# Practice Paper 2 Question 3

Find positive integers \(a,b,c,d\) such that

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

## Hints

- Hint 1Given \(\frac{1}{x}<1\) when \(x>1,\) how does \(a+\frac{1}{x}\) relate to \(\frac{15}{11}?\)
- Hint 2With that in mind, what can you say about \(b+\frac{1}{y}?\)
- Hint 3Can you write a fraction \(\frac{m}{n}\) in a different way?
- Hint 4... to resemble \(\frac{1}{z}\) perhaps (\(z\) may be a fraction)?
- Hint 5How about applying that approach to \(1+\frac{4}{11}\)?
- Hint 6... and continue successively?

## Solution

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 \[ \begin{align} \dfrac{15}{11} &= 1+\dfrac{4}{11} \\[2mm] &= 1+\dfrac{1}{\dfrac{11}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{3}{4}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{\dfrac{4}{3}}} \\[2mm] &= 1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3}}}. \end{align} \] And hence \(a=1,b=2,c=1,d=3.\)

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