2005 AMC 10A Problem 13

The exponents are huge, but they are all multiples of $50$. Rewrite each part as something to the $50$th power so they can be compared directly.

1. Rewrite with power $50$.
- $(130n)^{50}$ is already a $50$th power.
- $n^{100} = (n^2)^{50}$.
- $2^{200} = (2^4)^{50} = 16^{50}$.
2. Remove the power. All three bases are positive. For positive numbers, a bigger base gives a bigger $50$th power and the other way round. So the condition is the same as
$$130n > n^2 > 16.$$
3. Right part: $n^2 > 16$. Since $4^2 = 16$ and $5^2 = 25$, this means $n \ge 5$.
4. Left part: $130n > n^2$. Divide both sides by the positive number $n$: $130 > n$. So $n \le 129$.
5. Count. The numbers $5, 6, \ldots, 129$. There are $129 - 5 + 1 = 125$ of them.

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

When every exponent in an inequality shares a common factor, rewrite everything as that power and compare the bases instead. For positive numbers this keeps the inequality pointing the same way. After that it is two small inequalities, and the only thing to be careful about is the plus one when counting from $5$ to $129$.

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 $\boxed{\textbf{(E)}\ 125}$.