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

Revision as of 17:42, 1 July 2025

Problem

How many positive integers $n$ satisfy the following condition:

$\left(130n\right)^{50} > n^{100} > 2^{200} \ \text{?}$

$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution 1

Since $n > 0$ , all $3$ terms of the inequality are positive, so we may take the $50$ th root, yielding

$\begin{align*}&\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} \\ \iff &130n > n^2 > 2^4 \\ \iff &130n > n^2 > 16.\end{align*}$

Solving each part separately, while noting that $n > 0$ , therefore gives $n^2 > 16 \iff n > 4$ and $130n > n^2 \iff n < 130$ .

Hence the solution is $4 < n < 130$ , and therefore the answer is the number of positive integers in the open interval $(4,130)$ , which is $129-5+1 = \boxed{\textbf{(E) } 125}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 How many positive integers <math>n</math> satisfy the following condition:
-<math> (130n)^{50} > n^{100} > 2^{200}\ ?</math>
+<cmath>\left(130n\right)^{50} > n^{100} > 2^{200} \ \text{?}</cmath>
-<math> \textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125 </math>
+<math>
+\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125
+</math>
-==Solution==
+==Solution 1==
-We're given <math> (130n)^{50} > n^{100} > 2^{200} </math>, so
+Since <math>n > 0</math>, all <math>3</math> terms of the inequality are positive, so we may take the <math>50</math>th root, yielding
-<math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math> (because all terms are positive) and thus
+<cmath>\begin{align*}&\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} \\ \iff &130n > n^2 > 2^4 \\ \iff &130n > n^2 > 16.\end{align*}</cmath>
-<math> 130n > n^2 > 2^4 </math>
+Solving each part separately, while noting that <math>n > 0</math>, therefore gives <math>n^2 > 16 \iff n > 4</math> and <math>130n > n^2 \iff n < 130</math>.
-<math> 130n > n^2 > 16 </math>
+Hence the solution is <math>4 < n < 130</math>, and therefore the answer is the number of positive integers in the open interval <math>(4,130)</math>, which is <math>129-5+1 = \boxed{\textbf{(E) } 125}</math>.
-Solving each part separately:
-<math> n^2 > 16 \Longrightarrow n > 4 </math>
-<math> 130n > n^2 \Longrightarrow 130 > n </math>
-So <math> 4 < n < 130 </math>.
-Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math>129-5+1 = \boxed{\textbf{(E) }125}</math>
-==Solution 2==
-We're given <math>\left(130n\right)^{50}>n^{100}>2^{200}</math>.
-Alternatively to solution 1, first deal with the first half: <math>\left(130n\right)^{50}>\left(n^{2}\right)^{50}</math>. Because the exponents are equal, we can ignore them and solve for <math>n</math>: <math>130n>n^{2}</math>, or <math>n<130</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\}</math> which contains the same number of elements as <math>\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}</math> which clearly contains <math>125</math> values or choice <math>\boxed{\textbf{(E) } 125}</math>.
-~JH. L
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Difference between revisions of "2005 AMC 10A Problems/Problem 13"

Revision as of 17:42, 1 July 2025

Problem

Solution 1

See also