Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Discrete Mathematics
Modular Arithmetic
For any two integers a, n with n > 0 there is only one pair (q, r)
such that
• a = qn + r
• 0≤r<n
By the definition of mod we have
a mod n = r
Note that 0 ≤ (a mod n) < n, in other words
(a mod n) ∈ {0, 1, 2, . . . , n − 1} = Zn .
Recall: For all m ∈ Z and all numbers k such that 0 ≤ k ≤ n − 1
we have
= |{z}
m · |{z}
n + |{z}
0 )
0
= (m · n) mod n
(because
m
| {z· n}
1
= (1 + m · n) mod n
(because
1+m·n
| {z }
= |{z}
m · |{z}
n + |{z}
1 )
2
= (2 + m · n) mod n
(because
2+m·n
| {z }
= |{z}
m · |{z}
n + |{z}
2 )
k
..
.
= (k + m · n) mod n
k+m·n
| {z }
= |{z}
m · |{z}
n + |{z}
k )
a
q
a
n−1+m·n
|
{z
}
a
n
q
a
..
.
n − 1 = (n − 1 + m · n) mod n (because
r
q
a
(because
n
r
n
q
r
n
r
= |{z}
m · |{z}
n + n − 1)
| {z }
q
n
Definition (Modular addition, multiplication) Let n be a positive integer. Let a, b ∈ Zn = {0, 1, . . . , n − 1}. We define
a ⊕ b = (a + b)mod n and a ⊗ b = (a · b)mod n
r
37.1 The following numbers are from Z10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
a. 3 ⊕ 3 = (3 + 3) mod 10 = 6 mod 10 = 6,
because 6 = 0 · 10 + |{z}
6
r
b. 6 ⊕ 6 = (6 + 6) mod 10 = 12 mod 10 = 2,
because 12 = 1 · 10 + |{z}
2
r
c. 7 ⊕ 3 = (7 + 3) mod 10 = 10 mod 10 = 0,
because 10 = 1 · 10 + |{z}
0
r
d. 9 ⊕ 8 = (9 + 8) mod 10 = 17 mod 10 = 7,
because 17 = 1 · 10 + |{z}
7
r
e. 12 ⊗ 4 is undefined, because 12 6∈ Z10 .
f. 3 ⊗ 3 = (3 · 3) mod 10 = 9 mod 10 = 9,
because 9 = 0 · 10 + |{z}
9
r
g. 4 ⊗ 4 = (4 · 4) mod 10 = 16 mod 10 = 6,
because 16 = 1 · 10 + |{z}
6
r
h. 7 ⊗ 3 = (7 · 3) mod 10 = 21 mod 10 = 1,
because 21 = 2 · 10 + |{z}
1
r
i. 5 ⊗ 2 = (5 · 2) mod 10 = 10 mod 10 = 0,
because 10 = 1 · 10 + 0
j. 6 ⊗ 6 = (6 · 6) mod 10 = 36 mod 10 = 6,
because 36 = 3 · 10 + 6
k. 4 ⊗ 6 = (4 · 6) mod 10 = 24 mod 10 = 4,
because 24 = 2 · 10 + 4
l. 4 ⊗ 1 = (4 · 1) mod 10 = 4 mod 10 = 4,
because 4 = 0 · 10 + 4
m. 12 ⊗ 5 is undefined, because 12 6∈ Z10 .
37.2. Solve the following equations for x in the Zn
a. 3 ⊗ x = 4 in Z11 . We want to find an element x ∈ Z11 , such
that 3 ⊗ x = (3 · x) mod 11 = 4.
• Since 4 = (4 + m · 11) mod 11 for all m ∈ Z, we have that
(3 · x) mod 11 = 4 if 3 · x = 4 + m · 11 for some m ∈ Z.
• Since x ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = Z11 , we have
3 · x ∈ {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30}.
{z
}
|
A
2
There is the only one number in the set A of the form 4+m·11,
namely 15 = 4 + 1 · 11, therefore 3 · x = 15 implies x = 5.
c. 3 ⊗ x ⊕ 8 = 1 in Z10 . We want to find an element x ∈ Z10 , such
that 3 ⊗ x ⊕ 8 = (3 · x + 8) mod 10 = 1.
• Since 1 = (1 + m · 10) mod 10 for all m ∈ Z, we have that
(3 · x + 8) mod 10 = 1 if 3 · x + 8 = 1 + m · 10 for some
m ∈ Z.
• Since x ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = Z10 , we have
3 · x + 8 ∈ {8, 11, 14, 17, 20, 23, 26, 29, 32, 35}.
|
{z
}
A
There is the only one number in the set A of the form 1+m·10,
namely 11 = 1 + 1 · 10, therefore 3 · x + 8 = 11 implies x = 1.
d. 342 ⊗ x ⊕ 448 = 73 in Z1003 . We want to find an element x ∈
Z1003 , such that 342 ⊗ x ⊕ 448 = (342 · x + 448) mod 1003 = 73.
• Since 73 = (73 + m · 1003) mod 1003 for all m ∈ Z, we
have that (342 · x + 448) mod 1003 = 73 if
342 · x + 448 = 73 + m · 1003 for some m ∈ Z.
• Since x ∈ {0, 1, 2, . . . , 1002} = Z1003 , we have
342 · x + 448 ∈ {342 · n + 448 : n ∈ N, 0 ≤ x ≤ 1002}.
|
{z
}
A
There is the only one number in the set A of the form 73 + m ·
1003, namely 73 + 177 · 1003 = 342 · |{z}
518 +448. Thus x = 518.
x
37.3. Solve the following equations for x in the Zn specified.
a. 2 ⊗ x = 4 in Z10
3
2⊗x=4
=⇒ (2 · x) mod 10 = 4
=⇒ 2 · x = 4 + m · 10
(because in Z10 we have 4 = (4 + m · 10) mod 10 for m ∈ Z )
=⇒ 2 · x = 4 + m · 10 ∈ {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}
(because x ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9})
=⇒ m · 10 ∈ {|{z}
−4 , |{z}
−2 , |{z}
0 , |{z}
2 , |{z}
4 , |{z}
6 , |{z}
8 , |{z}
10 , |{z}
12 , |{z}
14 }
0−4
2−4
4−4
6−4
8−4
10−4 12−4 14−4 16−4 18−4
=⇒ m = 0 or m = 1
2 · x = 4 + m · 10 =⇒ 2 · x = 4 + 0 · 10 or 2 · x = 4 + 1 · 10
=⇒ x = 2 or x = 7
b. 2 ⊗ x = 3 in Z10 if (2 · x) mod 10 = 3 (i.e. 2 · x = 3 + m · 10 for
some m ∈ Z)
2 ⊗ x = 3 =⇒ (2 · x) mod 10 = 3
=⇒ 2 · x = 3 + m · 10
(because in Z10 we have 3 = (3 + m · 10) mod 10 for m ∈ Z )
=⇒ 2 · x = 3 + m · 10 ∈ {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}
(because x ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9})
=⇒ m · 10 ∈ {|{z}
−3 , |{z}
−1 , |{z}
1 , |{z}
3 , |{z}
5 , |{z}
7 , |{z}
9 , |{z}
11 , |{z}
13 , |{z}
15 }
0−3
=⇒
2−3
4−3
6−3
8−3
10−3 12−3 14−3 16−3 18−3
no solutions, because no numbers in this set is a multiple of 10.
c. 9 ⊗ x = 4 in Z12 (no solutions).
d. 9 ⊗ x = 6 in Z12 is true for x = 2, or x = 6, or x = 10.
37.4. Solve the following equations for x in the Zn specified.
• Consider the set {x⊗x : x ∈ Z13 } = {(x2 ) mod 13 : x ∈ Z13 }.
We have
{|{z}
1 , |{z}
4 , |{z}
9 , |{z}
3 , |{z}
12 , |{z}
10 , |{z}
10 , |{z}
12 , |{z}
3 , |{z}
9 , |{z}
4 , |{z}
1 }
1⊗1
2⊗2
3⊗3
4⊗4
5⊗5
6⊗6
7⊗7
8⊗8
9⊗9
10⊗10 11⊗11 12⊗12
a. x ⊗ x = 1 in Z13 is true for x = 1, and x = 12.
b. x ⊗ x = 11 in Z13 no solutions.
c. x ⊗ x = 12 in Z13 is true for x = 5, and x = 8.
• Consider the set {x⊗x : x ∈ Z15 } = {(x2 ) mod 15 : x ∈ Z15 }.
We have
{|{z}
1 , |{z}
4 , |{z}
9 , |{z}
1 , |{z}
10 , |{z}
6 , |{z}
4 , |{z}
4 , |{z}
6 , |{z}
10 , |{z}
1 , |{z}
9 , |{z}
4 , |{z}
1 }
1⊗1
2⊗2
3⊗3
4⊗4
5⊗5
6⊗6
4
7⊗7
8⊗8
9⊗9
10⊗10 11⊗11 12⊗12 13⊗13 14⊗14
d. x ⊗ x = 4 in Z15 is true for x = 2, x = 7, and x = 8 .
e. x ⊗ x = 10 in Z15 is true for x = 5, and x = 10 .
f. x ⊗ x = 14 in Z15 , no solutions.
5