>. We're all familiar with what <math>6^3</math> is, it's <math>256</math>, barely smaller. 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 10:29, 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 
, barely smaller. 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 |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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