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AMC8 2018 Test Paper
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Sample Question 25:

How many perfect cubes lie between \(2^8+1\) and \(2^{18}+1\), inclusive?

\(\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58\)




Answer Keys

Question 25: E


Solutions

Question 25
Step 1: First, we need to calculate the value of \(2^8+1\), which equals 257.

Step 2: We know that \(6^3\) equals 216, which is less than 257. The smallest cube greater than this is \(7^3\), which equals 343. Therefore, 7 is the smallest number whose cube is in the given range.

Step 3: Next, consider \(2^{18}\). \(2^{18}\) is the same as \((2^6)^3\), or \(64^3\). This will be the largest cube less than \(2^{18}+1\), so 64 is the largest number whose cube is in the given range.

Step 4: Given that the numbers whose cubes fall within the range are from 7 to 64 inclusive, the number of such numbers (or cubes) can be found by subtracting the smallest number (7) from the largest (64) and adding 1. This gives us \(64-7+1=58\).

So, the answer is \(\boxed{\textbf{(E) }58}\).