Discrete Mathematics with Graph Theory (3rd Edition) 83

# Discrete Mathematics with Graph Theory (3rd Edition) 83 -...

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Section 4.2 81 (c) Repeat the preceding arguments, but instead, writing the second term as 9(al + 11a2 + ... + ~an). ntimes 33. (a) Let 91 = gcd(a, b) and 9 = gcd(9l, c). We have to show that 9 = gcd(a, b, c). Since 9 I 91 and 91 I a we see that 9 I a. Similarly, 9 I b. Since also 9 I c, we see that 9 is a common divisor of a, b, c. Next suppose that d is any common divisor of c. Since in particular, d is a common divisor of a and b, d I 91 (by Corollary 4.2.11). Since also d I c, we see that d is a common divisor of 91 and c, so d ::; 9. This proves that 9 is the largest of the common divisors of a, b and c. (b) There exist integers m and n such that gcd(a, b) = ma + nb and integers r B such that gcd(gcd(a, b), c) = r gcd(a, b)+BC. Since gcd(a, b, c) = gcd(gcd(a, b), c), we get gcd(a, b, c) = r(ma + nb) + BC = (rm)a + (rn)b + which expresses gcd(a, b, c) as an integral linear com- bination of a, b and c. (c) gcd(231, 165, 105) = gcd(gcd(231, 165), 105) = gcd(33, 105) = 3. Since 33 = -2(231) + 3(165) and 3 = -5(105) + 16(33), we have 3 = -5(105) + 16[-2(231) + 3(165)) - 32(231) + 48(165). (d) gcd(8580,6279) = 39 = -30(8580)+41(6279) andgcd(2873, 39) = 13 = -1(2873)+74(39). So 13 = gcd(6279, 8580, 2873) ~ -1(2873) + 74[-30(8580) + 41(6279)) = -1(2873) - 2220(8580) + 3034(6279). (e) gcd(5577, 18837, 25740) = gcd(gcd(25740, 18837), 5577)
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