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2005 AMC 10A Problem 13

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2005 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 2 of 5 · about 3 min · position P11-15

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

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

A)
$0$
B)
$7$
C)
$12$
D)
$65$
E)
$125$

Answer

Key insight

Take the 50th root of every part: the chain becomes 130n > n^2 > 16, that is, 4 < n < 130.

Solution

All three quantities are positive, so we may take $50$ th roots without changing the direction of either inequality. Since $n^{100} = (n^2)^{50}$ and $2^{200} = (2^4)^{50} = 16^{50}$ , the condition becomes
$130n > n^2 > 16.$

Right part: $n^2 > 16$ with $n$ positive means $n \geq 5$ .

Left part: $130n > n^2$ ; dividing by the positive $n$ gives $n < 130$ , so $n \leq 129$ .

The valid integers are $5, 6, \dots, 129$ , and there are $129 - 5 + 1 = 125$ of them.

The answer is $\boxed{\textbf{(E)}\ 125}$ .

Why this works

Huge exponents in a chain of inequalities are usually a common power in disguise; find the shared exponent ( $50$ here) and strip it off. After that the problem is two elementary inequalities, and the only care needed is the fencepost count at the end.

The trap

Counting the integers strictly between 4 and 130 as 126 or 124 instead of 125 (129 - 5 + 1).

Common mistakes

Counting the integers strictly between 4 and 130 as 126 or 124 instead of 125 (129 - 5 + 1).
Turning $2^{200}$ into $2^4 = 16$ incorrectly (say $2^2$ or $2^{50}$ ) after taking the $50$ th root.

Techniques

Set up the equation/formula and compute; no special trick needed