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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>. 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==
 
  ==Solution 2==
−
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>
 
  Fun Fact: <math>2^{18}+1=262145</math> ~ xxsc
 
  Fun Fact: <math>2^{18}+1=262145</math> ~ xxsc
  

Revision as of 16:03, 12 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 $2^{18}+1$. So, the required number of cubes is $64-7+1= \boxed{\textbf{(E) }58}$ 

==Solution 2==

First, $2^8+1=257$. Then, $2^{18}+1=262145$. Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ending with $64$. Now, by counting how many numbers are between these, we find the answer to be $\boxed{\textbf{(E) }58}$ 

Fun Fact: $2^{18}+1=262145$ ~ xxsc

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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