14. (GDR 2) Which fraction $p / q$, where $p, q$ are positive integers less than 100, is closest to $\sqrt{2}$? Find all digits after the decimal point in the decimal representation of this fraction that coincide with digits in the decimal representation of $\sqrt{2}$ (without using any tables).
14. We have that 
$$ \left|\frac{p}{q}-\sqrt{2}\right|=\frac{|p-q \sqrt{2}|}{q}=\frac{\left|p^{2}-2 q^{2}\right|}{q(p+q \sqrt{2})} \geq \frac{1}{q(p+q \sqrt{2})} $$ 
because $\left|p^{2}-2 q^{2}\right| \geq 1$. The greatest solution to the equation $\left|p^{2}-2 q^{2}\right|=1$ with $p, q \leq 100$ is $(p, q)=(99,70)$. It is easy to verify using (1) that $\frac{99}{70}$ best approximates $\sqrt{2}$ among the fractions $p / q$ with $p, q \leq 100$. Second solution. By using some basic facts about Farey sequences one can find that $\frac{41}{29}100$ because $99 \cdot 29-41 \cdot 70=1$. Of the two fractions $41 / 29$ and $99 / 70$, the latter is closer to $\sqrt{2}$.