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

Revision as of 10:59, 2 August 2006

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

$(130n)^{50} > n^{100} > 2^{200}$ ?

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

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 seperatly:

$n^2 > 16 \Longrightarrow n > 4$

$130n > n^2 \Longrightarrow 130 > n$

So $4 < n < 130$ .

Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $125 \Longrightarrow \mathrm{(E)}$ .

@@ Line 7: / Line 7: @@
 ==Solution==
-<math> (130n)^{50} > n^{100} > 2^{200} </math>
+We're given <math> (130n)^{50} > n^{100} > 2^{200} </math>, so
-<math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math>
+<math> \sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}} </math> (because all terms are positive) and thus
 <math> 130n > n^2 > 2^4 </math>
-<math> 130n < n^2 < 16 </math>
+<math> 130n > n^2 > 16 </math>
 Solving each part seperatly:
-<math> n^2 > 16 </math>
+<math> n^2 > 16 \Longrightarrow n > 4 </math>
-<math> n < 4 </math>
+<math> 130n > n^2 \Longrightarrow 130 > n </math>
-<math> 130n > n^2 </math>
-<math> 130 < n </math>
 So <math> 4 < n < 130 </math>.
-Therefore the answer is the number of positive integers over the interval <math> (4,130) </math> which is <math> 125 \Rightarrow E </math>.
+Therefore the answer is the number of [[positive integer]]s over the interval <math> (4,130) </math> which is <math> 125 \Longrightarrow \mathrm{(E)} </math>.
 ==See Also==
@@ Line 35: / Line 31: @@
 *[[2005 AMC 10A Problems/Problem 14|Next Problem]]
+ [[Category:Introductory Algebra Problems]]