Practice problems for finding sums and products using modular arithmetic.

Problem Analysis

The task involves solving sums and products using modular arithmetic. Modular arithmetic deals with the remainders of division operations. For a given modulus \( n \), two numbers \( a \) and \( b \) are congruent modulo \( n \) if their difference \( a - b \) is divisible by \( n \). This is denoted as:
\[
a \equiv b \pmod{n}
\]

#### Key Steps:
1. Sum in Modular Arithmetic: Compute the sum of the numbers, then find the remainder when the sum is divided by the modulus.
2. Product in Modular Arithmetic: Compute the product of the numbers, then find the remainder when the product is divided by the modulus.
3. Negative Numbers: If a number is negative, add the modulus to it until it becomes non-negative before performing the operation.

Solution

#### Part 2: Sums Using Modular Arithmetic

a. \( 4 + 5 \mod 3 \)
\[
4 + 5 = 9
\]
\[
9 \div 3 = 3 \text{ remainder } 0
\]
\[
4 + 5 \equiv 0 \pmod{3}
\]

b. \( -14 + 3 \mod 8 \)
\[
-14 + 3 = -11
\]
Since \(-11\) is negative, add the modulus \(8\) repeatedly until it becomes non-negative:
\[
-11 + 8 = -3
\]
\[
-3 + 8 = 5
\]
\[
-14 + 3 \equiv 5 \pmod{8}
\]

c. \( 3 + 8 \mod 4 \)
\[
3 + 8 = 11
\]
\[
11 \div 4 = 2 \text{ remainder } 3
\]
\[
3 + 8 \equiv 3 \pmod{4}
\]

d. \( -52 + 29 \mod 9 \)
\[
-52 + 29 = -23
\]
Since \(-23\) is negative, add the modulus \(9\) repeatedly until it becomes non-negative:
\[
-23 + 9 = -14
\]
\[
-14 + 9 = -5
\]
\[
-5 + 9 = 4
\]
\[
-52 + 29 \equiv 4 \pmod{9}
\]

e. \( 2 + 19 \mod 5 \)
\[
2 + 19 = 21
\]
\[
21 \div 5 = 4 \text{ remainder } 1
\]
\[
2 + 19 \equiv 1 \pmod{5}
\]

f. \( 97 + 5 \mod 10 \)
\[
97 + 5 = 102
\]
\[
102 \div 10 = 10 \text{ remainder } 2
\]
\[
97 + 5 \equiv 2 \pmod{10}
\]

g. \( -3 + 15 \mod 6 \)
\[
-3 + 15 = 12
\]
\[
12 \div 6 = 2 \text{ remainder } 0
\]
\[
-3 + 15 \equiv 0 \pmod{6}
\]

h. \( 83 + (-5) \mod 11 \)
\[
83 + (-5) = 78
\]
\[
78 \div 11 = 7 \text{ remainder } 1
\]
\[
83 + (-5) \equiv 1 \pmod{11}
\]

i. \( -12 + 5 \mod 7 \)
\[
-12 + 5 = -7
\]
Since \(-7\) is negative, add the modulus \(7\) repeatedly until it becomes non-negative:
\[
-7 + 7 = 0
\]
\[
-12 + 5 \equiv 0 \pmod{7}
\]

j. \( -4 + 39 \mod 13 \)
\[
-4 + 39 = 35
\]
\[
35 \div 13 = 2 \text{ remainder } 9
\]
\[
-4 + 39 \equiv 9 \pmod{13}
\]

#### Part 3: Products Using Modular Arithmetic

a. \( 4 \cdot 5 \mod 3 \)
\[
4 \cdot 5 = 20
\]
\[
20 \div 3 = 6 \text{ remainder } 2
\]
\[
4 \cdot 5 \equiv 2 \pmod{3}
\]

b. \( -14 \cdot 3 \mod 8 \)
\[
-14 \cdot 3 = -42
\]
Since \(-42\) is negative, add the modulus \(8\) repeatedly until it becomes non-negative:
\[
-42 + 8 = -34
\]
\[
-34 + 8 = -26
\]
\[
-26 + 8 = -18
\]
\[
-18 + 8 = -10
\]
\[
-10 + 8 = -2
\]
\[
-2 + 8 = 6
\]
\[
-14 \cdot 3 \equiv 6 \pmod{8}
\]

c. \( 3 \cdot 8 \mod 4 \)
\[
3 \cdot 8 = 24
\]
\[
24 \div 4 = 6 \text{ remainder } 0
\]
\[
3 \cdot 8 \equiv 0 \pmod{4}
\]

d. \( -52 \cdot 29 \mod 9 \)
First, simplify \(-52 \mod 9\):
\[
-52 \div 9 = -6 \text{ remainder } 2 \quad (\text{since } -52 + 54 = 2)
\]
So, \(-52 \equiv 2 \pmod{9}\).

Next, simplify \(29 \mod 9\):
\[
29 \div 9 = 3 \text{ remainder } 2
\]
So, \(29 \equiv 2 \pmod{9}\).

Now compute:
\[
2 \cdot 2 = 4
\]
\[
-52 \cdot 29 \equiv 4 \pmod{9}
\]

e. \( 2 \cdot 19 \mod 5 \)
\[
2 \cdot 19 = 38
\]
\[
38 \div 5 = 7 \text{ remainder } 3
\]
\[
2 \cdot 19 \equiv 3 \pmod{5}
\]

f. \( 97 \cdot 5 \mod 10 \)
\[
97 \cdot 5 = 485
\]
\[
485 \div 10 = 48 \text{ remainder } 5
\]
\[
97 \cdot 5 \equiv 5 \pmod{10}
\]

g. \( -3 \cdot 15 \mod 6 \)
\[
-3 \cdot 15 = -45
\]
Since \(-45\) is negative, add the modulus \(6\) repeatedly until it becomes non-negative:
\[
-45 + 6 = -39
\]
\[
-39 + 6 = -33
\]
\[
-33 + 6 = -27
\]
\[
-27 + 6 = -21
\]
\[
-21 + 6 = -15
\]
\[
-15 + 6 = -9
\]
\[
-9 + 6 = -3
\]
\[
-3 + 6 = 3
\]
\[
-3 \cdot 15 \equiv 3 \pmod{6}
\]

h. \( 83 \cdot (-5) \mod 11 \)
First, simplify \(83 \mod 11\):
\[
83 \div 11 = 7 \text{ remainder } 6
\]
So, \(83 \equiv 6 \pmod{11}\).

Next, simplify \(-5 \mod 11\):
\[
-5 + 11 = 6
\]
So, \(-5 \equiv 6 \pmod{11}\).

Now compute:
\[
6 \cdot 6 = 36
\]
\[
36 \div 11 = 3 \text{ remainder } 3
\]
\[
83 \cdot (-5) \equiv 3 \pmod{11}
\]

i. \( -12 \cdot 5 \mod 7 \)
\[
-12 \cdot 5 = -60
\]
Since \(-60\) is negative, add the modulus \(7\) repeatedly until it becomes non-negative:
\[
-60 + 7 = -53
\]
\[
-53 + 7 = -46
\]
\[
-46 + 7 = -39
\]
\[
-39 + 7 = -32
\]
\[
-32 + 7 = -25
\]
\[
-25 + 7 = -18
\]
\[
-18 + 7 = -11
\]
\[
-11 + 7 = -4
\]
\[
-4 + 7 = 3
\]
\[
-12 \cdot 5 \equiv 3 \pmod{7}
\]

j. \( -4 \cdot 39 \mod 13 \)
First, simplify \(-4 \mod 13\):
\[
-4 + 13 = 9
\]
So, \(-4 \equiv 9 \pmod{13}\).

Next, simplify \(39 \mod 13\):
\[
39 \div 13 = 3 \text{ remainder } 0
\]
So, \(39 \equiv 0 \pmod{13}\).

Now compute:
\[
9 \cdot 0 = 0
\]
\[
-4 \cdot 39 \equiv 0 \pmod{13}
\]

Final Answers

#### Part 2: Sums
\[
\boxed{
\begin{aligned}
&\text{a. } 0, &\text{b. } 5, &\text{c. } 3, &\text{d. } 4, &\text{e. } 1, \\
&\text{f. } 2, &\text{g. } 0, &\text{h. } 1, &\text{i. } 0, &\text{j. } 9
\end{aligned}
}
\]

#### Part 3: Products
\[
\boxed{
\begin{aligned}
&\text{a. } 2, &\text{b. } 6, &\text{c. } 0, &\text{d. } 4, &\text{e. } 3, \\
&\text{f. } 5, &\text{g. } 3, &\text{h. } 3, &\text{i. } 3, &\text{j. } 0
\end{aligned}
}
\]