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MATH 135
Fall 2005
Assignment 6 Solutions
HandIn Problems
Exercise 334:
Solve the congruence
5
x
≡
7
(mod 15)
.
Solution:
Here, gcd(5
,
15) = 5, but 5
6 
7, so there is no solution by Theorem 3.54.
Exercise 344:
Find the inverse of [23] in
Z
41
.
Solution:
We must solve 23
x
≡
1 (mod 41). Apply the Extended Euclidean Algorithm to 41 and 23.
41
y
+ 23
x
=
r
q
i
1
0
41
0
1
23
1

1
18
1

1
2
5
1
4

7
3
1

5
9
2
3
9

16
1
1

23
41
0
1
Hence 23(

16) + 41(9) = 1. Since gcd(23
,
41) = 1 the solution to the linear congruence is unique.
Therefore 23(

16)
≡
1 (mod 41) and so
[23]

1
= [

16] = [25]
∈
Z
41
.
Check:
23
·
25 = 575 = 14
·
41 + 1.
Exercise 351:
Solve the following simultaneous congruences.
x
≡
1
(mod 2)
x
≡
2
(mod 3)
x
≡
3
(mod 7)
Solution:
By the Generalized Chinese Remainder Theorem, this system of linear congruences will have a unique solution modulo
2(3)(7) = 42, since the moduli (2, 3 and 7) are pairwise coprime (that is, gcd(2
,
3) = gcd(3
,
7) = gcd(2
,
7) = 1).
Now,
x
≡
1 (mod 2) and
x
≡
2 (mod 3) are equivalent to
x
≡ 
1 (mod 2) and
x
≡ 
1 (mod 3) respectively.
Hence, by Proposition 3.64, the solution to the ﬁrst two is
x
≡ 
1 (mod 6).
(We have could have also obtained
x
≡ 
1 (mod 6) (or
x
≡
5 (mod 6)) using the usual Chinese Remainder Theorem
technique).
We have reduced the three simultaneous congruences to two simultaneous congruences.
x
≡ 
1
(mod 6)
x
≡
3
(mod 7)
.
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View Full DocumentAn integer
x
satisﬁes the second congruence if and only if
x
= 3 + 7
y
for
y
∈
Z
.
Substitute this into the ﬁrst congruence.
3 + 7
y
≡ 
1
(mod 6)
7
y
≡ 
4
(mod 6)
y
≡ 
4
(mod 6)
y
≡
2
(mod 6)
This is equivalent to
y
= 2 + 6
z
for
z
∈
Z
.
Since
x
= 3 + 7
y
, then
x
= 3 + 7(2 + 6
z
) = 17 + 42
z
.
The solution to the original system of congruences is therefore
x
≡
17 (mod 42).
Check:
17 + 42
z
≡
1 (mod 2), 17 + 42
z
≡
2 (mod 3), 17 + 42
z
≡
3 (mod 7).
Exercise 354:
Solve the following simultaneous congruences.
161
x
≡
49
(mod 200)
74
x
≡
1
(mod 53)
Solution:
The ﬁrst congruence is equivalent to the Diophantine Equation 161
x
+ 200
y
= 49 for some
y
∈
Z
.
Apply the Extended Euclidean Algorithm to 200 and 161:
200
y
+ 161
x
=
r
q
i
1
0
200
0
1
161
1

1
39
1

4
5
5
4
29

36
4
7

33
41
1
1
200

161
0
7
The gcd(200
,
161) = 1, so the congruence has a solution. From the next to last row we see that 161(41)+200(

33) = 1.
Multiplying by 49 gives
161(2009) + 200(

1617) = 49
.
Therefore
x
= 2009 and
y
=

1617 is a particular solution, and by Theorem 3.54, the complete solution to the
congruence is
x
≡
2009 (mod 200) (in other words,
x
≡
9 (mod 200)).
An integer
x
satisﬁes the ﬁrst congruence if and only if
x
= 9 + 200
y
for some
y
∈
Z
.
Substitute this into the second congruence and determine if there is a solution.
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 Spring '08
 Vanderburgh

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