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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>. Therefore, 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>. Therefore, the required number of cubes is <math>64-7+1= \boxed{\textbf{(E) }58}</math>
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Video Solution: https://youtu.be/-ukVt81krhE
  
 
==See Also==
 
==See Also==

Revision as of 12:08, 9 November 2019

Problem 25

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

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 clearly will be the largest cube less than $2^{18}+1$. Therefore, the required number of cubes is $64-7+1= \boxed{\textbf{(E) }58}$


Video Solution: https://youtu.be/-ukVt81krhE

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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