Continuable fractions

Question

Let $X$ and $Y$ be positive integers such that the first few digits of the fractions $\dfrac{29}{{\color{red}{70}}},\ \dfrac{{\color{red}{70}}}{{\color{blue}{169}}},\ \dfrac{{\color{blue}{169}}}{{\color{green}{408}}},\ \dfrac{{\color{green}{408}}}{{\color{brown}{X}}},\ \dfrac {{\color{brown}{X}}}{{\color{grey}{Y}}} $ are all equal to $0.4142$. What is ${{\color{brown}{X}}}+{{\color{grey}{Y}}}?$

Note: The fractions are not equal.

Answer 1

A useful trick for approximating the fractional part of $\sqrt{2}$ is the following sequence $\frac{p}{q}\implies \frac{q}{p+2q}$ Clearly, the sequence in question follows this sequence as shown below $\frac{29}{70} \implies \frac{70}{29 + (2)(70)} = \frac{70}{169} \implies \frac{169}{70 + (2)(169)} = \frac{169}{408} $ Therefore, the sequence containing $X$ and $Y$ is as follows $\frac{169}{408}, \frac{408}{985}, \frac{985}{2378}$

Thus, $X + Y = 985 + 2378 = \boxed{3363}$

Answer 2

There is a relatively trivial solution to this problem: we know that $\frac{408}X\approx .4142$; this means that $X\approx \frac{408}{.4142}=985.0313\ldots$ etc. Now, any $X$ larger than this will lead to a ratio $\frac{408}X$ that's less than $.4142$ and thus can't start with those digits. What's more, by explicit calculation, $\frac{408}{984}=.4146\ldots$ is clearly too large; thus $985$ is the only possible value for $X$. Once $X$ has been found this way, $Y$ can be computed using exactly the same method.