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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 Summer '10
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 Graph Theory, Addition, Integers

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