3. All the bus tickets for the Rijeka-Zagreb route departing at 8 AM have been sold. The seats on the bus are numbered in order from 1, 2, 3, and so on, up to the last seat on the bus. By mistake, two tickets were sold for one seat. The sum of the seat numbers on all the sold tickets is 1219. For which seat were two tickets sold?
## Solution.

First, we need to determine the total number of seats. The sum of the numbers of all seats must be less than 1219. To estimate the number of seats, let's sum the first 50 numbers:

$$
1+2+\cdots+49+50=(1+50)+(2+49)+\cdots+(25+26)=25 \cdot 51=1275
$$

This sum is greater than 1219, so we conclude that there are fewer than 50 seats.

The sum of the first 48 numbers is

$$
1+2+\cdots+47+48=(1+48)+(2+47)+\cdots+(24+25)=24 \cdot 49=1176
$$

If the number of seats were less than 48, the difference between 1219 and the sum of the numbers of all seats would be greater than the largest seat number. If the number of seats were greater than 48, the sum of the seat numbers would be greater than 1219.

Therefore, the bus has exactly 48 seats. The sum of the seat numbers on all sold tickets, if no two tickets were mistakenly sold for the same seat, would be 1176.

Since $1219-1176=43$, two tickets were mistakenly sold for seat 43.