Assignment #1 Solutions January 10, 2006

Problem #1 Show that ((a mod m) + b) mod m = (a + b) mod m and Practical Aspects of Modern Cryptography February 28, 2019

Answer #1 Let r1 = a mod m, r2 = (a + b) mod m, and r3 = ((a mod m) + b) mod m. Then there exist integers q1, q2, and q3 such that a = q1m + r1, a + b = q2m + r2, and ((a mod m) + b) = q3m + r3. Practical Aspects of Modern Cryptography February 28, 2019

Answer #1 (cont.) r2 – r3 = ((a + b) – q2m) – ((a mod m) + b) – q3m) = ((a + b) – q2m) – ((r1 + b) – q3m) = ((a + b) – q2m) – (((a – q1m) + b) – q3m) = (q1 – q2 + q3)m But 0 ≤ r2 < m and 0 ≤ r3 < m, so |r2 – r3| < m. Thus, (q1 – q2 + q3) = 0, and therefore r2 = r3. Practical Aspects of Modern Cryptography February 28, 2019

Answer #1 (cont.) Let r1 = a mod m, r2 = (a  b) mod m, and r3 = ((a mod m)  b) mod m. Then there exist integers q1, q2, and q3 such that a = q1m + r1, a  b = q2m + r2, and ((a mod m)  b) = q3m + r3. Practical Aspects of Modern Cryptography February 28, 2019

Answer #1 (cont.) r2 – r3 = ((a  b) – q2m) – ((a mod m)  b) – q3m) = ((a  b) – q2m) – ((r1  b) – q3m) = ((a  b) – q2m) – (((a – q1m)  b) – q3m) = (bq1 – q2 + q3)m But 0 ≤ r2 < m and 0 ≤ r3 < m, so |r2 – r3| < m. Thus, (bq1 – q2 + q3) = 0, and therefore r2 = r3. Practical Aspects of Modern Cryptography February 28, 2019

Problem #2 Find the mod 11 multiplicative inverse of each value x in the range 0 < x < 11. Practical Aspects of Modern Cryptography February 28, 2019

Answer #2  1 2 3 4 5 6 7 8 9 10 Practical Aspects of Modern Cryptography February 28, 2019

Answer #2 (cont.)  1 2 3 4 5 6 7 8 9 10 Practical Aspects of Modern Cryptography February 28, 2019

Answer #2 (cont.) The same cannot be done mod 12. For example, 2Y mod 12 is even regardless of the value of Y. Therefore 2 does not have a mod 12 multiplicative inverse. Practical Aspects of Modern Cryptography February 28, 2019

Problem #3 Compute 31415926535897932384626433832795028841971693993751058209749445923078164 mod 11. Practical Aspects of Modern Cryptography February 28, 2019

Answer #3 31 mod 11 = 3 36 mod 11 = 3 32 mod 11 = 9 37 mod 11 = 9 Practical Aspects of Modern Cryptography February 28, 2019

Problem #4 Compute 27182818284590452353602874713526624977572470936999595749669676277240766 mod 33. Practical Aspects of Modern Cryptography February 28, 2019

Answer #4 21 mod 33 = 2 26 mod 33 = 31 22 mod 33 = 4 27 mod 33 = 29 Practical Aspects of Modern Cryptography February 28, 2019

Problem #5 If p is a prime, then for integers a such that 0 < a < p, then a p - 1 mod p = 1. Use this fact to show that 65 is not prime. Practical Aspects of Modern Cryptography February 28, 2019

Answer #5 Thus 65 is not prime. 21 mod 65 = 2 216 mod 65 = 16 Practical Aspects of Modern Cryptography February 28, 2019