Solutions to Section 2.1 Homework Problems S. F. Ellermeyer

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1 Solutions to Section 21 Homework Problems S F Ellermeyer 1 [13] 9 = f13; 22; 31; 40; : : :g [ f4; 5; 14; : : :g [3] 10 = f3; 13; 23; 33; : : :g [ f 7; 17; 27; : : :g [4] 11 = f4; 15; 26; : : :g [ f 7; 18; 29; 40; : : :g mod 5 = 3 mod mod 5 = 3 2 mod 5 = 9 mod 5 = 4 mod mod 5 = mod 5 = 4 2 mod 5 = 16 mod 5 = 1 mod mod 5 = mod 5 = 1 11 mod 5 = 1 mod mod 5 = (103 mod 5) mod 5 = (3) (1) = mod 11 = 3 mod mod 11 = 3 2 mod 11 = 9 mod mod 11 = (3 9) mod 11 = 5 mod mod 11 = (3 5) mod 11 = 4 mod mod 11 = (58 4) mod 11 = 232 mod 11 = 1 mod mod 11 = mod 11 = 1 5 mod 11 = 1 mod mod 11 = mod 11 = mod mod 11 = (1 mod 11) (4 mod 11) = (1 4) mod 11 = 4 mod 11 = 4 1

2 7 Suppose that gcd (a; 5) = 1 Prove that a 4i 1 mod 5 for all i 2 N Proof: We want to show that (a 4 ) i 1 mod 5 for all i 2 N Since gcd (a; 5) = 1, then we know that a is not a multiple of 5 In other words, a is not congruent to 0 modulo 5 This means that either a 1 mod 5 or a 2 mod 5 or a 3 mod 5 or a 4 mod 5 We will consider each of these possibilities separately: If a 1 mod 5; then a mod 5 a 4 1 mod 5 If a 2 mod 5; then a mod 5 a 4 1 mod 5 If a 3 mod 5; then a mod 5 a 4 1 mod 5 If a 4 mod 5; then a mod 5 a 4 1 mod 5 Thus we have shown that it must be the case that a 4 1 mod 5 We can immediately conclude that (a 4 ) i 1 i mod 5 for any integer i 0 Hence a 4i 1 mod 5 for all i 2 N 9 (a) Since gcd (2; 4) = 2 and 2 does not divide 1, then the equation 2x 1 mod 4 does not have any lutions (b) Since gcd (4; 8) = 4 and 4 does divide 4, then the equation 4x 4 mod 8 does have lutions This equation is equivalent to x 1 mod 2, the lution set is [1] 2 We can write this lution set as a union of equivalence classes modulo 8 as follows: [1] 2 = [1] 8 [ [3] 8 [ [5] 8 [ [7] 8 (c) Since gcd (6; 12) = 6 and 6 does not divide 3, then the equation 6x 3 mod 12 does not have any lutions (d) Since gcd (8; 5) = 1 and 1 does divide 4, then the equation 8x 4 mod 5 does have lutions By inspection, we see that the lution set is [3] 5 (e) Since gcd (3; 11) = 1 and 1 does divide 7, then the equation 3x 7 mod 11 does have lutions The lution set is [6] (a) Since gcd (6; 15) = 3 and 3 does divide 9, then the equation 6x 9 mod 15 does have lutions This equation is equivalent to 2x 3 mod 5, the lution set is [4] 5 We can write this lution set as a union of equivalence classes modulo 15 as follows: [4] 5 = [4] 15 [ [9] 15 [ [14] 15 2

3 From this, we see that there are 3 distinct congruence classes (modulo 15) that comprise the lution set, that the smallest nonnegative lution is 4 and that the only lutions, x, such that 0 x < 15 are x = 4; x = 9, and x = 14 (b) Since gcd (15; 35) = 5 and 5 does divide 20, then the equation 15x 20 mod 35 does have lutions This equation is equivalent to 3x 4 mod 7, the lution set is [6] 7 We can write this lution set as a union of equivalence classes modulo 35 as follows: [6] 7 = [6] 35 [ [13] 35 [ [20] 35 [ [27] 35 [ [34] 35 From this, we see that there are 5 distinct congruence classes (modulo 35) that comprise the lution set, that the smallest nonnegative lution is 6 and that the only lutions, x, such that 0 x < 35 are x = 6; x = 13, x = 20, x = 27, and x = 34 (c) Since gcd (15; 19) = 1 and 1 does divide 16, then the equation 15x 16 mod 19 does have lutions By inspection, we see that the lution set is [15] 19 From this, we see that there is just 1 distinct congruence classes (modulo 19) that comprises the lution set, that the smallest nonnegative lution is 15 and that the only lution, x, such that 0 x < 19 is x = (a) To lve 31x 1 mod 53, we must lve the Diophantine equation By Euclid s Algorithm, we have which gives 1 = 9 2 (4) 31x + 53y = 1 53 = 1 (31) = 1 (22) = 2 (9) = 2 (4) + 1 = 9 2 (22 2 (9)) = 5 (9) 2 (22) = 5 (31 22) 2 (22) = 5 (31) 7 (22) = 5 (31) 7 (53 31) = 12 (31) 7 (53) 3

4 Thus (12; 7) is a lution of 31x + 53y = 1 and we conclude from this that (276; 161) is a lution of 31x + 53y = 23 Since 31 (276) = 23 + (161) (53), we see that x = 276 is a lution of the congruence equation 31x 23 mod 53 Al, since gcd (31; 53) = 1, the entire lution set of this congruence is [276] 53 (b) To lve 23x 1 mod 31, we must lve the Diophantine equation By Euclid s Algorithm, we have which gives 1 = x + 31y = 1 31 = 1 (23) = 2 (8) = 1 (7) + 1 = 8 (23 2 (8)) = 3 (8) 23 = 3 (31 23) 23 = 3 (31) 4 (23) Thus ( 4; 3) is a lution of 23x + 31y = 1 and we conclude from this that ( 60; 45) is a lution of 23x + 31y = 15 Since 23 ( 60) = 15+( 45) (31), we see that x = 60 is a lution of the congruence equation 23x 15 mod 31 Al, since gcd (23; 31) = 1, the entire lution set of this congruence is [ 60] 31 (which is equal to [2] 31 ) 15 (a) (b) 2 4 mod 7 = 16 mod 7 = 2 mod mod 7 = mod 7 = 2 5 mod 7 = 32 mod 7 = mod 13 = 25 mod 13 = mod 13 = mod 13 = (25) 2 mod 13 = 1 mod 13 4

5 gives us mod 13 = mod 13 = 1 25 mod 13 = 1 mod 13 = 1 (c) To nd the multiplicative inverse (modulo 7) of the number 2 20, we must nd the number x such that 1 x 6 such that (2 20 x) mod 7 = 1 mod 7 Note that (2 20 x) mod 7 = (2 20 mod 7 x mod 7) mod 7 and al note that x mod 7 = x (because we are requiring that 1 x 6) The problem here is to compute 2 20 mod 7 that we can simplify the equation that we have to lve: 2 4 mod 7 = 16 mod 7 = 2 mod mod 7 = mod 7 = 2 5 mod 7 = 4 Hence, we must lve the equation 4x mod 7 = 1 mod 7 (for a value of x such that 1 x 6) It is easy to see that the lution is x = 2 17 (a) The lution set of the system of congruences is x 1 mod 5 x 2 mod 6 x 3 mod 7 [1] 5 \ [2] 6 \ [3] 7 = [x 0 ] 567 = [x 0 ] 210 where x 0 is me number that we need to nd To nd x 0, we rst look for a number z 2 [1] 5 \ [2] 6 This z must satisfy z = 1 + 5s = 2 6t for me integers s and t We must therefore lve the Diophantine equation 5s + 6t = 1 An obvious lution is (s; t) = ( z = 4 2 [1] 5 \ [2] 6 1; 1) From this, we see that 5

6 Now we know that our number x 0 must lie in [ leads us to study the Diophantine equation 4] 30 \ [3] 7 This or s = 3 7t 30s + 7t = 7 Since an obvious lution of the above equation is (s; t) = (0; 1), we see that x 0 = 4 2 [ 4] 30 \ [3] 7 Thus the lution set of the above system of congruence equations is [ 4] The lution set of the system of congruences is x 3 mod 7 x 7 mod 11 x 2 mod 13 [3] 7 \ [7] 11 \ [2] 13 = [x 0 ] = [x 0 ] 1001 where x 0 is me number that we need to nd To nd x 0, we rst look for a number z 2 [3] 7 \ [7] 11 This z must satisfy z = 3 + 7s = 7 11t for me integers s and t We must therefore lve the Diophantine equation 7s + 11t = 4 A lution is (s; t) = ( 12; 8) From this, we see that z = 81 2 [3] 7 \ [7] 11 (a) Now we know that our number x 0 must lie in [ leads us to study the Diophantine equation 81] 77 \ [2] 13 This s = 2 13t or 77s + 13t = 83 Since a lution of the above equation is (s; t) = ( 83; 498), we see that x 0 = [ 81] 77 \ [2] 13 Thus the lution set of the above system of congruence equations is [ 6472] 1001 (which is the same as [535] 1001 ) 6

7 19 If x is a possible number of coins that the pirates could have been distributing, then x must satisfy the system of congruence equations x 3 mod 17 x 10 mod 16 x 0 mod 15 The lution set of this system of congruences is for me x 0 [3] 17 \ [10] 16 \ [0] 15 = [x 0 ] 4080 Solving s = 10 16t, which is equivalent to 17s + 16t = 7, we obtain a lution (s; t) = (7; 7) which means that [3] 17 \ [10] 16 and hence [3] 17 \ [10] 16 = [122] 272 Solving s = 0 15t, which is equivalent to 272s + 15t = 122 we obtain a lution (s; t) = ( 61; 1098) which means that 16; [122] 272 \ [0] 15 and hence [122] 272 \ [0] 15 = [ 16; 470] 4080 The smallest positive member of this equivalence class is 3; 930 Therefore, the pirates must have been ghting over at least 3; 930 coins 7