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Your name: Arvind Borde MATH 1: Homework 4 Solutions A] Are the following true or false? 1. 12 ≡ 5 (mod 5) 2. 12 ≡ 5 (mod 7) 3. 12 ≡ 7 (mod 5) 4. −1 ≡ 5 (mod 6) 5. 397 ≡ 2 (mod 3) 6. 397 ≡ 2 (mod 5) Answers 1. 2. 3. 4. 5. 6. False. 12 6≡ 5 (mod 5), because 5 − 12 = −7 is not divisible by 5. True. 12 ≡ 5 (mod 7), because 5 − 12 = −7 is divisible by 7. True. 12 ≡ 7 (mod 5), because 7 − 12 = −5 is divisible by 5. True. −1 ≡ 5 (mod 6), because 5 − (−1) = 6 is divisible by 6. False. 397 6≡ 2 (mod 3), because 2 − 397 = −395 is not divisible by 3. True. 397 ≡ 2 (mod 5), because 2 − 397 = −395 is divisible by 5. Note: Understanding how these equivalence questions are settled is crucial to understanding the rest of the homework. See pages 11 through 14 in the class notes. B] Solve 1. 2x + 1 ≡ 3 (mod 5) 2. 2x + 1 ≡ 13 (mod 5) 3. 2x + 3 ≡ 1 (mod 5) 4. x − 1 ≡ 3 (mod 7) 5. 1 − 3x ≡ 16 (mod 3) 6. y + 3 ≡ 3 (mod 3) 7. 3x ≡ 1 (mod 5) 8. 3x ≡ 1 (mod 7) 9. 4x ≡ 1 (mod 5) 10. 4x ≡ 0 (mod 5) 11. 4x ≡ 0 (mod 6) 12. 6x ≡ 0 (mod 10) Answers 1. 2x + 1 ≡ 3 (mod 5) i) Solve 2x + 1 = 3 normally: x = 1. ii) Is 0 ≤ x < 5 (the base)? Yes. ii) Answer: x ≡ 1. 2. 2x + 1 ≡ 13 (mod 5) i) Solve 2x + 1 = 13 normally: x = 6. ii) Is 0 ≤ x < 5 (the base)? No. iii) Since x ≥ 5, you repeatedly subtract the base: 6 − 5 = 1: in the correct range. iv) Answer: x ≡ 1. 3. 2x + 3 ≡ 1 (mod 5) i) Solve 2x + 3 = 1 normally: x = −1. ii) Is 0 ≤ x < 5 (the base)? No. iii) Since x < 0, you repeatedly add the base: −1 + 5 = 4: in the correct range. iv) Answer: x ≡ 4. 4. x − 1 ≡ 3 (mod 7) i) Solve x − 1 = 3 normally: x = 4. ii) Is 0 ≤ x < 7 (the base)? Yes. ii) Answer: x ≡ 4. 5. 1 − 3x ≡ 16 (mod 3) i) Solve 1 − 3x = 16 normally: x = −5. ii) Is 0 ≤ x < 3 (the base)? No. iii) Since x < 0, you repeatedly add the base: −5 + 3 = −2: Still negative. −2 + 3 = 1: in the correct range. iv) Answer: x ≡ 1. 6. y + 3 ≡ 3 (mod 3) i) Solve y + 3 = 3 normally: y = 0. ii) Is 0 ≤ y < 3 (the base)? Yes. ii) Answer: y ≡ 0. 7. 3x ≡ 1 (mod 5) Solve 3x = 1 normally: x = 1/3. Fraction. So, plug in values for x in 3x: 3(0) = 0. Is 0 ≡ 1 (mod 5)? No. 3(1) = 3. Is 3 ≡ 1 (mod 5)? No. 3(2) = 6. Is 6 ≡ 1 (mod 5)? Yes. x ≡ 2 is your whole number answer. 9. 4x ≡ 1 (mod 5) Solve 4x = 1 normally: x = 1/4. Fraction. So, plug in values for x in 3x: 4(0) = 0. Is 0 ≡ 1 (mod 5)? No. 4(1) = 4. Is 4 ≡ 1 (mod 5)? No. 4(2) = 8. Is 8 ≡ 1 (mod 5)? No. 4(3) = 12. Is 12 ≡ 1 (mod 5)?No. 4(4) = 16. Is 16 ≡ 1 (mod 5)? Yes. x ≡ 4 is your whole number answer. 8. 3x ≡ 1 (mod 7) Solve 3x = 1 normally: x = 1/3. Fraction. So, plug in values for x in 3x: 3(0) = 0. Is 0 ≡ 1 (mod 7)? No. 3(1) = 3. Is 3 ≡ 1 (mod 7)? No. 3(2) = 6. Is 6 ≡ 1 (mod 7)? No. 3(3) = 9. Is 9 ≡ 1 (mod 7)? No. 3(4) = 12. Is 12 ≡ 1 (mod 7)? No. 3(5) = 15. Is 15 ≡ 1 (mod 7)?Yes. x ≡ 5 is your whole number answer. Questions 10–12 in this set all have x ≡ 0 as an obvious answer. But two of them also have an additional, non-zero answer: 10. 4x ≡ 0 (mod 5) Plug in nonzero values for x in 4x: 4(1) = 4. Is 4 ≡ 0 (mod 5)? No. 4(2) = 8. Is 8 ≡ 0 (mod 5)? No. 4(3) = 12. Is 12 ≡ 0 (mod 5)? No. 4(4) = 16. Is 16 ≡ 0 (mod 5)? No. There is no alternative nonzero answer. 12. 6x ≡ 0 (mod 10) Plug in nonzero values for x in 6x: 6(1) = 6. Is 6 ≡ 0 (mod 5)? No. 6(2) = 12. Is 12 ≡ 0 (mod 5)? No. 6(3) = 18. Is 18 ≡ 0 (mod 5)? No. 6(4) = 24. Is 24 ≡ 0 (mod 5)? No. 6(5) = 30. Is 30 ≡ 0 (mod 5)? Yes. So x ≡ 5 is an alternative nonzero answer. C ] What are the additive and multiplicative inverses of 1. 5 (mod 7)? 2. 2 (mod 7)? Answers 1a. Additive inverse of 5 (mod 7): Set up the equation 5 + x ≡ 0 (mod 7). i) Solve 5 + x = 0 normally: x = −5. ii) Is 0 ≤ x < 7 (the base)? No. iii) Since x < 0, you repeatedly add the base: −5 + 7 = 2: in the correct range. So 2 is the additive inverse of 5 (mod 7). 11. 4x ≡ 0 (mod 6) Plug in nonzero values for x in 4x: 4(1) = 4. Is 4 ≡ 0 (mod 6)? No. 4(2) = 8. Is 8 ≡ 0 (mod 6)? No. 4(3) = 12. Is 12 ≡ 0 (mod 6)? Yes. So x ≡ 3 is an alternative nonzero answer. NOTE: How the coefficient of x in these equations relates to the base tells you if a nonzero answer will exist. Can you guess at the connection? 3. 2 (mod 3)? 1b. Multiplicative inverse of 5 (mod 7): Set up the equation 5x ≡ 1 (mod 7). Plug in values for x in 5x: 5(0) = 0. Is 0 ≡ 1 (mod 7)? No. 5(1) = 5. Is 5 ≡ 1 (mod 7)? No. 5(2) = 10. Is 10 ≡ 1 (mod 7)? No. 5(3) = 15. Is 15 ≡ 1 (mod 7)? Yes. So 3 is the multiplicative inverse of 5 (mod 7). 2a. Additive inverse of 2 (mod 7): Set up the equation 2 + x ≡ 0 (mod 7). i) Solve 2 + x = 0 normally: x = −2. ii) Is 0 ≤ x < 7 (the base)? No. ii) Since x < 0, you repeatedly add the base: −2 + 7 = 5: in the correct range. So 5 is the additive inverse of 2 (mod 7). 2b. Multiplicative inverse of 2 (mod 7): Set up the equation 2x ≡ 1 (mod 7). Plug in values for x in 5x: 2(0) = 0. Is 0 ≡ 1 (mod 7)? No. 2(1) = 2. Is 2 ≡ 1 (mod 7)? No. 2(2) = 4. Is 4 ≡ 1 (mod 7)? No. 2(3) = 6. Is 6 ≡ 1 (mod 7)? No. 2(4) = 8. Is 8 ≡ 1 (mod 7)? Yes. So 4 is the multiplicative inverse of 2 (mod 7). 3a. Additive inverse of 2 (mod 3): Set up the equation 2 + x ≡ 0 (mod 3). i) Solve 2 + x = 0 normally: x = −2. ii) Is 0 ≤ x < 3 (the base)? No. ii) Since x < 0, you repeatedly add the base: −2 + 3 = 1: in the correct range. So 1 is the additive inverse of 2 (mod 3). 3b. Multiplicative inverse of 2 (mod 3): Set up the equation 2x ≡ 1 (mod 3). Plug in values for x in 5x: 2(0) = 0. Is 0 ≡ 1 (mod 3)? No. 2(1) = 2. Is 2 ≡ 1 (mod 3)? No. 2(2) = 4. Is 4 ≡ 1 (mod 3)? Yes. So 2 is the multiplicative inverse of 2 (mod 3). D] Write down the addition and multiplication tables for integers mod 5. Answer Set up normal addition and multiplication tables: + 0 1 2 3 4 × 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 0 1 2 3 4 0 0 0 0 0 0 1 2 3 4 0 0 2 3 4 6 6 9 8 12 0 4 8 12 16 For each number that’s ≥ 5 in these tables, find a number that’s equivalent to it and < 5. + 0 1 2 3 4 × 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3 0 1 2 3 4 0 0 0 0 0 0 1 2 3 4 0 2 4 1 3 0 3 1 4 2 0 4 3 2 1