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Homework 9 Solutions Section 4.4 1b) To solve 5x ≡ 2 (mod 26) we solve 5x + 26y = 2 (but we only need x.) By Euclid’s algorithm we get 26 = 5 · 5 + 1 5 = 5·1 So 26 − 5 · 5 = 1 (duh). Multiply by 2 to get that 2 · 26 − 10 · 5 = 2. We then see that (−10)5 ≡ 2 (mod 26), thus 16 · 5 ≡ 2 (mod 26). Hence x = 16. 4a) This is just like Example 4.9 in the book. We get the same values for x1 , x2 and x3 . Hence a solution is given by x = 1 · 35 · 2 + 2 · 21 · 1 + 3 · 15 · 1 = 157 ≡ 52 (mod 105). 4c) Here we have that n = 6 · 11 · 17 = 1122 We use the Chinese Remainder Theorem. Now N6 = 11 · 17 = 187, N1 1 = 6 · 17 = 102, and N17 = 6 · 11 = 66. We need to solve 187x6 ≡ 1 (mod 6), 102x11 ≡ 1 (mod 11), and 66x6 ≡ 1 (mod 17). The first is just 1 · x6 ≡ 1 (mod 6), the second reduces to 3x11 ≡ 1 (mod 11), which you can do in your head to get x11 ≡ 4 (mod 11). The third equation reduces to 66x6 ≡ 1 (mod 17) or 15x6 ≡ 1 (mod 17) or (−2)x6 ≡ 1 (mod 17), multiply by -1 to get 2x6 ≡ −1 (mod 17), so we see that x6 ≡ 8 (mod 17). Hence we get that x = 5 · 187 · 1 + 4 · 102 · 4 + 3 · 66 · 8 = 4151 (mod 1122) Thus x ≡ 785 (mod 1122) 6) We see that we are required to find the smallest a > 2 such that a ≡ 0 (mod 2), a ≡ 2 (mod 3), a ≡ 2 (mod 4), a ≡ 2 (mod 5) and a ≡ 2 (mod 6). We see that obviously 2 is a solution, we seek to find the next smallest solution. But a ≡ 2 (mod 4) implies that a ≡ 0 (mod 2). Also if a ≡ 0 (mod 2) and a ≡ 2 (mod 3), then this implies that a ≡ 2 (mod 6). So the set of 5 congruences reduces to just three congruences, namely a ≡ 2 (mod 3), a ≡ 2 (mod 4), and a ≡ 2 (mod 5). So by the Chinese Remainder Theorem we know that all solutions will congruent modulo 60, hence since 2 is a solution, the next smallest one will be 62. 9) We are required to find the smallest a such that a ≡ 1 (mod 2), a ≡ 1 (mod 3), a ≡ 1 (mod 4), a ≡ 1 (mod 5), a ≡ 1 (mod 6) and a ≡ 0 (mod 7). Now a ≡ 1 (mod 4) implies that a ≡ 1 (mod 2) and a ≡ 1 (mod 6) implies that a ≡ 1 (mod 3). Hence we see that the first 5 congruences reduce to the three congruences a ≡ 1 (mod 4), a ≡ 1 (mod 5), a ≡ 1 (mod 6) which has an obvious solution, namely a ≡ 1 (mod 120). So a ≡ 0 (mod 7) and a ≡ 1 (mod 120) and hence the solution will be in mod 840. We just list the numbers that are congruent to 1 (mod 120) until we get one that is a multiple of 7. We find that 721 ≡ 0 (mod 7). Hence the solution is 721. 1 18) We need to obtain two incongruent solutions to the three congruences 2x ≡ 3 (mod 5), 4x ≡ 2 (mod 6) and 3x ≡ 2 (mod 7). The solution to the first equation is x ≡ 4 (mod 5), the solution to the third equation is x ≡ 3 (mod 7). The reason there are two incongruent solution is that the second equation above has two solutions, namely x ≡ 2 (mod 6) and x ≡ 5 (mod 6). So we get two different systems of equations, (1) x≡4 (mod 5), x ≡ 3 (mod 7), and x ≡ 2 (mod 6) (2) x≡4 (mod 5), x ≡ 3 (mod 7), and x ≡ 5 (mod 6) and Solving the first equation we see that n = 5 · 6 · 7 = 210 with N5 = 42, N6 = 35, and N7 = 30. So we need to solve 42x5 ≡ 1 (mod 5), 35x6 ≡ 1 (mod 6), and 30x7 ≡ 1 (mod 7). We can do each of these in our head to get that x5 ≡ 3 (mod 5), x6 ≡ 5 (mod 6) and x7 ≡ 4 (mod 7). We get that x = 4 · 42 · 3 + 2 · 35 · 5 + 3 · 30 · 4 = 1214 (mod 210) So x ≡ 164 (mod 210). The only thing that changes for the second equation is the congruence modulo 6, but the setup is all the same and we get the same x5 , x6 and x7 . So now we get that x = 4 · 42 · 3 + 5 · 35 · 5 + 3 · 30 · 4 = 1529 So x ≡ 59 (mod 210). So our two solutions are x = 59 and x = 164 2 (mod 210)