I am trying to figure out how to complete RSA manually. I am trying to encode a simple block message (Mi). I used CrypTool to determine the encryption. When I "manually" computed the plaintext, I obtained a different number than what CrypTool provided. Can someone guide me? Am I doing the manual encryption for RSA correct?

RSA Manually

  • $\begingroup$ What is the integer $A$? $\endgroup$
    – Ievgeni
    Jun 16, 2021 at 15:03
  • $\begingroup$ It could be also $Z$... $\endgroup$
    – Ievgeni
    Jun 16, 2021 at 15:09
  • 1
    $\begingroup$ there are many textbook RSA example on internet, please refer to them. your Z is called ${\phi(n)}$ . and public exponent e is chosen as ${1<e<\phi(n)}$, and private exponent d is selected such that ${ed \equiv 1(mod(\phi(n))}$ $\endgroup$
    – SSA
    Jun 16, 2021 at 15:24
  • $\begingroup$ Just used the Cryptool Online RSA with your setup and got the $\color{red}{Red}$ numbers. What is your problem, then? $\endgroup$
    – kelalaka
    Jun 16, 2021 at 18:20
  • $\begingroup$ @kelalaka Thank you for the online resource! When I used CrypTool RSA Encryption and input the text MI, I receive the following: Numbers input in base 10 format: 360 / Encryption into ciphertext: 11,807. Why is there only one number? $\endgroup$
    – Jame
    Jun 16, 2021 at 23:50

1 Answer 1


As kelalaka mentioned in the comments, the reason for the difference is that the tool turns the message "MI" into a single number 360, while you have encrypted each character "M", "I" individually in your question.

The way that the tool turns the message "MI" into the number 360 is because it uses the alphabet of 26 capital letters plus the space character (number 0), giving 27 in total. So it turns each character into a digit between 0 and 26, and then convert from base 27 to base 10. In your case this gives: $$ \text{"M"} = 13 \\ \text{"I"} = 9\\ 13*27 + 9 = 360$$

It then encrypts this number 360 in the same way, using $$360^{11} \pmod{40741} = 11807.$$ You can decode the base 10 number (360) back into a message by turning it back into base 27, giving digits $(13, 9)$, and then computing their corresponding position in the alphabet ("M", "I").


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