Qin Jiushao 6
All letters will denote integers
Least Common Multiples. Let l= m ∨ n , the least
common multiple of m and n . Then for any a and b ,
a ≡ b mod m and a ≡ b mod n if and only if
a ≡ b mod l .
Chinese Remainder Theorem I. Let m and n be
positive integers, let d= m ∧ n and l= m ∨ n . Let a and
b be arbitrary. Then the congruences x ≡ a mod m and
x ≡ b mod n have a common solution if and only if
a ≡ b mod d . If this is the case, then the two
congruences are equivalent to a unique congruence of the
form x ≡ c mod l .
Equivalencies. Consider the congruences x ≡ a mod m
and x ≡ b mod n . Then for any integers i and j , x ≡ a
mod i and x ≡ b mod j is an equivalent system of
congruences when and only when m ∧ n =i ∧ j and
m ∨ n =i ∨ j .
Chinese Remainder Theorem II. Consider the system of
congruences
x ≡ a1 mod n1 ,
x ≡ a 2 mod n2 ,

x ≡ a k mod nk .
Then the following are equivalent:

Every pair of them has a solution;
ai ≡ a j mod ni ∧ n j for all i, j = 1, , k ;


The system is equivalent to a system of the form:
x ≡ a1 mod m1 ,
x ≡ a 2 mod m2 ,

x ≡ a k mod mk .
1,
re mi | ni for i = 1,..., k , and for i, j = 1, , k , mi ∧ m j =
and m1 ∨ m2 ∨  ∨ mk = n1 ∨ n2 ∨  ∨ nk .
The system has a common solution, that is, there exists
an integer which satisfies all the congruences.
If this is the case, then there exists a unique solution mod
n1 ∨ n2 ∨  ∨ nt .
xy105 ≡ 0 mod 495 is equivalent to xy105 ≡ 0 mod 5,
xy105 ≡ 0 mod 9 and xy105 ≡ 0 mod 11, since
495 = 5 ∨ 9 ∨ 11 . This in turn forces x = 3 and y = 9 .
Consider the following congruences:
x ≡ 11 mod 60
x ≡ 43 mod 52
Note 43 ≡ 11 mod 4, and =
4 52 ∧ 60 . Also
780
= 52 ∨ 60 . Starting with the larger clock,
x = 11 + 60t for some t . Substituting in the second
congruence, 11 + 60t ≡ 43 mod 52 , which becomes
2t ≡ 8 mod 13 which is easily solved as t ≡ 4 mod
13, and x =
11 + 60(4 + 13k ) =
251 + 780k ≡ 251 mod
780 is the congruence equivalent to the pair.
The following pairs are equivalent:
x ≡ 11 mod 60 & x ≡ 43 mod 52
x ≡ 11 mod 30 & x ≡ 43 mod 52
x ≡ 11 mod 15 & x ≡ 43 mod 52
x ≡ 11 mod 60 & x ≡ 43 mod 26
x ≡ 11 mod 60 & x ≡ 43 mod 13
Consider now the congruences
x ≡ 13 mod 24,
x ≡ 41 mod 50
x ≡ 31 mod 90
This is equivalent to:
x ≡ 13 mod 24,
x ≡ 41 mod 25
x ≡ 31 mod 90
Which in turn is equivalent to:
x ≡ 13 mod 24,
x ≡ 41 mod 25
x ≡ 31 mod 18
And again to
x ≡ 13 mod 8,
x ≡ 41 mod 25
x ≡ 31 mod 9
This in turn reduces to x ≡ 5 mod 8,
x ≡ 16 mod 25
x ≡ 4 mod 9
Solving: x = 16 + 25t , so 16 + 25t ≡ 4 mod 9, so
t = 6 , so x ≡ 166 mod 225 is equivalent to the last
two congruences. Now x = 166 + 225t , so mod 8,
6 + t ≡ 5 , so we can let t = −1 , and so our original
system reduces to the single congruence:
x ≡ −59 mod 1800.