Homework #2 solutions Due: June 15, 2012


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1 All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is write d = 475s + 385t for some integers s and t. Solution. The greatest common divisor d is computed by applying the Euclidean algorithm: 475 = = = = = = Hence d = gcd(475, 385) = 5. To write d = 5 as a linear combination, d = 475s + 385t, reverse the above Euclidean algorithm computation: 5 = Thus 5 = Alternate Solution: = 15 (25 15) = = 2( ) 25 = = ( ) = = 30( ) = Use the matrix method illustrated on Page 11. The notation k between two matrices means that the second matrix is obtained from the first by subtracting k times the second row from the first, while, if the k is on top (as in k ), then k times the first row is subtracted from the second. [ ] Hence d = (475, 385) = 5 = (a) Calculate d = gcd(4307, 1121) and express it as a linear combination of 4307 and Math
2 Solution. The simplest method is to use the Euclidean algorithm. The following is the matrix format of this algorithm: [ ] Hence d = (4307, 1121) = 59 = (b) Calculate m = lcm[4307, 1121]. Hint: You may want to refer to the formula on page 22 of the handout. Solution. By the formula on page 22 of the handout, we have (a, b)[a, b] = ab for any positive integers a and b. Hence, using part (a): [4307, 1121] = (4307, 1121) = = Which integers can be expressed in the form 12m + 20n, where m and n are integers? Solution. We proved in class that if a and b are natural numbers with at least one not zero, then {ax + by : x, y Z} = az + bz = dz, where d = (a, b). This is essentially the same argument that is used to prove Theorem and Theorem in the handout. Thus, since (12, 20) = 4 it follows that {12m + 20n : m, n Z} = (12, 20)Z = 4Z. Thus the integers that can be expressed in the form 12m + 20n are all of the multiples of Give a proof by induction to show that 5 2n 1 is divisible by 24, for all positive integers n. Solution. Let P (n) be the statement 5 2n 1 is divisible by 24. We will prove that P (n) is true for all positive integers n by induction. The statement P (1) is the statement is divisible by 24. Since = 24, this is certainly a true statement. For the induction step, assume that the statement P (k) is true. That is, assume that 5 2k 1 is divisible by 24. Using this, we must show that it follows that P (k + 1) is also Math
3 true. That is, we must show that 5 2(k+1) 1 is divisible by 24, provided that 5 2k 1 is divisible by 24. Assuming that 5 2k 1 = 24m for some m P, we conclude that 5 (2(k+1) 1 = 5 2k+2 1 = 5 2k = 5 2k 5 2 (5 2 24) (since 1 = ) = 5 2 (5 2k 1) + 24 = 5 2 (24m) + 24 (by the induction hypothesis) = 24(5 2 m + 1). Hence 24 divides 5 2(k+1) 1 provided 24 divides 5 2k 1. This is the statement that P (k + 1) is true, provided P (k) is true, and by the principle of induction, we conclude that P (n) is a true statement for all n P. 5. Determine if each of the following statements is true or false. (a) mod 9 Solution. True since = 3 9. (b) 29 1 mod 7 Solution. False since 29 1 = 30 = ( 5) 7+5, so 29 1 is not divisible by 7. (c) 29 6 mod 7 Solution. True since 29 6 = 35 = ( 5) 7. (d) mod 11 Solution. True since 132 = Find all the solutions (when there are any) of the following linear congruences: (a) 8x 6 mod 14 Solution. gcd(8, 14) = 2 so there are 2 distinct solutions modulo 14. Divide the congruence by 2 to get 4x 3 mod 7. Since [4] 1 7 = [2] 7, it follows that x 6 mod 7. Hence, x 6 mod 14 or x 13 mod 14. (b) 66x 100 mod 121 Solution. Since gcd(66, 121) = 11 and , there are no solutions to this congruence. (c) 21x 14 mod 91 Math
4 Solution. gcd(21, 91) = 7 and 7 14 so there are 91/7 = 13 distinct solutions mod 91. Dividing by 7 gives an equivalent congruence 3x 2 mod 13, which has the solution x 5 mod 13. Thus, the solutions mod 91 are x r mod 91, where r {5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89}. (d) 21x 14 mod 89 Solution. Since gcd(21, 89) = 1 there is a unique solution mod 89. Applying the Euclidean algorithm gives = 1, so that the inverse of 21 modulo 89 is 17. Multiplying the congruence by 17 gives x 17 21x mod Find the inverse of 13 modulo 35 and use it to solve the equation [13] 35 x = [9] 35 in Z 35. Solution. From the Euclidean algorithm, = 1. Hence, [13] 1 35 = [ 8] 35 = [27] 35. Multiplying the equation [13] 35 x = [9] 35 by [13] 1 35 = [ 8] 35 gives x = [ 8] 35 [9] 35 = [ 72] 35 = [ ] 35 = [33] Find the multiplicative inverses of the given elements (if possible). (a) [91] 2565 in Z 2565 Solution. Use the Euclidean algorithm to write = 1. Hence, [91] = [451] (b) [423] 6087 in Z 6087 Solution. Use the Euclidean algorithm to check that (423, 6087) = 3 = 1 so there is not multiplicative inverse of 423 modulo Solve the following system of linear congruences: x 4 (mod 24) x 7 (mod 11) Solution. By inspection or using the Euclidean algorithm, find the equation ( 5) = 1. Then the solutions to the simultaneous congruence are given by x 4 (11 11) + 7(( 5)24) = 356 (mod 264). Math
5 10. In Z 18 find all units and all zero divisors. Solution. The units consist of all [a] 18 such that (a, 18) = 1, while the zero divisors consist of all [a] 18 such that (a, 18) > 1. Thus the units are and the zero divisors are 11. Find {[1] 18, [5] 18, [7] 18, [11] 18, [13] 18, [17] 18 } {[0] 18, [2] 18, [3] 18, [4] 18, [6] 18, [8] 18, [9] 18, [10] 18, [12] 18, [14] 18, [15] 18, [16] 18 }. (a) 6 76 mod 13, Solution mod 13 by Euler s theorem, so write 76 = to get 6 76 = = (6 12 ) (6 2 ) mod 13. (b) mod 11, Solution. By Euler, mod 11. Since 1001 = it follows that = 7 mod 11. (c) 2 25 mod 21, Solution. By Euler, φ(21) = 8 so mod 21. Since 25 = , we have = 2 mod 21. (d) 7 66 mod 120. Solution. Since φ(120) = 32, Euler gives mod 120. Since 66 = it follows that = 49 mod Calculate φ(32), φ(33), φ(120), and φ(384). (φ(n) is the Euler φ function evaluated at n.) Solution. 32 = 2 5 so 4φ(32) = = 16; 33 is prime so φ(33) = ; φ(120) = φ(2 3 )φ(3)φ(5) = ( ) 2 4 = 32; 384 = so φ(284) = φ(2 7 )φ(3) = ( )(3 1) = 128. Math
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