1. M ODULAR A RITHMETIC Lesson 4.6

2. This section, we are grouping numbers, based on their reminders. 18 ÷ 4 = 4 r. 2 22 ÷ 4 = 5 r. 2 78 ÷ 4 = 19 r. 2 These are said to be in the same congruence class. 18 ≡ 22 (mod 4)  bc their remainders are the same

3. H OW MANY CONGRUENCE CLASSES ARE THERE IN MODULO 7? 77 ÷ 7 = 11 r. 0 78 ÷ 7 = 11 r. 1 79 ÷ 7 = 11 r. 2 80 ÷ 7 = 11 r. 3 81 ÷ 7 = 11 r. 4 82 ÷ 7 = 11 r. 5 83 ÷ 7 = 11 r. 6 84 ÷ 7 = 12 r. 0 R0R0 R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 77 ≡ 84 (mod 11) There are 7 congruence classes!

4. C ONGRUENCE T HEOREM For all integers a and b and all positive integers m, a ≡ b (mod m) iff m is a factor of a – b. ISBN and credit card check numbers are determined in some way by modular arithmetic.

5. F IND THE CONGRUENCE CLASSES MOD 7: In what class does 427 fall? 426 ≡ x (mod 7) 426 ÷ 7 = 60 r. 6 426 ≡ 6 (mod 7) R6R6 In what class does -17 fall? -17 ≡ y (mod 7) -17 ÷ 7 = -2 r. -3 but r cannot be negative! change to -3 r. 4 R4R4

6. E XAMPLE 3 Find the smallest positive value of n for which n – 32 ≡ 75 (mod 11) n ≡ 107 (mod 11) Congruence class of R 8 n = 8

7. E XAMPLE 4 If m ≡ 4 (mod 9) and n≡ 7(mod 9), what is the remainder when 5m + n 2 is divided by 9? 5m + n 2 ≡ ?? (mod 9) Plug in for n and m 5(4) + 7 2 = 69 5m + n 2 ≡ 69 (mod 9) 69 ÷ 9 = 7 r. 6 So, 6!

8. E XAMPLE 2 Use modular arithmetic to explain why a date that falls on Friday this year will fall on Wednesday four years from now. 365 days in a year + 1 leap year! 1461 days ÷ 7 days in a week = 208 weeks r. 5 Friday – r. 0Tuesday – r. 4 Sat. – r.1Wednesday – r. 5 Sun. – r.2Thursday – r. 6 Monday – r.3

9. E XAMPLE 5 The final digit of a 12-digit Universal Product Code (UPC) is a check digit. Suppose the digits of a UPS are represented by X 1 X 2 X 3 X 4 X 5 X 6 -X 7 X 8 X 9 X 10 X 11 X c To calculate the check digit X c, first compute 3(X 1 +X 3 +X 5 +X 7 +X 9 +X 11 ) + (X 2 +X 4 +X 6 +X 8 +X 10 ) (mod 10). If this number is zero, then X c =0, If not, subtract the number from 10. Determine the check digit of 300219 – 09529 3(3 + 0 + 1 + 0 + 5 + 9) + (0 + 2 + 9 + 9 + 2) = 3(18) + (22) = 76 ÷ 10 = 7 r. 6 So X c =4

10. E XAMPLE 6 Prove by contradiction: With one exception, If n is a prime number, then n + 1 is not a prime number. Thought: Cannot be true, because if n is a prime number, it must be odd. Therefore, n + 1 is even.  Assume n + 1 IS a prime number. n ≡ 1 (mod 2) n + 1 ≡ 1 (mod 2) Subtracting one from both sides, n ≡ 0 (mod 2) CONTRADICTION!!! Therefore, n + 1 is not a prime number.

11. H OMEWORK Pages 261 – 262 1-3, 5-11, 19