A base $\displaystyle 10$ three digit numeral $n$ is selected at random.What is the probability that both the base $\displaystyle 9$ and base $\displaystyle 11$ representation of $n$ are both three digit numeral?

If the probability can be expressed in the form $\dfrac{a}{b}$ where $\gcd(a,b)=1$ ,then find $a+b$ .

The answer is 377.

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Note: Here all numbers are written in base ten. The largest number that has a 3 digit base 9 representation is 728 $(8\cdot9^2+8\cdot9^1+8\cdot9^0)$ . The smallest number that has a 3 digit base 11 representation is 121 $(1\cdot11^2+0\cdot11^1+0\cdot11^0)$ . Thus 608 numbers satisfy the given constraint. There are 900 3 digit numbers in base 10. $\frac{608}{900}=\frac{152}{225}$ . This is in lowest terms, so the answer is 152+225=377.