The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # 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 1
Given $$\frac{1}{x}<1$$ when $$x>1,$$ how does $$a+\frac{1}{x}$$ relate to $$\frac{15}{11}?$$
• Hint 2
With that in mind, what can you say about $$b+\frac{1}{y}?$$
• Hint 3
Can 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 5
How 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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