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

Revision as of 16:38, 8 December 2020

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 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 $125 \Longrightarrow \mathrm{(E)}$ .

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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 ==See also==
-{{AMC10 box|year=2005|ab=A|num-b=12|num-a=13}}
+{{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}}
 [[Category:Introductory Number Theory Problems]]
 {{MAA Notice}}