At the elementary school U Tří lip, where Lukáš also attends, they are holding a knowledge competition with pre-arranged tasks. Each correctly solved task is scored as many points as its order. Each unsolved or partially solved task is not scored at all. Lukáš correctly solved the first 12 tasks. If he had correctly solved the last 12 tasks, he would have scored 708 more points.

How many competition tasks were there?

(M. Petrová)
To solve the problem, it is not necessary to enumerate the sums; it is sufficient to recognize the relationships between them. Let the sum of the first twelve numbers be denoted by $s$, i.e.,

$$
1+2+\cdots+11+12=s.
$$

The sum of the 2nd to 13th numbers is 12 greater than $s$, because each of the addends is 1 greater than in the previous case:

$$
2+3+\cdots+12+13=s+12 \text{. }
$$

The sum of the 3rd to 14th numbers is again 12 greater than the previous sum, i.e., 12 greater than $s$:

$$
3+4+\cdots+13+14=(s+12)+12=s+2 \cdot 12 \text{. }
$$

The sum of twelve consecutive numbers starting from $k$ is $s+(k-1) \cdot 12$. Therefore, if the sum of twelve consecutive numbers is $s+12n$, then the last of these numbers is $12+n$.

Lukáš could have earned 708 points more, and since $708=12 \cdot 59$, the last of the added numbers would have to be $12+59=71$. There were 71 competition tasks.

Another solution. The sum of the first twelve numbers is

$$
1+2+\cdots+11+12=13 \cdot 6=78 \text{. }
$$

The sum of twelve consecutive numbers starting from $k$ is

$$
k+(k+1)+\cdots+(k+10)+(k+11)=(2k+11) \cdot 6
$$

Lukáš could have earned an additional 708 points, totaling $78+708=786$ points. By manipulating the equation $(2k+11) \cdot 6=786$, we get $k=60$. The last of the added numbers on the left side of equation (2) would have to be $60+11=71$. There were 71 competition tasks.

Note. In the manipulations (1) and (2), we use the fact that the sum of the first and last number is the same as the sum of the second and second-to-last, etc., and that there are 6 such pairs. Without this insight, the sum (2) can be expressed using (1) as $12k+78-12=12k+66$.

Evaluation. 2 points for partial insights related to the sum of twelve consecutive numbers; 2 points for formulating the problem using an unknown ($12n=708$ in the first solution, or $(2k+11) \cdot 6=786$ in the second); 2 points for solving and the quality of the commentary.