Find positive integers $a,b,c,d$ such that

$\displaystyle a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}=\frac{15}{11}.$

Given $\frac{1}{x}<1$ when $x>1,$ how does $a+\frac{1}{x}$ relate to $\frac{15}{11}?$

With that in mind, what can you say about $b+\frac{1}{y}?$

Can you write a fraction $\frac{m}{n}$ in a different way?

… to resemble $\frac{1}{z}$ perhaps ($z$ may be a fraction)?

How about applying that approach to $1+\frac{4}{11}$?

… and continue successively?

Let $x=\frac{1}{b+\cdots}.$ Consider the value of $x.$ Since $b \geq 1$ and the numerator is $1,$ it follows that $x<1$. Now, consider the value of $a$. Since $x<1,$ $1 \leq \frac{15}{11} < 2$ and $a$ is integral, $a$ must be $1.$ So we now have $\frac{15}{11} = 1 + \frac{4}{11}.$ Since we require the numerator of each nested fraction to be $1,$ we can use the fact that $\frac{m}{p} = \frac{1}{\frac{p}{m}}$ to get $\frac{15}{11} = 1 + \frac{1}{\frac{11}{4}}.$ Using this technique successively, we can decompose the expression and get \begin{aligned} \dfrac{15}{11} &= 1+\dfrac{4}{11} \\[2mm] &= 1+\dfrac{1}{\dfrac{11}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{3}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{\dfrac{4}{3}}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3}}}. \end{aligned} And hence $a=1,b=2,c=1,d=3.$