2005 AMC 10A Problem 13

P11-15~3 minPrint

Exponents, Logarithms & Radicals Absolute Value & Inequalities Set up the equation/formula and compute; no special trick needed

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

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

Show solution without answering

Key insight

Every exponent is a multiple of 50. Take the 50th root of each part to get 130n > n squared > 16, which means n is from 5 to 129.

Solution

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

Why this works

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 $129$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)