Section 4.5
(d) [BB] We wish to solve the system of congruences
x
0
(mod 4)
x

2
(mod 3)
x
10
(mod 25).
101
Solving the first two congruences, we obtain
x
==
8 (mod 12). Since 1(25) 
2(12) = 1,
x =
8(25) 
10(2)(12)
=
40
==
260 (mod 300).
(e) We wish to solve the system of congruences
x
3
(mod 4)
x

2
(mod 3)
x
24
Solving the first two congruences, we obtain
x
==
11 (mod 12). Since 1(25) 
2(12) = 1,
x
= 11(25)  24(2)(12) = 301 == 299 (mod 300).
20. [BB] Call the number
x.
Then
x
==
4 (mod 12) and
x
==
15 (mod 25). By the Chinese Remainder
Theorem, this system has a solution which is unique mod 12(25) = 300. There is exactly one such
number in the interval 200
<
x
<
500 so the given congruences determine the number uniquely. Since
1(25) 
2(12) = 1, the method suggested in the text gives
x
==
4(1)(25) 
15(2)(12) = 260
==
40
(mod 300). The number was 340.
21. (a) [BB]
ab
==
3(3)
=
1 (mod 4)
ab
==
0(8) = 0 (mod 9)
ab
==
4(18) = 22 (mod 25).
Since 1 = 9 
2(4), we get
ab
==
1(9) 
0(2)(4) = 9 (mod 36) as the solution to the first
two congruences. Then, since 1 = 9(36)
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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.
 Summer '10
 any
 Graph Theory, Congruence

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