1. Assignment #1 Solutions
January 10, 2006

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

3. 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.
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4. Answer #1 (cont.)
r2 – r = ((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.
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February 28, 2019

5. 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

6. Answer #1 (cont.)
r2 – r = ((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

7. Problem #2
Find the mod 11 multiplicative inverse of each value x in the range 0 < x < 11.
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8. Answer #2  1 2 3 4 5 6 7 8 9 10
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9. Answer #2 (cont.)  1 2 3 4 5 6 7 8 9 10
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10. 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.
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February 28, 2019

11. Problem #3
Compute mod 11.
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12. Answer #3 31 mod 11 = 3 36 mod 11 = 3 32 mod 11 = 9 37 mod 11 = 9
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13. Problem #4
Compute mod 33.
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14. Answer #4 21 mod 33 = 2 26 mod 33 = 31 22 mod 33 = 4 27 mod 33 = 29
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15. 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.
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February 28, 2019

16. Answer #5 Thus 65 is not prime. 21 mod 65 = 2 216 mod 65 = 16
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February 28, 2019