Solution. Let $a$ and $b$ be two natural numbers such that $a+b=168, a=24 n_{1}, b=24 n_{2}$, where $n_{1}$ and $n_{2}$ are natural numbers. Then we have $24 n_{1}+24 n_{2}=168$, from which $n_{1}+n_{2}=7$. Considering that $n_{1}$ and $n_{2}$ are natural numbers summing to 7, it is easy to find the numbers $a$ and $b$:

$$
\begin{array}{ll}
a=24\left(n_{1}=1\right), & b=144\left(n_{2}=6\right) \\
a=48\left(n_{1}=2\right), & b=120\left(n_{2}=5\right) \\
a=72\left(n_{1}=3\right), & b=96\left(n_{2}=4\right)
\end{array}
$$

Answer: 24 and 144, or 48 and 120, or 72 and 96.
39(1281). Find two natural numbers whose sum is 168, and their greatest common divisor is 24.