# 2003 AMC 10A Problems/Problem 20

## Problem 20

A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals? $\mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7$

## Solution

To be a three digit number in base-10: $10^{2} \leq n \leq 10^{3}-1$ $100 \leq n \leq 999$

Thus there are $900$ three-digit numbers in base-10

To be a three-digit number in base-9: $9^{2} \leq n \leq 9^{3}-1$ $81 \leq n \leq 728$

To be a three-digit number in base-11: $11^{2} \leq n \leq 11^{3}-1$ $121 \leq n \leq 1330$

So, $121 \leq n \leq 728$

Thus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11.

Therefore the desired probability is $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}$.

~ pi_is_3.14

~IceMatrix

## Video Solution by WhyMath

~savannahsolver

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