2. Determine the natural numbers $a, b, c$ and $d$ such that:

$$
a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}=\frac{599}{121}
$$
2. Determine the natural numbers $a, b, c$ and $d$ such that:

$$
a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}=\frac{599}{121}
$$

## Solution.

Since

$a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}=\frac{599}{121}=4+\frac{115}{121}$

we conclude that $a=4$.

2 POINTS

Further, we have:

$\frac{1}{b+\frac{1}{c+\frac{1}{d}}}=\frac{115}{121}$

$\frac{b+\frac{1}{c+\frac{1}{d}}}{1}=\frac{121}{115}$

1 POINT

$b+\frac{1}{c+\frac{1}{d}}=1+\frac{6}{115}$

We conclude that $b=1$.

2 POINTS

By an analogous procedure, we get

$\frac{c+\frac{1}{d}}{1}=\frac{115}{6}$

1 POINT

i.e., $c+\frac{1}{d}=19+\frac{1}{6}$

so $c=19$.

Finally, from $\frac{1}{d}=\frac{1}{6}$ we conclude that $d=6$.