Someone always bets on 4 numbers in a $2 \times 2$ square and one number in a square that only touches the corner of this square. (For example, the numbers 40, 41, 55, 56, 69.) How long does it take for them to play every possible combination once, if they fill out 4 tickets per week?
A pre-marked 5 squares can always fit into a $3 \times 3$ square, which we call a frame. For a given set of numbers, there are 4 possible combinations according to which corner square of the frame the standalone square selects. In the given example, the other 3 combinations are: $39,55,56,70,71$; $41,54,55,69,70$; $71,39,40,54,55$. Therefore, it does not change the essence of the question if our player completes their plan as follows: every week, they choose 4 appropriate quintets from a different frame. Thus, they need to play for as many weeks as there are frames available on the lottery ticket.

To determine the frame, it is sufficient to specify the number in the top-left field, which in the example is 39; we call this the leading number of the frame. The leading number can only be chosen from the first 4 rows and the first 13 columns of the ticket. Thus, there are $4 \cdot 13=52$ leading numbers, meaning our player needs to play for exactly 1 year according to the plan.

![](https://cdn.mathpix.com/cropped/2024_05_02_a13952fbc87a4b42a233g-1.jpg?height=200&width=957&top_left_y=613&top_left_x=600)

Remarks. 1. It is somewhat more complicated to determine the number of possibilities by first counting the $2 \times 2$ "small frames" and then separating them based on the number of "free" vertices suitable for connecting the fifth square. Most solutions solved the problem this way or even more complicatedly.

According to a similar calculation, there are $5 \cdot 14=70$ small frames. Among them, $2 \cdot 2=4$ are in the corners of the ticket, $2(14-2) + 2(5-2)=30$ touch the boundary on one side, and the rest $(14-2)(5-2)=36$ are inside. The number of free vertices is 1, 2, and 4, respectively, so the number of fillings is $4 \cdot 1 + 30 \cdot 2 + 36 \cdot 4 = 208$. The player fills out this many tickets in 52 weeks. In the latter method, the identity

$$
(a-2)(b-2) \cdot 2 \cdot 2 = (a-3)(b-3) \cdot 2 \cdot 2 + (a-3) \cdot 1 \cdot 2 \cdot 2 + 1 \cdot (b-3) \cdot 2 \cdot 2 + 1 \cdot 1 \cdot 2 \cdot 2
$$

is used, where $a$ is the number of rows and $b$ is the number of columns of the ticket. The more complicated side was calculated in the above.

2. Many solutions start like this: "The number to be found must be calculated by ...". According to the above, there is no "must" in this problem. However, it is worth keeping in mind that often there are multiple ways to choose from, and it is advisable to look for the simpler one. Let's live with the rich vocabulary of our language!