Discrete Mathematics with Graph Theory (3rd Edition) 104

# Discrete Mathematics with Graph Theory (3rd Edition) 104 -...

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Unformatted text preview: 102 Solutions to Exercises Since 1 = 3(27) - 10(8), we get ab == 6(3)(27) - 12(10)(8) = -474 == 174 (mod 216). Then, since 1 = 11(216) - 19(125), we get ab == 39(11)(216) - 174(19)(125) = -320,586 == 3414 (mod 27,000). Since there are two integers less than 50,000 which are congruent to 3414 (mod 27,000), we cannot determine ab exactly. (e) [BB] ab == 3 (mod 4) ab == 12 (mod 25) ab == 104 (mod 343). Since 1 = 25 - 6(4), we get ab == 3(25) - 12(6)(4) = -213 == 87 (mod 100). Since 1 = 7(343)-24(100), wegetab == 87(7)(343)-104(24)(100) = -40,713 == 27,887 (mod 34,300). Since 27,887 is the only integer less than 50,000 which is congruent to 27,887 (mod 34,300), it must be that ab = 27,887. (f) ab == 3 (mod 4) ab == 2 (mod 25) ab == 255 (mod 343). Since 1 = 25 - 6(4), we get ab == 3(25) - 2(6)(4) = 27 (mod 100). Then, since 1 = 7(343) - 24(100), we get ab == 27(7)(343) - 255(24)(100) = -547,173 == 1627 (mod 34,300). Since there are two integers less than 50,000 which are congruent to 1627 (mod 34,300), we cannot...
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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