[ Cryptography $\quad]$

Each letter of the original message was replaced by its two-digit ordinal number in the Russian alphabet according to the table:

| A | Б | B | $\Gamma$ | D |  | $\mathrm{E}$ | Ë | Ж | 3 | V |  | Й | K | Л | M | $\mathrm{H}$ | $\mathrm{O}$ | $\Pi$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 01 | 02 | 03 | 04 | 05 |  | 06 | 07 | 08 | 09 | 10 |  | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| $\overrightarrow{\mathrm{P}}$ | $\mathrm{C}$ | $\overline{7}$ |  | $\bar{y}$ | $\Phi$ | $\mathrm{X}$ | $\overrightarrow{\mathrm{L}}$ | $\bar{\tau}$ |  | $\stackrel{\text { II }}{ }$ | ШЩ | Ђ | $\overline{\mathrm{b}}$ | $\overline{\mathrm{l}}$ |  | $\vartheta$ | Ю | Я |
| 18 | 19 | 20 |  | 21 | 22 | $\overline{23}$ | 24 | 25 |  | 26 | 27 | 28 | 29 |  | 30 | 31 | 32 | 33 |

The resulting numerical sequence was divided (from right to left) into three-digit numerical groups without overlaps and gaps. Then each of the resulting three-digit numbers was multiplied by 77, and only the last three digits of the product were kept. As a result, the following sequence of digits was obtained:

317564404970017677550547850355. Restore the original message.
Use arithmetic of remainders modulo 1000.

## Solution

Let's agree to write $a \equiv b$ instead of $a \equiv b(\bmod 1000)$. To find the last letter of the original message, we need to solve the congruence $77 n \equiv 355$. Notice that

$n \equiv 1001 n = 13 \cdot 77 n \equiv 13 \cdot 355 = 4165$, so $n \equiv 615$. Similarly, we solve the equations $77 n \equiv ***$, where the right-hand side consists of other three-digit numerical groups of the cipher message (850, 547, 550, etc.). The desired numerical sequence is 121332252610221801150111050615.

## Answer

THE CIPHER KEY IS FOUND.