Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Modular Arithmetic
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Modular Arithmetic
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Subject: General Mathematics
Class: Senior Secondary 1
Term: 1st Term
Week: 1
Theme: Number and Numeration
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Recall and carry out the basic operations of addition, subtraction,multiplication. Carryout the operations in modular Arithmetic Apply modulararithmetic in daily life.
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. It focuses on the remainder when one integer is divided by another.
A. Congruence Modulo n: Two integers, a and b, are said to be congruent modulo n if their difference (a - b) is an integer multiple of n. This is equivalent to saying that a and b have the same remainder when divided by n. Mathematically, this is written as: `a ≡ b (mod n)` where `n` is the modulus (a positive integer). The modulus `n` must be greater than 1. `a mod n` represents the remainder when `a` is divided by `n`. This remainder is always an integer from `0` to `n-1`.
Example: `17 ≡ 5 (mod 12)` because `17 - 5 = 12`, and `12` is a multiple of `12`. Also, `17 ÷ 12` gives a remainder of `5`. `25 ≡ 1 (mod 4)` because `25 - 1 = 24`, and `24` is a multiple of `4`. Also, `25 ÷ 4` gives a remainder of `1`. * `40 ≡ 0 (mod 8)` because `40 - 0 = 40`, and `40` is a multiple of `8`. Also, `40 ÷ 8` gives a remainder of `0`.
B. Basic Operations in Modular Arithmetic: The results of operations in modular arithmetic are always the remainders obtained when the standard arithmetic result is divided by the modulus. The final answer must always be less than the modulus and non-negative (i.e., between 0 and `n-1`).
1. Addition in Modular Arithmetic: To add two numbers `a` and `b` modulo `n`: ` (a + b) (mod n) ` First, perform the standard addition `a + b`. Then, divide the sum by `n` and the remainder is the answer.
Worked Example 1 (Nigerian Context: Market Days) A local market in a Nigerian community operates on a 5-day cycle. If today is the 3rd day of the cycle, what day will it be after 4 more days?
Solution: Let the days be numbered 0, 1, 2, 3,
4. Current day = 3 Days to add = 4 Sum = 3 + 4 = 7 Modulus = 5 (since it's a 5-day cycle) `7 (mod 5)`: Divide 7 by 5. `7 = 1 × 5 + 2` The remainder is
2. So, `3 + 4 ≡ 2 (mod 5)`. After 4 more days, it will be the 2nd day of the cycle.
Worked Example 2 (Nigerian Context: Time)
If a meeting starts at 9:00 AM and lasts for 7 hours, what time will the meeting end on a 12-hour clock?
Solution: Starting hour = 9 Duration = 7 hours Sum = 9 + 7 = 16 Modulus = 12 (for a 12-hour clock) `16 (mod 12)`: Divide 16 by 12. `16 = 1 × 12 + 4` The remainder is
4. So, `9 + 7 ≡ 4 (mod 12)`.
The meeting will end at 4:00 PM.
2. Subtraction in Modular Arithmetic: To subtract two numbers `b` from `a` modulo `n`: ` (a - b) (mod n) ` First, perform the standard subtraction `a - b`. If the result is negative, add multiples of `n` until the result is positive and less than `n`. If the result is positive, divide by `n` and the remainder is the answer.
Worked Example 3 (Nigerian Context: 24-hour Clock)
A flight is scheduled to depart at 23:00 hours (11 PM) and arrive 5 hours earlier than planned. What is the new departure time on a 24-hour clock?
Solution: Scheduled departure = 23 Hours earlier = 5 Difference = 23 - 5 = 18 Modulus = 24 (for a 24-hour clock) `18 (mod 24)`: Since 18 is already less than 24 and positive, the remainder is
1
8. So, `23 - 5 ≡ 18 (mod 24)`.
The new departure time is 18:00 hours (6 PM).
Worked Example 4 (Nigerian Context: Negative Result) Calculate `3 - 7 (mod 5)`.
Solution: Difference = 3 - 7 = -4 Modulus = 5 Since -4 is negative, add multiples of 5 until the result is positive and less than 5. `-4 + 5 = 1` So, `3 - 7 = 18 Modulus = 24 (for a 24-hour clock) `18 (mod 24)`: Since 18 is already less than 24 and positive, the remainder is
1
8. So, `23 - 5 ≡ 18 (mod 24)`.
The new departure time is 18:00 hours (6 PM).
Worked Example 4 (Nigerian Context: Negative Result) Calculate `3 - 7 (mod 5)`.
Solution: Difference = 3 - 7 = -4 Modulus = 5 Since -4 is negative, add multiples of 5 until the result is positive and less than 5. `-4 + 5 = 1` So, `3 - 7 ≡ 1 (mod 5)`. (Alternatively: `3 - 7 = -4`. To make it positive, we can add 5 to -4 repeatedly until it falls within [0, 4]. -4 + 5 =
1. So the answer is 1.)
3. Multiplication in Modular Arithmetic: To multiply two numbers `a` and `b` modulo `n`: ` (a × b) (mod n) ` First, perform the standard multiplication `a × b`. Then, divide the product by `n` and the remainder is the answer.
Worked Example 5 (Nigerian Context: Production Cycles) A factory produces a certain product in batches of 6, and it takes 3 hours to complete one batch. If the factory operates on an 8-hour shift, what is the 'remainder' of hours left after producing 5 batches?
Solution: Time per batch = 3 hours Number of batches = 5 Total time = 3 × 5 = 15 hours Modulus = 8 (for an 8-hour shift cycle) `15 (mod 8)`: Divide 15 by 8. `15 = 1 × 8 + 7` The remainder is
7. So, `3 × 5 ≡ 7 (mod 8)`. After producing 5 batches, there will be 7 hours 'left' in the sense of the cycle's position, implying the factory would be 7 hours into a new 8-hour cycle if it continued.
Worked Example 6: Calculate `4 × 6 (mod 7)`.
Solution: Product = 4 × 6 = 24 Modulus = 7 `24 (mod 7)`: Divide 24 by 7. `24 = 3 × 7 + 3` The remainder is
3. So, `4 × 6 ≡ 3 (mod 7)`.
4. Division in Modular Arithmetic (Introduction): Division in modular arithmetic is not as straightforward as the other operations. It typically involves finding a multiplicative inverse. If we want to solve for `x` in `ax ≡ b (mod n)`, we are looking for a number `x` such that when `a` is multiplied by `x`, the result is congruent to `b` modulo `n`. The multiplicative inverse of `a` modulo `n`, denoted `a−1`, exists if and only if `a` and `n` are coprime (i.e., their greatest common divisor, `gcd(a, n) = 1`). If `a−1` exists, then `x ≡ b × a−1 (mod n)`. For SS1, focus on simple cases where the unknown can be found through trial and error or inspection. Worked Example 7 (Simple Division Concept for SS1): Solve for `x`: `3x ≡ 2 (mod 5)`.
Solution: We are looking for a number `x` (between 0 and 4) such that when multiplied by 3, gives a remainder of 2 when divided by
5. Let's test values for `x`: If `x = 0`: `3 × 0 = 0`, `0 (mod 5) = 0`. (Not 2) If `x = 1`: `3 × 1 = 3`, `3 (mod 5) = 3`. (Not 2) If `x = 2`: `3 × 2 = 6`, `6 (mod 5) = 1`. (Not 2) If `x = 3`: `3 × 3 = 9`, `9 (mod 5) = 4`. (Not 2) If `x = 4`: `3 × 4 = 12`, `12 (mod 5) = 2`. (Yes!) So, `x ≡ 4 (mod 5)`. (Teacher's
Note: For SS1, extensive treatment of multiplicative inverses might be too advanced for an introductory lesson. Emphasize that division means finding a missing factor, and simple problems can be solved by testing values or by recognizing patterns in remainders. More complex inverse problems are typically for higher levels.)* Phase 1: Introduction and Recall (10-15 minutes)
Teacher Activity: Begins by asking students about their understanding of remainders from primary school mathematics. Asks questions related to time (e.g., "If it's 10 AM, what time will it be in 5 hours? What if it's 10 PM and 5 hours pass?"). Introduces the concept of "counting in cycles" using a physical clock face or drawing one on the board. Explains that modular arithmetic is simply formalizing this concept of remainders and cycles. Defines "congruence modulo n" and illustrates with simple examples.
Student Activity: Responds to questions on remainders and time calculations. Observes the clock face demonstration and participates in discussions. Takes notes on definitions and examples.
Phase 2: Developing Key Concepts and Operations (30-40 minutes)
Teacher Activity: Systematically explains addition in modular arithmetic using clear steps and Worked Example 1 &
2. Demonstrates how to handle results greater than or equal to the modulus by finding the remainder. Explains subtraction in modular arithmetic, including how to handle negative results by adding the modulus. Uses Worked Example 3 &
4. Explains multiplication in modular arithmetic with Worked Example 5 &
6. Briefly introduces the concept of 'division' in modular arithmetic as finding a missing factor, demonstrating with Worked Example
7. Emphasizes its difference from standard division. Encourages students to ask questions and clarifies misconceptions. Writes all steps clearly on the board.
Student Activity: Actively listens and observes the teacher's explanations and worked examples. Copies notes and worked examples into their notebooks. Participates by suggesting steps or answers during the demonstration. Asks clarifying questions. Attempts simple calculations as guided by the teacher.
Phase 3: Guided Practice (15-20 minutes)
Teacher Activity: Presents 3-5 guided practice questions (see section 4) one by one. Guides students through solving each problem, asking them to contribute steps. Provides immediate feedback and corrects errors.
Student Activity: Works on the guided practice questions individually or in pairs. Presents solutions and explains their reasoning. Actively participates in discussions about the solutions.
Phase 4: Independent Practice and Consolidation (10-15 minutes)
Teacher Activity: Assigns a set of independent practice questions (see section 5) for students to solve. Monitors students' progress, providing individual support as needed. Circulates around the classroom to observe and assess understanding.
Student Activity: Works independently on the assigned practice questions. Seeks clarification from the teacher if encountering difficulties.
Phase 5: Wrap-up and Assignment (5 minutes)
Teacher Activity: Recap the key concepts learned: congruence, and basic operations (addition, subtraction, multiplication, and simple division) in modular arithmetic. Highlights the real-life applications discussed. Assigns homework (additional problems from a textbook or self-created, aligning with the lesson's objectives).
Student Activity: Participates in the recap session. Records homework assignment.
Question 1 (Addition): A trader at a local Nigerian market accepts payments in a specific currency. If she has collected ₦17 from a customer and then receives another ₦9, what is the total amount (in naira) modulo 10? (Assume a simplified scenario where we are interested in the last digit of the total amount).
Solution: Identify the numbers and modulus: Amounts are 17 and
9. Modulus is
1
0. Perform standard addition: 17 + 9 =
2
6. Find the remainder modulo 10: 26 ÷ 10 = 2 with a remainder of
6. Result: `17 + 9 ≡ 6 (mod 10)`.
Commentary: This illustrates how modular arithmetic can simplify large numbers to their "units digit" representation, which is useful in various checks and balances.
Question 2 (Subtraction): A driver in Lagos is planning a trip. He needs 7 hours for the journey. If he estimates he has only 3 hours available on a particular day (using a 24-hour clock concept for daily cycles), how many 'hours short' is he in modulo 24?
Solution: Identify the numbers and modulus: Needed hours = 7, Available hours =
3. Modulus =
2
4. Perform standard subtraction: 3 - 7 = -
4. Adjust for negative result modulo 24: Since -4 is negative, add 24 to it. -4 + 24 =
2
0. Result: `3 - 7 ≡ 20 (mod 24)`.
Commentary: This shows that the driver is 20 hours "short" in the current 24-hour cycle, meaning he would need to accumulate 20 more hours to complete his desired journey if he started from 3 hours. Alternatively, it represents the time on a clock if you go back 4 hours from 3 A
M. Question 3 (Multiplication): A tailor makes traditional Nigerian attire. Each piece of attire requires 4 meters of fabric. If she makes 7 pieces of attire, and fabric is supplied in rolls of 5 meters, how many meters of fabric will be "left over" in terms of the 5-meter cycle (i.e., modulo 5) from the total required?
Solution: Identify the numbers and modulus: Meters per attire = 4, Number of attires =
7. Modulus =
5. Perform standard multiplication: 4 × 7 = 28 meters.
Find the remainder modulo 5: 28 ÷ 5 = 5 with a remainder of
3. Result: `4 × 7 ≡ 3 (mod 5)`.
Commentary: This helps in managing resources or understanding the pattern of remainders. After accounting for all 5-meter rolls, 3 meters of fabric would effectively be leftover in the sense of the modular cycle. Question 4 (Simple Division/Finding Unknown): In a village that observes a 6-day market cycle (days 0, 1, 2, 3, 4, 5), if 2 times a certain number of days results in the 4th day of the cycle, what is that number of days? i.e., solve `2x ≡ 4 (mod 6)`.
Solution: Identify the equation and modulus: `2x ≡ 4 (mod 6)`. Test values for x (from 0 to 5, inclusive): If `x = 0`: `2 × 0 = 0`, `0 (mod 6) = 0`. (Not 4) If `x = 1`: `2 × 1 = 2`, `2 (mod 6) = 2`. (Not 4) If `x = 2`: `2 × 2 = 4`, `4 (mod 6) = 4`. (Yes!) If `x = 3`: `2 × 3 = 6`, `6 (mod 6) = 0`. (Not 4) If `x = 4`: `2 × 4 = 8`, `8 (mod 6) = 2`. (Not 4) If `x = 5`: `2 × 5 = 10`, `10 (mod 6) = 4`. (Yes!)
Result: `x ≡ 2 (mod 6)` and `x ≡ 5 (mod 6)`.
Commentary: This example shows that in modular arithmetic, there can be multiple solutions for `x` (when the `gcd(a, n) > 1` and `b` is divisible by `gcd(a, n)`). For SS1, this problem demonstrates how to find solutions by checking values within the modulus range.
Time Management and Calendars (Nigerian Context): Clock Arithmetic: Students regularly use 12-hour and 24-hour clocks. Modular arithmetic explains why adding 5 hours to 10 AM results in 3 PM (`10+5 = 15 ≡ 3 (mod 12)`), or why 19:00 hours plus 7 hours becomes 02:00 hours the next day (`19+7 = 26 ≡ 2 (mod 24)`). This helps in understanding flight schedules, bus departures, and daily planning.
Market Cycles: In many Nigerian communities, market days follow a fixed cycle (e.g., 4-day, 5-day, or 7-day cycles). Modular arithmetic can predict when the next market day will occur, given the current day and the number of days passed. For example, if a market holds every 5 days and today is the 2nd day, in 8 days it will be `2+8 = 10 ≡ 0 (mod 5)` (the starting day of the cycle).
Day of the Week: Predicting the day of the week after a certain number of days (e.g., if today is Tuesday (day 2), what day will it be in 100 days? `2 + 100 = 102 (mod 7)` which is `102 = 14 7 + 4`, so day 4, which is Thursday).
Digital Systems and Traffic Control: Digital Systems: Modular arithmetic is fundamental in computer science, especially in cryptography (for secure communication), hashing algorithms (for data storage and retrieval), and error detection codes. While complex for SS1, it can be explained that these systems rely on operations that "wrap around" similar to clocks.
Traffic Lights: The sequence of traffic lights (Red-Amber-Green) operates on a modular cycle. Understanding the timing of these lights (e.g., how many seconds before the next change) involves modular thinking, ensuring smooth traffic flow in bustling cities like Lagos or Abuja.
Pattern Recognition and Codes: Modular arithmetic helps in identifying repeating patterns. This can be seen in number sequences, musical scales, or even weaving patterns in traditional Nigerian fabrics. Basic understanding of modulo operations can be an introduction to simple coding or decoding activities, where letters or numbers are shifted by a fixed amount (Caesar cipher being a simple example) providing a practical application in information security.