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

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The second half: <math>n^{100}>2^{200}</math>, or <math>n^{100}>4^{100}</math>, which means <math>n>4</math>.
 
The second half: <math>n^{100}>2^{200}</math>, or <math>n^{100}>4^{100}</math>, which means <math>n>4</math>.
  
−
Therefore <math>4<n<130</math> and <math>n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\} which contains </math>125<math> values or choice </math>\boxed{\textbf{(E) } 125}$.
+
Therefore <math>4<n<130</math> and <math>n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}</math> which contains <math>125</math> values or choice <math>\boxed{\textbf{(E) } 125}</math>.
  
 
~JH. L
 
~JH. L

Revision as of 06:17, 16 July 2022

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

See also

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

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