Master of crime missing girl puzzle 1-3 answer book

The answer to the mystery of the missing girl of the crime master is 1-3. I believe many friends are not very clear about this piece. Next, Bian Xiao will tell you the answer to the mystery of the missing girl of the crime master. Interested friends can come and find out.

Master of crime missing girl puzzle 1-3 answer book

1, Zhang Guimei

2、cnaefb 163

3. Gansu

The first level of analysis:

According to the tip of the topic, the name "Zhang Guimei" can be obtained by winding the note on the pencil of Zhonghua brand 2B. Because the result is 1 1 characters, there is only one answer. Interested users can search Zhang Guimei to learn about her life story.

The second level of analysis:

Morse code is hidden in the shoe print pattern of the second level, with short horizontal representing and long horizontal representing one. The letters of Morse code corresponding to the left and right heels are T and F respectively, suggesting that the left shoe print is the "correct step" to follow. According to Morse password comparison table, the final result is CNEFB163. CNAEF is the abbreviation of China Teaching Union. If someone discovers this at this time, it can be inferred that the missing Beryl probably went to teach somewhere. There are 163 words, and this string of numbers probably represents an email account.

The third level of analysis:

The puzzle content of the third level needs to be obtained by the user himself. Cnaefb means CNA ef+ beryl. Send any content to cnaefb@ 163.com to get an automatic reply with puzzles. The contents of the automatic reply mailbox are:

I have been lonely and neglected, and I really want to give you double feelings. In a world without me, can you finally find me?

FGPOA

The bold word "double" implies that this problem should be decrypted by multiplication password. "In a world without me" means that the list of plaintext elements should be the 26 letters left after removing beryl (girl's name). Reorder the remaining letters and compare them one by one. The corresponding numbers of FGPOA are 4, 5, 13, 12 and 1 respectively.

Decryption multiplication password is as follows

Step 1: Make sure that the total number of characters of plaintext information is n, and n=2 1 (the total number of remaining letters in the alphabet after beryl is removed).

Step 2: Determine the value of the K key ... The value of the key is not clearly given in the clue, so it is assumed that the length of the character FGPOA or Beryl is the key, so K=5.

Step 3: We need to calculate the inverse of k- modulo n. In the multiplication cipher, the total number of plaintext n, the key k and the inverse modulo n of k always satisfy the following relations.

Mod (the inverse of k×k modulo n, n)= 1, or it can be expressed by the general division formula (the inverse of k× k modulo n) -n= quotient 1.

By calculating the above formula, we can get the inverse n = 17 of the k-module.

That is, mod (5× 1 7,21) =1or 5× 17-2 1= quotient1) can be decrypted after inversion.

Step 4: Calculate and decrypt. Let a letter rank a in the plaintext alphabet, and the decryption formula is mod(a×k modulo n, the inverse of n). The remainder calculated by this formula represents the ranking of password letters in the alphabet. (Through the previous calculation, we can know that in this cryptographic problem, n=2 1, and the inverse of k-module n = 17).

Take the first letter f in FGPOA as an example. F ranks fourth in the new plaintext alphabet, so it is calculated by the number 4, that is, a=4. Through calculation, it can be obtained that mod (the inverse of a× k module n, n) = mod (4×17,21) = 5. Or use the general division formula 4× 17-2 1=35 to calculate the remainder is 5.

This shows that the ciphertext corresponding to plaintext F should be ranked in the fifth place of the alphabet, that is, the letter G.

After calculating the remaining letters in the same way, we can get the final result of Gansu, which is Gansu. (Note: When the dividend in the remainder calculation is less than the divisor, the result of the remainder is the dividend itself. )