# Problem 4.

Functions $f_{0}(x), f_{1}(x), \ldots, f_{6}(x)$ with domain $\{0,1,2,3\}$ and range $\{0,1,2,3\}$ are defined by tables (Table 2). Table:

| $x$ | $f_{0}(x)$ | $f_{1}(x)$ | $f_{2}(x)$ | $f_{3}(x)$ | $f_{4}(x)$ | $f_{5}(x)$ | $f_{6}(x)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 0 | 3 | 2 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 3 | 3 | 2 | 2 | 1 |
| 2 | 2 | 1 | 0 | 2 | 1 | 3 | 3 |
| 3 | 3 | 2 | 1 | 0 | 3 | 1 | 2 |

(a) For the function $f(x)$ defined by the equations $f(0)=3, f(1)=1, f(2)=2, f(3)=0$, find different numbers $a, b, c \in\{0,1, \ldots, 6\}$ such that the relation

$$
=f_{c}\left(f_{b}\left(f_{a}(x)\right)\right)
$$

holds for all $x=0,1,2,3$;

(b) Prove that for any function $f(x)$ with domain $\{0,1,2,3\}$ and range $\{0,1,2,3\}$, mapping different elements to different values, there exist numbers $a, b, c \in\{0,1, \ldots, 6\}$ (not necessarily different) for which the equation (1) holds.
Answer: (a) $a=3, b=5, c=6$.

## Problem 5

Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The initial layout of the digits on the keyboard changes depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is replaced by the value $a_{i}$, which is the last digit of the number $i \cdot k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the unlock code if it is known that when entering the code:

| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
|  | 0 |  |$\rightarrow$| $a_{1}$ | $a_{2}$ | $a_{3}$ |
| :--- | :--- | :--- |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | $a_{9}$ |
|  |  | $a_{0}$ |

the user entered the digits as follows:

- when $a_{3}=3$: left, right, right, right;
- when $a_{3}=9$: right, right, left, left;
- when $a_{3}=1$: left, left, right, right;
- when $a_{3}=7$: right, right, left, right.

Answer: 7832.

## Problem 6

To encrypt a 13-letter message in Russian, the following steps were performed: 1) the letters of the message were sequentially converted using a table (Fig.1) into a sequence of numbers $\left.x_{1}, x_{2}, \ldots, x_{13} ; 2\right)$ a (secret) natural number $k_{1}$ was chosen and the sum $x_{14}=x_{1}+x_{2}+\cdots+x_{13}+k_{1}$ was appended to the sequence on the right; 3) in the extended sequence $x_{1}, x_{2}, \ldots, x_{13}, x_{14}$, the numbers $x_{i}$ were replaced by numbers $y_{i}$ according to the formulas: $y_{i}=2 x_{i}+x_{i+1}+(-1)^{\frac{i+1}{2}} k_{1}$ if $i$ is odd and $y_{i}=x_{i-1}+x_{i}+(-1)^{\frac{i}{2}} k_{2}$ if $i$ is even, where $k_{2}$ is another (secret) natural number; 4) each $y_{i}$ was replaced by its remainder when divided by 32. The result was: 20, 31, 12, 11, $\mathbf{6}, \mathbf{9}, \mathbf{5}, \mathbf{9}, \mathbf{1 4}, \mathbf{2 7}, \mathbf{9}, \mathbf{1 0}, \mathbf{1 1}, \mathbf{1 6}$. Find the original message.

## Answer: PROBE LANDED.

## Criteria for Determining Winners and Prize Winners of the Final Stage of the 2013-14 Academic Year

Each problem of the on-site stage of the olympiad was scored on a 4-point scale (from 0 to 3 points). A fully solved problem was scored 3 points. -1st place - no fewer than 5 fully solved problems; -2nd place - no fewer than 4 fully solved problems or no fewer than 14 points; -3rd place - no fewer than 3 fully solved problems or no fewer than 10 points.