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← 04ps-06 Rsa Encryption Solution

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Showing Revision 1 created 10/24/2012 by Amara Bot.

  1. The plaintext is 387.
  2. Recall that encryption involved raising a message to the power of e mod n
  3. and decryption involved raising a ciphertext to the power of d mod n.
  4. And so to decrypt to ciphertext 903,
  5. we just take it to the d power mod n and get 387,
  6. which we can see in Python.
  7. And, if we want to check our answer,
  8. we can encrypt 387 using a public key
  9. and we get 903--our ciphertext.