Difference between revisions of "2018 AMC 8 Problems/Problem 25"

Revision as of 14:30, 6 May 2020

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 1

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(Brute force)

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 because we have enough time to do this, 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 $\boxed{\textbf{(E) }58}$

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Video Solution

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.

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