# Difference between revisions of "2005 AMC 10A Problems/Problem 13"

## Problem

How many positive integers $n$ satisfy the following condition: $(130n)^{50} > n^{100} > 2^{200}\ ?$ $\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

## Solution

We're given $(130n)^{50} > n^{100} > 2^{200}$, so $\sqrt{(130n)^{50}} > \sqrt{n^{100}} > \sqrt{2^{200}}$ (because all terms are positive) and thus $130n > n^2 > 2^4$ $130n > n^2 > 16$

Solving each part separately: $n^2 > 16 \Longrightarrow n > 4$ $130n > n^2 \Longrightarrow 130 > n$

So $4 < n < 130$.

Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{\textbf{(E) }125}$

## Solution 2

We're given $\left(130n\right)^{50}>n^{100}>2^{200}$.

Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$. Because the exponents are equal, we can ignore them and solve for $n$: $130n>n^{2}$, or $n<130$.

The second half: $n^{100}>2^{200}$, or $n^{100}>4^{100}$, which means $n>4$.

Therefore $4 and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{\textbf{(E) } 125}$.

~JH. L

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