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Tutorial 3 Number Theory

. . Question 1(i) Solve the following linear congruences:(a) 4x ? 5 (mod 7)(b) 7 (3 ? x) ? 5 (mod 11)(c)8x ? 3 (mod 19)?Determine the least positive integer which simultaneously satisfies all three of the linear congruences in part (i).. Solution to question 1(a)4x ? 5 (mod 7) ?.

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Tutorial 3 Number Theory

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    1. Tutorial 3 Number Theory Please go to >Tools (at the top of the screen)> Audio > Audio Setup Wizard and follow the instructions.

    3. Question 1 (i) Solve the following linear congruences: (a) 4x ? 5 (mod 7) (b) 7 (3 x) ? 5 (mod 11) (c) 8x ? 3 (mod 19) Determine the least positive integer which simultaneously satisfies all three of the linear congruences in part (i).

    4. Solution to question 1 (a) 4x ? 5 (mod 7)

    5. Solution to question 1 (a) 4x ? 5 (mod 7) ? 25x ? 5 (mod 7)

    6. Solution to question 1 (a) 4x ? 5 (mod 7) ? 25x ? 5 (mod 7) ? 5x ? 1 ? 15 (mod 7)

    7. Solution to question 1 (a) 4x ? 5 (mod 7) ? 25x ? 5 (mod 7) ? 5x ? 1 ? 15 (mod 7) ? x ? 3 (mod 7)

    8. Solution to question 1 (a) 4x ? 5 (mod 7) ? 25x ? 5 (mod 7) ? 5x ? 1 ? 15 (mod 7) ? x ? 3 (mod 7) (b) 7 (3 x) ? 5 (mod 11) ? 7x ? 16 (mod 11) ? 4x ? 16 (mod 11) ? x ? 4 ? 7 (mod 11) (c) 8x ? 3 (mod 19) ? 27x ? 3 (mod 19) ? 9x ? 1 ? 18 (mod 19) ? x ? 2 ? 17 (mod 19)

    9. ii) x ? 3(mod 7), x ? 7 (mod 11), x ? 17 (mod 19) x ? 17 (mod 19) ? x = 17,

    10. ii) x ? 3(mod 7), x ? 7 (mod 11), x ? 17 (mod 19) x ? 17 (mod 19) ? x = 17, and x ? 3 (mod 7) ? x = 17,

    11. ii) x ? 3(mod 7), x ? 7 (mod 11), x ? 17 (mod 19) x ? 17 (mod 19) ? x = 17, and x ? 3 (mod 7) ? x = 17, 150, and x ? 7 (mod 11) ? x = 150,

    12. ii) x ? 3(mod 7), x ? 7 (mod 11), x ? 17 (mod 19) x ? 17 (mod 19) ? x = 17, and x ? 3 (mod 7) ? x = 17, 150, and x ? 7 (mod 11) ? x = 150, Hence x ? 150 (mod 1463) is the least positive integer satisfying all three congruences.

    13. Fermats Little Theorem (FLT) Question 2 Find the least positive residue of 2563 (mod 31)

    14. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31)

    15. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31) Hence 2563 ? (2530)2 x 253

    16. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31) Hence 2563 ? (2530)2 x 253 ? 12 x ( 6)3

    17. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31) Hence 2563 ? (2530)2 x 253 ? 12 x ( 6)3 ? 5 x ( 6)

    18. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31) Hence 2563 ? (2530)2 x 253 ? 12 x ( 6)3 ? 5 x ( 6) ? 30 ? 1 (mod 31)

    19. (i) Find the least positive residue of 2563 (mod 31) Use FLT with a = 25, p = 31 and gcd (25, 31) = 1 ? 2530 ?1 (mod 31) Hence 2563 ? (2530)2 x 253 ? 12 x ( 6)3 ? 5 x ( 6) ? 30 ? 1 (mod 31) Therefore the least positive residue of 2563 (mod 31) is 1.

    20. (ii) Solve 11x ? 1 (mod 31).

    21. (ii) Solve 11x ? 1 (mod 31). Explain why the solution of this congruence is equal to the least positive residue of 1129 (mod 31).

    22. (ii) Solve 11x ? 1 (mod 31). Explain why the solution of this congruence is equal to the least positive residue of 1129 (mod 31). Write down a linear congruence, modulo 31, whose solution is congruent to 1128 (mod 31) and hence determine the least positive residue of 1128 (mod 31).

    23. (ii) Solving 11x ? 1 (mod 31)

    24. (ii) Solving 11x ? 1 (mod 31) ? 20x ? 32 (mod 31) ? 5x ? 8 ? 70 (mod 31) ? x ? 14 ? 17 (mod 31)

    25. Explain why the solution of this congruence 11x ? 1 (mod 31) is equal to the least positive residue of 1129 (mod 31).

    26. Explain why the solution of this congruence 11x ? 1 (mod 31) is equal to the least positive residue of 1129 (mod 31). FLT gives 1130 ? 1 (mod 31) So 1130 = 11 x 1129 ? 1 (mod 31)

    27. Explain why the solution of this congruence 11x ? 1 (mod 31) is equal to the least positive residue of 1129 (mod 31). FLT gives 1130 ? 1 (mod 31) So 1130 = 11 x 1129 ? 1 (mod 31) Also 11x ? 1 (mod 31) has a unique solution because gcd (11, 31) = 1

    28. Explain why the solution of this congruence 11x ? 1 (mod 31) is equal to the least positive residue of 1129 (mod 31). FLT gives 1130 ? 1 (mod 31) So 1130 = 11 x 1129 ? 1 (mod 31) Also 11x ? 1 (mod 31) has a unique solution because gcd (11, 31) = 1 Hence 1129 ? 17 (mod 31), i.e. the solution of the congruence 11x ? 1 (mod 31) is equal to the least positive residue of 1129 (mod 31).

    29. Write down a linear congruence, modulo 31, whose solution is congruent to 1128 (mod 31) and hence determine the least positive residue of 1128 (mod 31).

    30. Write down a linear congruence, modulo 31, whose solution is congruent to 1128 (mod 31) and hence determine the least positive residue of 1128 (mod 31). A linear congruence whose solution is congruent to 1128 (mod 31) is 11x ? 17 (mod 31),

    31. Write down a linear congruence, modulo 31, whose solution is congruent to 1128 (mod 31) and hence determine the least positive residue of 1128 (mod 31). A linear congruence whose solution is congruent to 1128 (mod 31) is 11x ? 17 (mod 31), and solving this gives ? 51x ? 17 (mod 31) ? 3x ? 1 ? 30 (mod 31) ? x ? 10 (mod 31)

    32. Write down a linear congruence, modulo 31, whose solution is congruent to 1128 (mod 31) and hence determine the least positive residue of 1128 (mod 31). A linear congruence whose solution is congruent to 1128 (mod 31) is 11x ? 17 (mod 31), and solving this gives ? 51x ? 17 (mod 31) ? 3x ? 1 ? 30 (mod 31) ? x ? 10 (mod 31) Hence the least positive residue of 1128 (mod 31) is 10.

    33. (iii) Prove that a31 ? a (mod 231) for every integer a.

    34. (iii) alternative version of FLT used - a may not be relatively prime to the modulus Note that 231 = 3 x 7 x 11

    35. (iii) alternative version of FLT used - a may not be relatively prime to the modulus Note that 231 = 3 x 7 x 11 FLT (alternative version) gives a 3 ? a (mod 3) So a 31 ? (a 3 )10 ? a10 x a ? (a 3 )3 x a 2 ? a 3 x a 2 ? a x a 2 ? a 3 ? a (mod 3)

    36. (iii) alternative version of FLT used - a may not be relatively prime to the modulus Note that 231 = 3 x 7 x 11 FLT (alternative version) gives a 3 ? a (mod 3) So a 31 ? (a 3 )10 ? a10 x a ? (a 3 )3 x a 2 ? a 3 x a 2 ? a x a 2 ? a 3 ? a (mod 3) FLT (alternative version) gives a 7 ? a (mod 7) So a 31 ? (a 7 )4 x a 3 ? a 4 x a 3 ? a 7 ? a (mod 7)

    37. (iii) alternative version of FLT used - a may not be relatively prime to the modulus Note that 231 = 3 x 7 x 11 FLT (alternative version) gives a 3 ? a (mod 3) So a 31 ? (a 3 )10 ? a10 x a ? (a 3 )3 x a 2 ? a 3 x a 2 ? a x a 2 ? a 3 ? a (mod 3) FLT (alternative version) gives a 7 ? a (mod 7) So a 31 ? (a 7 )4 x a 3 ? a 4 x a 3 ? a 7 ? a (mod 7) FLT (alternative version) gives a 11 ? a (mod 11) So a 31 ? (a 11 ) 2 x a 9 ? a 2 x a 9 ? a 11 ? a (mod 11)

    38. (iii) alternative version of FLT used - a may not be relatively prime to the modulus Note that 231 = 3 x 7 x 11 FLT (alternative version) gives a 3 ? a (mod 3) So a 31 ? (a 3 )10 ? a10 x a ? (a 3 )3 x a 2 ? a 3 x a 2 ? a x a 2 ? a 3 ? a (mod 3) FLT (alternative version) gives a 7 ? a (mod 7) So a 31 ? (a 7 )4 x a 3 ? a 4 x a 3 ? a 7 ? a (mod 7) FLT (alternative version) gives a 11 ? a (mod 11) So a 31 ? (a 11 ) 2 x a 9 ? a 2 x a 9 ? a 11 ? a (mod 11) Hence by Corollary to Theorem 1.3 Unit 3, a31 ? a (mod 231) for every integer a

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