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Discrete Mathematics with Graph Theory (3rd Edition) 99

# Discrete Mathematics with Graph Theory (3rd Edition) 99 -...

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Section 4.5 97 (c) We must have 1(9) + 2(1) + 3(2) + 4(3) + 5(A) + 6(4) + 7(5) + 8(5) + 9(1) + 10(10) == 0 (mod 11); hence, 5A + 237 == 0 (mod 11), 5A + 6 == 0 (mod 11), 5A == 5 (mod 11) and A = 1. (d) [BB] We must have 1(2) + 2(7) + 3(2) + 4(9) + 5(1) + 6(8) + 7(8) + 8(A) + 9(6) + 10(2) == 0 (mod 11); hence, 8A + 241 == 0 (mod 11), 8A + 10 == 0 (mod 11), 8A == 1 (mod 11) and A = 7. (e) We must have 1(8) + 2(9) + 3(A) + 4(5) + 5(3) + 6(9) + 7(8) + 8(2) + 9(8) + 10(4) == 0 (mod 11); hence, 3A + 299 == 0 (mod 11), 3A + 2 == 0 (mod 11), 3A == -2 == 9 (mod 11) and A = 3. 3. The first digit A would have to satisfy I(A) + 2(3) + 3(1) + 4(5) + 5(2) + 6(6) + 7(6) + 8(7) + 9(8) + 10(2) == 0 (mod 11); hence, A + 265 == 0 (mod 11), A + 1 == 0 (mod 11), and there is no A in the range 0-9 with this property, so no such ISBN number exists. 4. [BB] at + 2a2 + ... + 9ag + 10alO == 0 (mod 11) ~ at + 2a2 + ... + 9ag == -10alO ~ at + 2a2 + ... + 9ag == alO since -1OalO == alO (mod 11). 5. (a) Suppose the ISBN numberata2 ... alO is correct. Thus a = at +2a2+" ·+iai+·· ·+1OalO == 0 (mod 11). If digit ai is miscopied as x, then the test computes b = at +2a2+ . +ix+· +1OalO. The difference a - b = iai - ix = i(ai - x), so b == i(x - ai) (mod 11). Now 11 I i(ai - x) implies 11 I i or 11 I (ai - x) (because 11 is prime). Since 1 ~ i ~ 10, the first possibility cannot occur. Since the numbers ai and x
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