13.374 Three candles have the same length but different thicknesses. The first candle was lit 1 hour earlier than the other two, which were lit simultaneously. At some point, the first and third candles had the same length, and 2 hours after this, the first and second candles had the same length. How many hours does it take for the first candle to burn out if the second candle burns out in 12 hours and the third in 8 hours?
Solution. Let $x$ be the burning time of the first candle, and $l$ be the length of the candles. Then $\frac{l}{x}, \frac{l}{12}, \frac{l}{8}$ are the burning rates of the candles (i.e., the rates at which the lengths of the candles decrease). According to the problem, after $\frac{\frac{l}{x} \cdot 1}{\frac{l}{8}-\frac{l}{x}}$ hours from when the first candle is lit, the third candle will "catch up" to the first in length, and after $\frac{\frac{l}{x} \cdot 1}{\frac{l}{12}-\frac{l}{x}}$ hours, the second candle will "catch up" to the first in length. Therefore, $\frac{\frac{1}{x}}{\frac{1}{12}-\frac{1}{x}}-\frac{\frac{1}{x}}{\frac{1}{8}-\frac{1}{x}}=2 \Rightarrow\left[\begin{array}{l}x=16, \\ x=6 .\end{array}\right.$

The second solution does not make sense in context, as $x>12$

Answer: 16 hours.