McCombs Math 381
The Chinese Remainder Theorem
1
Basic Idea:
We want to devise a strategy for finding simultaneous solutions to a
system of linear congruences.
Chinese Remainder Theorem:
Given pairwise relatively prime positive integers
1
2
3
,
,
...,
k
n
n
n
n
,
and integers
1
2
3
,
,
...,
k
a
a
a
a
. The system of congruences
(
)
(
)
(
)
1
1
2
2
mod
mod
.
.
.
mod
k
k
x
a
n
x
a
n
x
a
n
!
!
!
has a unique simultaneous solution
(
)
mod
x
n
, where
1
2
3
k
n
n
n
n
n
=
!
!
!!!
.
Moreover,
(
)
mod
x
n
is the smallest positive integer that solves the system.
Examples:
1.
Solve each system.
(a)
(
)
(
)
23 mod100
31 mod 49
x
x
!
!
(
)
23 mod100
100
23
x
x
k
!
"
=
+
(
)
100
23
31 mod 49
k
+
!
, now reduce
mod 49
to get
2
k
+
23
!
31 mod49
(
)
"
2
k
!
8 mod49
(
)
Since
(
)
gcd 2,49
1
=
, we can divide both sides by
2
.
(
)
4 mod 49
49
4
k
k
m
!
"
=
+
x
=
100
k
+
23
=
100 49
m
+
4
(
)
+
23
=
4900
+
423
Thus, the smallest positive integer solution is
423
x
=
.

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