blic key, e. This public key must be co-prime with : 16272. To be able to perform the encryption and decryption computations with a pocket calculator (just for this example's sake!), assume that Alice picked for her public key e = 5, a number that represents the smallest eligible value (3 and 4 divide into 16 272). Going to the public directory or Alice's website, Bob, like any other visitor, may read, "Alice's encryption instructions: please kindly use two-ASCII block encryption modulo 16 571 as the standard; my public key is number 5." No more instructions are needed for Bob to proceed to encryption. Bob then proceeds to encrypt his love declaration according to the formula Ci = me mod N, where Mi is the decimal block number i in Bob's message sequence M1M2M3M4. This gives C1 = m mod 16 571 = 94525 mod 16 571 = 3704 C2 = m5 mod 16 571 = 143265 mod 16 571 = 766 (20.39) C3 = m3 mod 16 571 = 13 0495 mod 16 571 = 475 C4 = ms mod 16 571 = 143255 mod 16 571 = 372. In the above computations, it is absolutely essential that Bob make no truncation errors. 19 Having finished these computations, Bob emails his cipher message C1C2C3C4 = 3704 766 - 475 372 to Alice; he may possibly include a few more words in order to identify himself (e.g., "Bob your classmate."), but this is not the point of the endeavor. Alice is then the only person who is able to decrypt Bob's message. Her private key, d, is defined by the formula ed = 1 mod , or d=e-1 mod , namely d = 5-1 mod 16 272. Thus, Alice needs to find the inverse of 5 modulo 16272. Since 16 272 is not the product of two primes (16 272 = 16 x 9 x 113), she can't use Euler's theorem. Instead, she 19 For instance, with a pocket calculator the function 94525 yields 7.544293311 X 1019, but the last ten digits are missing! To avoid truncations from the pocket calculator, the trick is to use the modulus- arithmetic formula |m5/N = |m2 X |m²1 X |m|n|n|n, in which the different terms and their suc- cessive products (after reducing each one to their modulo N residue) fit in the calculat