Maruška wrote one of ten consecutive natural numbers on each of ten cards. However, she lost one card. The sum of the numbers on the remaining nine cards was 2012. What number was written on the lost card?

(L. Hozová)
The smallest of the ten written numbers is denoted by $p$. The numbers on the cards were then:

$$
p, p+1, p+2, p+3, \ldots, p+9
$$

The sum of the numbers on all ten cards was $10 p+45$. The equation:

First, let's assume that the card with the number $p$ was lost. Then the following would apply:

$$
\begin{aligned}
(10 p+45)-p & =2012 \\
9 p & =1967 .
\end{aligned}
$$

From the previous line, it is clear that $p$ would not be a natural number in this case. Therefore, we reject the assumption that the card with the number $p$ was lost.

Now let's assume that the card with the number $p+1$ was lost. Similarly, we reject this possibility:

$$
\begin{aligned}
(10 p+45)-(p+1) & =2012 \\
9 p & =1968 .
\end{aligned}
$$

We also reject the assumption that the card with the number $p+2$ was lost:

$$
\begin{aligned}
(10 p+45)-(p+2) & =2012 \\
9 p & =1969 .
\end{aligned}
$$

The card with the number $p+3$ could not have been lost either:

$$
\begin{aligned}
(10 p+45)-(p+3) & =2012, \\
9 p & =1970 .
\end{aligned}
$$

Only when we assume that the card with the number $p+4$ was lost do we arrive at an integer value for $p$:

$$
\begin{aligned}
(10 p+45)-(p+4) & =2012 \\
9 p & =1971 \\
p & =219 .
\end{aligned}
$$

We still need to discuss five more possibilities, namely the loss of cards with numbers $p+5$ to $p+9$. In the previous discussions, we always arrive at an equation with only the number $9 p$ on the left side. The numbers on the right side of these equations increase by 1 each time. We have just obtained a number divisible by nine on the right side, and therefore in the next five discussions, a number divisible by nine cannot appear. The problem thus has a unique solution, which is that the card with the number $p+4=219+4=223$ was lost.

Another solution. The smallest of the ten written numbers is denoted by $p$. The numbers on the cards were then:

$$
p, p+1, p+2, p+3, \ldots, p+9
$$

The number on the lost card is denoted by $p+c$, where $c$ represents a single-digit natural number or zero. The sum of the numbers on all ten cards was $10 p+45$. The sum of the numbers on the nine remaining cards after the loss was $10 p+45-(p+c)$, which simplifies to $9 p+45-c$. We thus arrive at the equation, which we rearrange as follows:

$$
\begin{aligned}
9 p+45-c & =2012, \\
9 p & =1967+c .
\end{aligned}
$$

The sum on the right side of this equation must be divisible by nine. When dividing $1967: 9$, we get a result of 218 with a remainder of 5. Therefore, $1967=9 \cdot 218+5$. From this decomposition, it is clear that the sum $1967+c$ is divisible by nine only if $c=4$. The unknown $p$ is then equal to 219, and the number on the lost card was $p+c=219+4=223$.

Evaluation. 1 point for expressing the sum of the numbers on the ten cards, e.g., $10 p+45$; 2 points for the result; 3 points for describing the procedure.

If the contestant proceeds as we did in the first solution and stops checking further possibilities after finding an integer $p$ without any justification, deduct 1 point.

Note. Contestants can solve the problem by "estimation" using the arithmetic mean of the numbers on the nine remaining cards: $2012: 9=223$, remainder 5. They can then try writing, for example, nine consecutive natural numbers such that 223 is in the middle: $219+220+\ldots+227=$ $=2007$. They then consider whether any of the addends can be "increased" by 5 ... If such work is not justified to show that the problem has a unique solution, award a maximum of 5 points.