hw3sol - WEEK 3 HOMEWORK SOLUTIONS 3.4-9(a q 2 r 5(e q 0 r...

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WEEK 3 HOMEWORK SOLUTIONS 3.4-9 : (a) q 2, r 5 (b) q -11, r 10 (c) q 34, r 7 (d) q 77, r 0 (e) q 0, r 0 (f) q 0, r 3 (g) q -1, r 2 (h) q 4, r 0 3.4-11 : If a mod m = b mod m , then a = qm + r and b = q prime m + r for integers q, q prime , r by applying the division algorithm. Thus a - b = ( q - q prime ) m , a multiple of m . Hence a b (mod m ). 3.4-12 : If a b (mod m ), then we have a - b = dm , for some integer d . Apply the division algorithm to b , writing b = qm + r , with 0 r < m . Thus b mod m = r . Then we have a = dm + b = ( d + q ) m + r , and since 0 r < m , we see that a mod m = r as well, completing the proof. 3.4-17 : (a) 1 (b) 2 (c) 3 (d) 9 3.4-21 : By assumption we can write a - b = dm , for some integer d . In addition, since n divides m , we can write m = kn , for some integer k . Then a - b = dkn , showing that a - b is a multiple of n and hence that a b (mod n ). 3.4-29 : 2,6,7,10,8,2,6,7,10,8,2,6,7,10,8,... 3.4-31 : (a) GR QRW SDVV JR (b) QB ABG CNFF TB (c) QX UXM AHJJ ZX 3.4-32 : (a) BLUE JEANS (b) TEST TODAY (c) EAT DIM SUM 3.5-2 : (a) Yes (b) No, 27 = 3 3 (c) No, 93 = 3 · 31 (d) Yes (e) Yes (f) Yes 3.5-5 : We have 10! = 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 = 2 · 3 · 2 · 2 · 5 · 2 · 3 · 7 · 2 ·
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