13.233. A tourist was returning from vacation on a bicycle. On the first leg of the journey, which was 246 km, he traveled on average 15 km less per day than he did on the last leg of the journey, which was 276 km. He arrived home right on time at the end of his last vacation day. It is also known that it took him one day more than half the number of days remaining until the end of his vacation to complete the first leg of the journey. How many days before the end of his vacation did the tourist set out for home?
## Solution.

Let the tourist travel 267 km in $x$ days, and 246 km in $\frac{x}{2}+1$. On the first segment of 246 km, his speed was $v$ km/day, and on the second segment of 276 km, it was $(v+15)$ km/day.

Then $\left\{\begin{array}{l}\left(\frac{x}{2}+1\right) v=246, \\ x(v+15)=276,\end{array}\right.$

from which $x=2$.

Therefore, the tourist set off for home $x+\frac{x}{2}+1=4$ days before the end of the vacation.

Answer: 4 days.