There is an encrypted message. Each character was replaced by the same two-digit number. Spaces, letters, numbers. Different characters were replaced of course by two-digit numbers.

Constructed a table, counted each occurrence of a number in a line, but the frequency of using the letters of the Russian alphabet did not help. Tried to decrypt, by the method of decrypting a simple cipher also did not work.

I put the code:

17 51 42 33 42 34 24 13 23 45 27 14 33 50 18 27 50 51 42 18 49 13 23 45 31 33 14 35 35 13 45 35 33 50 22 15 29 13 42 27 45 18 13 39 17 12 42 27 17 39 45 29 17 45 29 27 17 33 24 50 49 45 26 41 45 13 29 31 14 18 27 13 45 29 45 26 37 45 51 13 18 17 29 45 41 32 45 39 50 24 14 27 27 51 49 45 17 22 33 13 27 24 15 42 45 22 50 12 42 27 15 45 40 13 22 33 17 24 50 33 14 27 42 45 24 13 45 35 17 42 40 34 45 37 26 45 13 29 31 14 18 27 13 27 51 49 45 17 27 35 33 13 29 12 42 24 17 45 32 41 45 13 29 31 14 18 27 13

I will be glad to any advice or attempt to help, maybe I started wrong, or I made a mistake somewhere, but this is the first and last option that came to mind.

1 answer 1

The most used symbol is 45 , and this is a space. If we put it, we get the most realistic placement of words:

 ......... ............. ...... ......... ......... .. ....... .. ....... . .. ..... .. ........ ........ ...... ........... .. ..... .. .......... .......... .. .......

It would be naive to believe that on such a small text we get the frequency of letters, which is close to the general case. We do not have very often used words as articles a/the in the English language, so there’s nothing to start from.

Let's go another way, take the longest word, it is in second place:

 27 14 33 50 18 27 50 51 42 18 49 13 23

Symbols 27 , 50 , 18 repeated 2 times. I took a dictionary of 100,000 words, and tried to find words with exactly the same repetition of letters. There were 2 words, the meaning of one was incomprehensible, the second - туристический . Given the possible cases, take the floor without ending: туристическ .

Add these letters to the dictionary, and see what happens:

 .чере.... туристическ.. .ру... .ри....ет с....ет.. .. .т.р.ик .. ...уст. . .. ч.с.. .. .и.уттчк ..р.т..е .и.ет. ...р..ируте .. ..е.. .. ...уст.тчк .т.р...е.. .. ...уст.

Using the dictionary again, you can see that in place of the first word, only something related to the queue fits. Add letters to the очеред in the dictionary. We get:

 очеред... туристическ.. .ру... .ри....ет с..о.ето. .о .тор.ик .. ...уст. . .. ч.со. .. .и.уттчк о.р.т..е .и.ет. ...ро.ируте .. .ое.д .. ...уст.тчк от.р...е.о .. ...уст.

Interesting picture here:

 .о .тор.ик [29]o [29]тор[24]ик

Two words start with one letter, obviously, here во вторник :

 очередн.. туристическ.. .ру... .ри..в.ет с..о.ето. во вторник .. .в.уст. в .. ч.сов .. .инуттчк о.р.тн.е .и.ет. ...ронируте н. .ое.д .. .в.уст.тчк от.р.в.ено .. .в.уст.

Now:

 очередн[13][23] = очередная туристическ[13][23] = туристическая ч[13]сов = часов

After adding letters to the dictionary:

 очередная туристическая .ру..а .ри..вает са.о.ето. во вторник .. ав.уста в .. часов .. .инуттчк о.ратн.е .и.ет. .а.ронируте на .ое.д .. ав.устатчк от.рав.ено .. ав.уста

Farther:

 ав[31]уста = августа гру[35][35]а = группа отправ[12]ено = отправлено пое[40]д = поезд са[39]олето[39] = самолетом очередная туристическая группа при..вает самолетом во вторник .. августа в .. часов .. минуттчк о.ратн.е .илет. за.ронируте на поезд .. августатчк отправлено .. августа очередная туристическая группа прибывает самолетом во вторник [26][41] августа в [26][37] часов [41][32] минуттчк обратные билеты забронируте на поезд [37][26] августатчк отправлено [32][41] августа

Finally, you probably need to apply logic regarding numbers in dates / hours / minutes.