$\triangle$ Let's find the distance that the father and son walked from one coincidence of steps to the next. It is equal to the least common multiple of the numbers 80 and 60, i.e., $240 \mathrm{~cm}=2.4$ m. Therefore, the entire distance they walked is

$$
600 \cdot 2.4 \mathrm{M}=1440 \mathrm{~m}=1 \mathrm{Km} 440 \mathrm{M}
$$

Answer: 1 km 440 m.
485. A father and son were walking one behind the other on a snow-covered road. The father's step length is $-80 \mathrm{~cm}$, the son's $-60 \mathrm{~cm}$. Their steps coincided 601 times, including at the beginning and the end of the journey. What distance did they cover?