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$a=1,b=2,c=1,d=3.$