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

Revision as of 01:14, 2 December 2018

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 $64-7+1= \boxed{\textbf{(E) }58}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem 25==
-How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive ?
+How many perfect cubes lie between <math>2^8+1</math> and <math>2^{18}+1</math>, inclusive?
 <math>\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58</math>

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

Revision as of 01:14, 2 December 2018

Problem 25

Solution

See Also