Best 1/2 answered by an example:
You want to trasmit the letter A disguised as the number 1.
You broadcast the number 22.
Your public key is 3. Anyone who has your public key can quickly figure out that the real message is 22 – (7 times * Public Key 3) = 1 == Letter A.
Your private key is actually 7. So to encrypt your message, you would have gone 22 – (Private Key 7 * 3 times) == 1 == A
The key to notice: The real message is the remainder of division, NOT the transmitted number itself.
When taken to very large numbers, the number of possible private keys to a public key is so high that it is practically unguessable.
Note that if you transmit 43 the end results are the same: 43 divided by 7 leaves remainder 1, with 7 as the encryption key.
For decryption: 43 divided by 3 again yields 1.
== ANYONE can decrypt if they have your public key. Only you can encrypt: 43 / 7 yields one as remainder.
Latest Answers