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Spring 2010
MATH 4320: Solutions to Prelim 1
Instructor: Yuri Berest
Problem 1.
a
. Solving the congruence 72
x
≡
36 (mod 376) is equivalent to solving
the equation 72
x
+ 376
y
= 36 . Now, using Euclid’s algorithm, we compute
(72
,
376) = 8 . Since 8 does not divide 36 , the equation 72
x
+ 376
y
= 36
(and hence the congruence) has no solutions in integers.
b
. The problem is to ﬁnd a common solution to the system of three
congruences:
x
≡
1 (mod 9)
,
(1)
x
≡
3 (mod 7)
,
(2)
x
≡
4 (mod 5)
.
(3)
To do this we use the Chinese Remainder Theorem as follows. First, we solve
the ﬁrst two congruences: it follows from (1) and (2) that
x
= 1+9
k
= 3+7
m
for some
m, k
∈
Z
. This gives 9
k

7
m
= 2 ; whence
m
=
k
= 1 and
x
= 10.
Thus, by the Chinese Remainder Theorem, a common solution of (1) and (2)
is given by
x
≡
10 (mod 63)
.
(4)
Now, we ﬁnd a common solution to the system of congruences (3) and
(4).
We have
x
= 4 + 5
s
= 10 + 63
t
, so that 5
s

63
t
= 6 .
Since
5
·
(

25) + 63
·
2 = 1 , we see that
s
=

150,
t
=

12 and
x
=

746. Thus,
a common solution to the system (1)(3) is
x
≡ 
746 (mod 315), or equiva
lently
x
∈ {
746 + 315
k
:
k
∈
Z
}
. The smallest positive integer in this last
set corresponds to
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 Spring '08
 CRISSINGER
 Congruence

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