Part I: Preliminaries
The a = b xor c, because b xor c = b xor (a xor b) = a xor (b xor b) = a xor 0 = a
# of bits to represent x is log2x, and log2(1012) = log10(1012) / log10(2) = 12 / 0.3 = 40 bits
inv(10) = 14 - 10 = 4, so that (4 + 10)%14 = 0
No, only 0 has no multiplicative inverse
Knowing that y = -5, x is clearly 12. Then 12*8 = 1 + 5*19, so (12*8) % 19 = 1
ap mod p = a (if a < p). 143737 mod 19 = 143737 mod 18 mod 19 = 14371 = 12
phi(15) = (3-1)*(5-1) = 8, so that 25802 mod 15 = 25802 mod 8 mod 15 = 252 mod 15 = 625 mod 15 = 10
n interations of the loop, with a square in each iteration, and a product for each 1 bit, so n/2 products on the average. Thus 3n/2 multiplications on the average.
n cannot be prime
Part II: Coding and Information Theory
H = (1/2)*log2(1/(1/2)) + (1/4)*log2(1/(1/4)) + (1/4)*log2(1/(1/4)) = (1/2)*1 + (1/4)*2 + (1/4)*2 = 3/2
One can get at most 1/4 of the channel bandwidth through, so that a 56K channel could transmit at most 14K bits/sec of information. That is a theoretical limit, but general codes to meet the limit have never been devised.
Suppose the data digits of an ISBN number are 014004656. Determine the value of the final digit, labeled a9 above, so that the check equation is true.
(9 + 32 + 20 + 24 + 15 + 12 + a9) % 11 = (112 + a9) % 11 = (2 + a9) % 11, and for this to be 0, one must have a9 = 9
Non-commutativity. No. 0#a = a#0 for any a, and any of 0, 1, 2, 3, and 4 commute
Part III: Introduction to Cryptography
No. They give word breaks and one can use letter frequencies or occurrences of double letters
Known: have corresponding plaintext/ciphertext pairs. Chosen: can choose plaintext and get the corresponding ciphertext
Trivial, very hard, impossible (if key used only once)
IV is xored in at the beginning
See the writeup about CBC mode
Alice obtains the puzzle sequence number by breaking the puzzle, but since the puzzles were sent in random order, Boris must break puzzles until he finds that number.
Commutativity of encryption: EB(EA(M)) = EA(EB(M)). Also, difficulty of solving discrete log problem.
Alice fetches Bob's public key and encrypts for Bob. Bob uses his secret key to decrypt.
They must not be divisible by 3
e*d mod((p-1)*(q-1)) = 1
Used to calculate the secret exponent d from e, p, and q
RSA is very slow, AES is very fast