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

$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (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 $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains $125$ values or choice $\boxed{\textbf{(E) } 125}$ .

~JH. L

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2005 AMC 10A Problems/Problem 13

Contents

Problem

Solution

Solution 2

See also