Misalkan
p = 3
q = 7
n = p.q
= 3 x 7
= 21
m = (p-1) (q-1)
= (3 -1)(7 – 1)
= 12
e * d mod 12 = 1
e = 5
d = 17
public key = (e,n) = (5,21)
private key = (d,n) = (17,21)
T O M I
84 79 77 73
Proses Enkripsi dan Deskripsi
T = 84
| Enkripsi | Deskripsi |
| C = M ^ e mod n | M = C ^ d mod n |
| 8 ^ 5 mod 21 = 8 | 8 ^ 17 mod 21 = 8 |
| 4 ^ 5 mod 21 = 16 | 16 ^ 17 mod 21 = 4 |
O = 79
| Enkripsi | Deskripsi |
| C = M ^ e mod n | M = C ^ d mod n |
| 7 ^ 5 mod 21 = 7 | 7 ^ 17 mod 21 = 7 |
| 9 ^ 5 mod 21 = 18 | 18 ^ 17 mod 21 = 9 |
M = 77
| Enkripsi | Deskripsi |
| C = M ^ e mod n | M = C ^ d mod n |
| 7 ^ 5 mod 21 = 7 | 7 ^ 17 mod 21 = 7 |
| 7 ^ 5 mod 21 = 7 | 7 ^ 5 mod 21 = 7 |
I = 73
| Enkripsi | Deskripsi |
| C = M ^ e mod n | M = C ^ d mod n |
| 7 ^ 5 mod 21 = 7 | 7 ^ 17 mod 21 = 7 |
| 3 ^ 5 mod 21 = 12 | 12 ^ 17 mod 21 = 3 |