 that the result comes to: =me) mod N = = med mod N = mk(p-1)(q-1)+1 mod N (20.38) mod N]k mod N = m x 1 * mod . The result in Eq. (20.38) shows that Alice has fully and unambiguously retrieved the original plaintext block m. Note that for the above demonstration, we used different properties of modular arithmetic, and in particular Euler's theorem, as described in Appendix T. We also used the pre-set property of Alice's key, ed = 1 mod , from which we have ed=k(p-1)(q-1)+1.Aliceis, - thus, able to decrypt any message sent to her from various "Bob" correspondents who may use her public key for this very purpose. For illustration, consider a full example of RSA encryption and decryption. 18 First, here is Bob's private and intimate declaration to Alice: Plaintext I PT-ASCII 1001001 1101100 1101111 1110110 1100101 1111001 1101111 1110101 Consistently with the RSA approach, Bob may encrypt his plaintext by blocks, which represent for instance two letters each. With the above ASCII message, a two-letter block is a number of 2 X 7 = 14 bits, corresponding to a maximum size of 214 1 = 16383. Let us assume that this is the standard. On her part, Alice also chose two primes p, q whose product is greater or equal to 16 383, for instance p = 73 and q = 227, which gives N = pq = 16 571, and = (p - 1)(q - 1) = 72 X 226 = 16272. To be effective, Alice made the number N = 16571 known to everyone, just like a phone number in the directory. Of course, the numbers p, q, remain only known to her. This example was previously published in E. Desurvire, Global Telecommunications, Broadband Access, Optical Components and Networks, and Cryptography (New York: J. Wiley and Sons, 2004). 20.7 Public-key cryptography Following the standard, Bob must convert his message into 2-letter blocks and the resulting blocks into decimal numbers, as follows: Plaintext PT-ASCII 1001001 1101100 1101111 1110110 1100101 1111001 1101111 1110101 Block 10 11 00 11 00 10 11 11 10 01 11 01 11 11 11 01 01 Decimal 14326 13 049 14325 Alice has also chosen her pu