Nested oh nested

The closest perfect square less than $27$ is $25$ , so we know that $a_0$ will be $5$ . Let $\sqrt{27} = 5 + x$ . Then

$\begin{array}{rl} 5 + x &= \ \ \sqrt{27} \\ \\ 25 + 10x + x^2 &= \ \ 27 \\ \\ x (10 + x) &= \ \ 2 \\ \\ x &= \ \ \dfrac{2}{10 + x} \end{array}$

Now we have an expression for $x$ , which we can use to replace the $x$ in itself, i.e.

$x = \dfrac{2}{10 + \dfrac{2}{10 + x}}$

We could continue this indefinitely and get a continued fraction representation for $\sqrt{27}$ , but we want the numerators of our continued fraction to be $1$ 's rather than $2$ 's. Luckily, we notice that in the above expression, the numerator, the first term in the denominator and the fraction in the denominator can all be divided by $2$ . Then

$\begin{array}{rl} x &= \ \ \dfrac{2}{10 + \dfrac{2}{10 + x}} \\ \\ &= \ \ \dfrac{\frac{2}{2}}{\frac{10}{2} + \dfrac{\frac{2}{2}}{10 + x}} \\ \\ &= \ \ \dfrac{1}{5 + \dfrac{1}{10 + x}} \end{array}$

Now we can substitute the entire expression on the right into $x$ , getting

$\begin{array}{rl} x &= \ \ \dfrac{1}{5 + \dfrac{1}{10 + x}} \\ \\ &= \ \ \dfrac{1}{5 + \dfrac{1}{10 + \dfrac{1}{5 + \dfrac{1}{10 + x}}}} \\ \\ &= \ \ \dfrac{1}{5 + \dfrac{1}{10 + \dfrac{1}{5 + \dfrac{1}{10 + \dfrac{1}{5 + \dfrac{1}{10 + \ldots}}}}}} \end{array}$

and so

$\begin{array}{rl} \sqrt{27} &= \ \ 5 + x \\ \\ &= \ \ 5 + \dfrac{1}{5 + \dfrac{1}{10 + \dfrac{1}{5 + \dfrac{1}{10 + \ldots}}}} \end{array}$

so we have $a_0 = 5, a_1 = 5$ and $a_2 = 10$ , giving the answer $a_0 + a_1 + a_2 = \boxed{20}$