2003 AMC 10A Problem 20

First work out which numbers have three digits in each base, then count the numbers that are in both ranges.

1. Base 10, as a warm-up. The smallest three-digit number is $100 = 10^2$, and the largest is $999$, one less than $1000 = 10^3$. So there are $999 - 100 + 1 = 900$ choices for $n$.
2. Base 9. The same pattern holds. The smallest three-digit base-9 numeral is $100_9 = 9^2 = 81$. The first four-digit one is $1000_9 = 9^3 = 729$. So three digits in base 9 means
$$81 \le n \le 728.$$
3. Base 11. The smallest three-digit numeral is $100_{11} = 11^2 = 121$, and the first four-digit one is $11^3 = 1331$. So three digits in base 11 means
$$121 \le n \le 1330.$$
4. All three at once. $n$ must be at least $100$, $81$ and $121$, so at least $121$. It must be at most $999$, $728$ and $1330$, so at most $728$. That is
$$121 \le n \le 728,$$
which contains $728 - 121 + 1 = 608$ numbers.
5. Probability.
$$\frac{608}{900} \approx 0.68,$$
which is closest to $0.7$. (A quick check: $0.7 \times 900 = 630$ and $0.6 \times 900 = 540$, and $608$ is much nearer $630$.)

The answer is $\boxed{\textbf{(E)}\ 0.7}$.

"Has three digits in base $b$" just means the number is at least $b^2$ and less than $b^3$, the same way three-digit base-10 numbers run from $100$ to $999$. Turning each condition into a range and then taking the overlap reduces the problem to counting whole numbers between two ends. The word "closest" tells you a rough estimate is enough at the end.