Explanation 21 , 758 , 623 + 17 , 123 , 055 , n = 6 We have, 21 , 758 , 623 = 1 ( mod 6 ) and 17 , 123 , 055 = 3 ( mod 6 ) . So, 21 , 758 , 623 + 17 , 123 , 055 = 1 ( mod 6 ) + 3 ( mod 6 ) = 4 ( mod 6 ) ( 21 , 758 , 623 ) ( 17 , 123 , 055 ) , n = 6 We have, 21 , 758 , 623 = 1 ( mod 6 ) and 17 , 123 , 055 = 3 ( mod 6 ) . So, ( 21 , 758 , 623 ) ( 17 , 123 , 055 ) = 1 ( mod 6 ) .3 ( mod 6 ) = 3 ( mod 6 ) ( 17 , 123 ) 50 , n = 6 We have, 17 , 123 = 5 ( mod 6 ) . So, ( 17 , 123 ) 50 = 5 50 ( mod 6 ) . As, 5 2 = 25 = 1 ( mod 6 ) This implies: 5 50 = ( 5 2 ) 25 = 1 25 ( mod 6 ) = 1 ( mod 6 ) 10 4 , 10 8 , 10 12 , 10 20 , 10 24 , n = 7 As, 10 = 3 ( mod 7 ) We have 10 4 = 3 4 ( mod 7 ) = 81 ( mod 7 ) = 4 ( mod 7 ) Now, as 10 4 = 4 ( mod 7 ) , ( 10 4 ) 2 = 4 2 ( mod 7 ) 10 8 = 16 ( mod 7 ) = 2 ( mod 7 ) Also, as 10 4 = 4 ( mod 7 ) ( 10 4 ) 3 = 4 3 ( mod 7 ) 10 12 = 64 ( mod 7 ) = 1 ( mod 7 ) Now, 10 20 = 10 8 10 12 = 2 ( mod 7 ) .1 ( mod 7 ) = 2 ( mod 7 ) Also, 10 24 = ( 10 12 ) 2 = 1 2 ( mod 7 ) = 1 ( mod 7 ) 2 , 2 2 , 2 3 ,

3rd Edition

ISBN: 9780131679955

Author: Edgar G. Goodaire

Publisher: Prentice Hall

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Chapter 4.4, Problem 6E

To determine

Each indicated calculations, giving the answer modn.

21,758,623+17,123,055,n=6

(21,758,623)(17,123,055),n=6

(17,123)50,n=6

104,108,1012,1020,1024,n=7

2,22,23,..........,210,n=11

4,42,43,44,45,46,47,48,49,410,n=11

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Chapter 4.4, Problem 5E

Chapter 4.4, Problem 7E

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Chapter 4 Solutions

Discrete Mathematics with Graph Theory

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Each indicated calculations, giving the answer modn. 21,758,623+17,123,055,n=6 (21,758,623)(17,123,055),n=6 (17,123)50,n=6 104,108,1012,1020,1024,n=7 2,22,23,..........,210,n=11 4,42,43,44,45,46,47,48,49,410,n=11

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Chapter 4 Solutions

Each indicated calculations, giving the answer modn.

21,758,623+17,123,055,n=6

(21,758,623)(17,123,055),n=6

(17,123)50,n=6

104,108,1012,1020,1024,n=7

2,22,23,..........,210,n=11

4,42,43,44,45,46,47,48,49,410,n=11