Class 6 Mathematics - Ch 1: Mastery Guide: Factors, Multiples, Divisibility Rules, Prime Factorization, HCF & LCM with Real-Life Applications (FBISE)
Instructional Guide: Unit 1 Factors & Multiples
- Recognize factors and multiples of whole numbers up to large values.
- Master Tests of Divisibility for $2, 3, 4, 5, 6, 8, 9, 10, 11$.
- Calculate squares of up to two-digit numbers ($n^2$).
- Express composite numbers as products of prime factors in index notation ($a^b$).
- Find HCF using Prime Factorization and Long Division methods.
- Find LCM using Listing Multiples, Prime Factorization, and Division methods.
- Solve real-life word problems using HCF and LCM principles.
- Apply the golden relationship: $\text{Product of Two Numbers} = \text{HCF} \times \text{LCM}$.
- Factor vs Multiple Rule: A factor is a small building block (divides cleanly without remainder). A multiple is a giant tower built by skip-counting!
- Divisibility by 11 Trick: Sum the odd-place digits, sum the even-place digits. If their difference is $0$ or a multiple of $11$, the whole number is divisible by $11$!
- HCF vs LCM Clue Words:
• Look for "greatest", "largest", "maximum size", "cutting equally" $\to$ HCF.
• Look for "smallest", "least", "flashing together", "ringing together" $\to$ LCM.
- Arranging School Events: Arranging $40$ or $48$ auditorium chairs into equal rows without leaving any empty seats.
- Maritime Navigation: Synchronizing lighthouse warning lights that flash at different time intervals.
- School & Metro Timetables: Scheduling ringing bells and public transport arrivals that repeat at fixed minute cycles.
- Architecture & Construction: Cutting equal lengths of metal pipes and boundary fence posts without wasting materials.
🌟 1. Understanding Factors & Multiples
Imagine you have $40$ chairs in a classroom and you want to arrange them into neat, equal rows. How can we do that?
- $1$ row of $40$ chairs $\implies 1 \times 40 = 40$
- $2$ rows of $20$ chairs $\implies 2 \times 20 = 40$
- $4$ rows of $10$ chairs $\implies 4 \times 10 = 40$
- $5$ rows of $8$ chairs $\implies 5 \times 8 = 40$
- $8$ rows of $5$ chairs $\implies 8 \times 5 = 40$ (Numbers start repeating! We stop here.)
Therefore, the factors of $40$ are: $1, 2, 4, 5, 8, 10, 20, 40$.
📌 Key Definition: Factor
A factor of a number is a whole number that divides into it exactly without leaving any remainder.
• $1$ is a factor of every number.
• The greatest factor of any number is the number itself.
• The number of factors of a whole number is finite (countable).
📌 Key Definition: Multiple
A multiple of a number is the product obtained by multiplying that number by any counting whole number ($1, 2, 3, 4, \dots$).
• For example, the first 5 multiples of $3$ are: $3, 6, 9, 12, 15$.
• Every number is a multiple of itself.
• Every number has an infinite number of multiples.
• Important Fact: If $A$ is a factor of $B$, then $B$ is a multiple of $A$! (e.g., $3$ is a factor of $21$, and $21$ is a multiple of $3$).
⚡ 2. Master Tests of Divisibility (Superfast Mental Rules)
Instead of doing time-consuming long division on huge numbers (like $639$ or $819291$), we use Tests of Divisibility to test divisibility instantly!
| Divisor | Divisibility Rule | Pass Example | Fail Example |
|---|---|---|---|
| 2 | The last digit (unit digit) is even ($0, 2, 4, 6, 8$). | $39604$ (ends in $4$) | $435$ (ends in $5$) |
| 3 | The sum of all digits is a multiple of $3$. | $7512 \implies 7+5+1+2=15$ (divisible by $3$) | $79 \implies 7+9=16$ (not div. by $3$) |
| 4 | The number formed by the last two digits is divisible by $4$ (or ends in $00$). | $679\text{ }320 \implies 20 \div 4 = 5$ | $1038 \implies 38 \div 4 = 9\text{ R }2$ |
| 5 | The last digit is either $0$ or $5$. | $4805, 5000$ | $6123$ |
| 6 | The number is divisible by BOTH 2 and 3 (it is even AND its digit sum is a multiple of $3$). | $4506 \implies$ even, and $4+5+0+6=15$ | $6123$ (odd, so fails rule for $2$) |
| 8 | The number formed by the last three digits is divisible by $8$ (or ends in $000$). | $13\text{ }592 \implies 592 \div 8 = 74$ | $7172 \implies 172 \div 8 = 21\text{ R }4$ |
| 9 | The sum of all digits is a multiple of $9$. | $812\text{ }754 \implies 8+1+2+7+5+4 = 27$ | $7170 \implies 7+1+7+0=15$ |
| 10 | The last digit is $0$. | $1120, 9900, 24120$ | $4455$ |
| 11 | The difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$. | $4563 \implies (4+6)-(5+3)=10-8=2$ (No) $918291 \implies (9+8+9)-(1+2+1)=26-4=22$ (Yes!) |
$43212 \implies (4+2+2)-(3+1)=8-4=4$ (No) |
🌳 3. Prime Factorization, Factor Trees & Index Notation
Prime vs Composite Numbers
- Prime Number: A number greater than $1$ that has exactly two factors: $1$ and itself (e.g., $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots$). $2$ is the ONLY even prime number!
- Composite Number: A number that has more than two factors (e.g., $4, 6, 8, 9, 10, 12, 14, 15, \dots$).
- Co-Prime Numbers: Two numbers are co-prime if their common factor is only $1$ (i.e. $\text{HCF}=1$). For example, $10$ and $17$ are co-prime.
Square of a Number ($n^2$)
When a number is multiplied by itself, the result is called its square:
$$7^2 = 7 \times 7 = 49$$
$$13^2 = 13 \times 13 = 169$$
$$25^2 = 25 \times 25 = 625$$
Factor Tree & Index Notation
Every composite number can be broken down into prime branches until only prime numbers remain at the leaves:
Example: Prime Factorization of $252$
$$252 = 2 \times 126 = 2 \times 2 \times 63 = 2 \times 2 \times 3 \times 21 = 2 \times 2 \times 3 \times 3 \times 7$$
In Index Notation (Base and Power): $$252 = 2^2 \times 3^2 \times 7^1$$
🎯 4. Highest Common Factor (HCF) / GCD
The Highest Common Factor (HCF) of two or more numbers is the largest factor that divides all of them without remainder.
Method 1: Prime Factorization Method
Step 1: Write the prime factorization of each number.
Step 2: Circle the common prime factors.
Step 3: Multiply the common prime factors.
Example: Find HCF of $84, 108, 132$:
$$84 = \mathbf{2} \times \mathbf{2} \times \mathbf{3} \times 7$$
$$108 = \mathbf{2} \times \mathbf{2} \times \mathbf{3} \times 3 \times 3$$
$$132 = \mathbf{2} \times \mathbf{2} \times \mathbf{3} \times 11$$
$$\text{Common factors} = 2, 2, 3 \implies \mathbf{\text{HCF} = 2 \times 2 \times 3 = 12}$$
Method 2: Long Division Method (Euclidean Algorithm)
Divide the larger number by the smaller number. Then take the remainder as the new divisor and divide the previous divisor. Continue until the remainder is $0$. The last non-zero divisor is the HCF!
Example: For $546, 616$:
$$616 \div 546 = 1\text{ remainder }70$$
$$546 \div 70 = 7\text{ remainder }56$$
$$70 \div 56 = 1\text{ remainder }14$$
$$56 \div 14 = 4\text{ remainder }0 \implies \mathbf{\text{HCF}(546, 616) = 14}$$
Now test with the third number $224$: $224 \div 14 = 16\text{ remainder }0 \implies \mathbf{\text{Final HCF} = 14}$.
🔄 5. Least Common Multiple (LCM)
The Least Common Multiple (LCM) of two or more numbers is the smallest non-zero common multiple.
Method 1: Prime Factorization Method
$$\text{LCM} = \text{Product of Common Factors} \times \text{Product of Non-Common Factors}$$
(Alternatively: Take the highest power of every prime factor present across all numbers).
Example: For $36, 48, 56$:
$$36 = 2^2 \times 3^2$$
$$48 = 2^4 \times 3^1$$
$$56 = 2^3 \times 7^1$$
$$\text{LCM} = 2^4 \times 3^2 \times 7^1 = 16 \times 9 \times 7 = \mathbf{1008}$$
Method 2: Common Division / Ladder Method
Divide all numbers simultaneously by common prime divisors. If a number is not divisible, carry it down unchanged. Continue until no two numbers share a common prime factor.
$$\text{LCM} = \text{Product of all Divisors} \times \text{Remaining numbers}$$
⚖️ 6. The Golden Rule Connecting HCF and LCM
✨ Fundamental Formula
$$\text{First Number} \times \text{Second Number} = \text{HCF} \times \text{LCM}$$
• $\text{LCM} = \dfrac{\text{Product of two numbers}}{\text{HCF}}$
• $\text{HCF} = \dfrac{\text{Product of two numbers}}{\text{LCM}}$
• $\text{Second Number} = \dfrac{\text{HCF} \times \text{LCM}}{\text{First Number}}$
• Special Case: If two numbers are co-prime, their $\text{HCF} = 1$, so $\text{LCM} = \text{Product of the two numbers}$!
📝 Unit 1 Solved Exercises (Complete, Step-by-Step & Exhaustive)
Exercise 1.1
Q1. Identify the numbers that have $18$ as a factor: $54, 126, 198, 240, 320$.
• $126 \div 18 = 7$ (Remainder $0$) $\implies$ Yes ($126$)
• $198 \div 18 = 11$ (Remainder $0$) $\implies$ Yes ($198$)
• $240 \div 18 = 13\text{ R }6$ $\implies$ No
• $320 \div 18 = 17\text{ R }14$ $\implies$ No
Answer: $54, 126, 198$
Q2. Identify the numbers below which are factors of $144$: $1, 2, 3, 4, 6, 8, 9, 12, 14, 18, 32, 48, 144$.
$144 \div 1=144$, $144 \div 2=72$, $144 \div 3=48$, $144 \div 4=36$, $144 \div 6=24$, $144 \div 8=18$, $144 \div 9=16$, $144 \div 12=12$, $144 \div 14=10\text{ R }4$ (No), $144 \div 18=8$, $144 \div 32=4\text{ R }16$ (No), $144 \div 48=3$, $144 \div 144=1$.
Answer: $1, 2, 3, 4, 6, 8, 9, 12, 18, 48, 144$
Q3. Find all factors of:
(ii) $21$: $1 \times 21 = 21, 3 \times 7 = 21 \implies$ $1, 3, 7, 21$
(iii) $32$: $1 \times 32, 2 \times 16, 4 \times 8 \implies$ $1, 2, 4, 8, 16, 32$
(iv) $29$: $29$ is prime $\implies$ $1, 29$
(v) $40$: $1 \times 40, 2 \times 20, 4 \times 10, 5 \times 8 \implies$ $1, 2, 4, 5, 8, 10, 20, 40$
Q4. Write down the first 4 multiples of:
(ii) $6$: $6, 12, 18, 24$
(iii) $7$: $7, 14, 21, 28$
(iv) $9$: $9, 18, 27, 36$
(v) $11$: $11, 22, 33, 44$
(vi) $14$: $14, 28, 42, 56$
Q5. Write down the multiples of $5$ between $4$ and $31$:
Answer: $5, 10, 15, 20, 25, 30$
Q6. Identify the multiples of $8$ from: $14, 24, 30, 54, 56, 36, 72, 30, 64, 18, 48, 78, 96, 108, 120$.
Answer: $24, 48, 56, 64, 72, 96, 120$.
Exercise 1.2
Q1. State which of the following are divisible by $2, 3, 4$ or $5$ and give a reason for each:
• (ii) $63$: Divisible by $3$ because digit sum $6+3=9$ is a multiple of $3$.
• (iii) $110$: Divisible by $2$ (unit digit $0$ is even) and by $5$ (unit digit is $0$).
• (iv) $435$: Divisible by $3$ ($4+3+5=12$) and by $5$ (ends in $5$).
• (v) $6123$: Divisible by $3$ because digit sum $6+1+2+3=12$ is divisible by $3$.
• (vi) $1036$: Divisible by $2$ (even digit $6$) and by $4$ (last two digits $36 \div 4 = 9$).
• (vii) $3000$: Divisible by $2, 3, 4, 5$ (ends in $0$, digit sum $3$, last two digits $00$, ends in $0$).
Q2. State whether the following are divisible by $6, 8, 9, 10, 11$ and give a reason for each:
• (ii) $4128$: Divisible by $6$ (even & sum $4+1+2+8=15$) and by $8$ (last 3 digits $128 \div 8 = 16$).
• (iii) $9900$: Divisible by $6, 9, 10, 11$ (even, sum $18$, ends in $0$, alternate sum difference $(9+0)-(9+0)=0$).
• (iv) $43212$: Divisible by $6$ (even & sum $4+3+2+1+2=12$).
• (v) $4455$: Divisible by $9$ ($4+4+5+5=18$) and by $11$ ($(4+5)-(4+5)=0$).
• (vi) $7172$: Divisible by $11$ ($(7+7)-(1+2)=14-3=11$).
• (vii) $819291$: Divisible by $11$ ($(8+9+9)-(1+2+1)=26-4=22$, multiple of $11$).
• (viii) $24120$: Divisible by $6, 8, 9, 10$ (even, sum $9$, last 3 digits $120 \div 8 = 15$, ends in $0$).
Q3. From the list: $5, 8, 81, 85, 60, 26, 54, 45$. Write numbers divisible by $5$.
Q4. Is $6$ a factor of $432$? Give reason.
Exercise 1.3
Q1. Find the squares of the following numbers:
(v) $16^2 = 256$ • (vi) $19^2 = 361$ • (vii) $30^2 = 900$ • (viii) $100^2 = 10000$
(ix) $150^2 = 22500$ • (x) $21^2 = 441$ • (xi) $23^2 = 529$ • (xii) $25^2 = 625$
Q2. Complete each of the following factor trees:
(ii) $54 = 6 \times 9 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3$
(iii) $56 = 8 \times 7 = 2 \times 4 \times 7 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7$
(iv) $48 = 6 \times 8 = 2 \times 3 \times 2 \times 4 = 2 \times 3 \times 2 \times 2 \times 2 = 2^4 \times 3$
Q3. Find prime factors of these numbers:
(ii) $20 = 2 \times 2 \times 5$
(iii) $48 = 2 \times 2 \times 2 \times 2 \times 3$
(iv) $76 = 2 \times 2 \times 19$
(v) $88 = 2 \times 2 \times 2 \times 11$
(vi) $256 = 2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
(vii) $576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$
(viii) $1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
Q4. Express the following in index notation:
(ii) $7 \times 7 \times 7 \times 7 \times 7 = \mathbf{7^5}$
(iii) $3 \times 3 \times 3 \times 4 \times 4 \times 4 \times 4 = \mathbf{3^3 \times 4^4}$
(iv) $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 4 \times 4 \times 4 \times 5 = \mathbf{2^3 \times 3^3 \times 4^3 \times 5^1}$
Q5. Express as prime factors and write answer in index notation:
(ii) $275 = 5 \times 5 \times 11 = \mathbf{5^2 \times 11}$
(iii) $300 = 2 \times 2 \times 3 \times 5 \times 5 = \mathbf{2^2 \times 3 \times 5^2}$
(iv) $162 = 2 \times 3 \times 3 \times 3 \times 3 = \mathbf{2 \times 3^4}$
Q6. Find prime factors using the "factor tree" and write answers in index notation:
(ii) $136 = 2 \times 68 = 2 \times 2 \times 34 = 2 \times 2 \times 2 \times 17 = \mathbf{2^3 \times 17}$
(iii) $80 = 8 \times 10 = 2 \times 4 \times 2 \times 5 = 2 \times 2 \times 2 \times 2 \times 5 = \mathbf{2^4 \times 5}$
(iv) $130 = 10 \times 13 = 2 \times 5 \times 13 = \mathbf{2 \times 5 \times 13}$
Exercise 1.4
Q1. Find the HCF of the following numbers by using prime factors:
• (ii) $108, 315$: $108 = 2^2 \times 3^3$, $315 = 3^2 \times 5 \times 7 \implies \text{HCF} = 3^2 = \mathbf{9}$
• (iii) $180, 250, 126$: $180 = 2^2 \times 3^2 \times 5$, $250 = 2 \times 5^3$, $126 = 2 \times 3^2 \times 7 \implies \text{HCF} = \mathbf{2}$
• (iv) $735, 1050, 455$: $735 = 3 \times 5 \times 7^2$, $1050 = 2 \times 3 \times 5^2 \times 7$, $455 = 5 \times 7 \times 13 \implies \text{HCF} = 5 \times 7 = \mathbf{35}$
• (v) $288, 300, 216$: $288 = 2^5 \times 3^2$, $300 = 2^2 \times 3 \times 5^2$, $216 = 2^3 \times 3^3 \implies \text{HCF} = 2^2 \times 3 = \mathbf{12}$
Q2. Find the HCF of the following numbers by the long division method:
• (ii) $105, 90$: $105 \div 90 = 1\text{ R }15 \to 90 \div 15 = 6\text{ R }0 \implies \mathbf{\text{HCF} = 15}$
• (iii) $360, 444$: $444 \div 360 = 1\text{ R }84 \to 360 \div 84 = 4\text{ R }24 \to 84 \div 24 = 3\text{ R }12 \to 24 \div 12 = 2\text{ R }0 \implies \mathbf{\text{HCF} = 12}$
• (iv) $288, 600, 936$: $\text{HCF}(288, 600) = 24$, $\text{HCF}(24, 936) = 24 \implies \mathbf{\text{HCF} = 24}$
• (v) $1078, 693, 847$: $\text{HCF}(1078, 693) = 77$, $\text{HCF}(77, 847) = 77 \implies \mathbf{\text{HCF} = 77}$
• (vi) $1320, 840, 500, 650$: $\text{HCF}(1320, 840) = 120$, $\text{HCF}(120, 500) = 20$, $\text{HCF}(20, 650) = 10 \implies \mathbf{\text{HCF} = 10}$
Q3. Find the highest common factor of $36, 45$ and $72$:
Q4. Find the greatest number which exactly divides $595, 357$ and $102$:
Q5. Find the highest common factor of $56$ and $84$ using both methods:
• Division Method: $84 \div 56 = 1\text{ R }28 \to 56 \div 28 = 2\text{ R }0 \implies \mathbf{\text{HCF} = 28}$.
Q6. Three strings with length of $140\text{ cm}, 168\text{ cm}$ and $210\text{ cm}$ are cut into pieces of same length. Find the greatest length of pieces if no string is left over:
$140 = 2^2 \times 5 \times 7$, $168 = 2^3 \times 3 \times 7$, $210 = 2 \times 3 \times 5 \times 7$
$\text{Common factors} = 2 \times 7 = 14\text{ cm}$.
Answer: $14\text{ cm}$
Q7. Find the greatest sum of money that is exactly contained in $\text{Rs. }162, \text{Rs. }252$ and $\text{Rs. }342$:
$162 = 2 \times 3^4$, $252 = 2^2 \times 3^2 \times 7$, $342 = 2 \times 3^2 \times 19$
$\text{HCF} = 2 \times 3^2 = 2 \times 9 = 18$.
Answer: $\text{Rs. }18$
Exercise 1.5
Q1. Find the LCM of the following by taking multiples:
• (ii) $3, 6, 9$: Multiples of $3$: $3, 6, 9, 12, 15, 18, \dots$; Multiples of $6$: $6, 12, 18, \dots$; Multiples of $9$: $9, 18, \dots \implies \mathbf{\text{LCM} = 18}$.
Q2. Find the LCM by prime factorization:
• (ii) $18, 24, 36$: $18 = 2 \times 3^2, 24 = 2^3 \times 3, 36 = 2^2 \times 3^2 \implies \text{LCM} = 2^3 \times 3^2 = 8 \times 9 = \mathbf{72}$
• (iii) $84, 72, 112$: $84 = 2^2 \times 3 \times 7, 72 = 2^3 \times 3^2, 112 = 2^4 \times 7 \implies \text{LCM} = 2^4 \times 3^2 \times 7 = 16 \times 9 \times 7 = \mathbf{1008}$
• (iv) $90, 45, 75$: $90 = 2 \times 3^2 \times 5, 45 = 3^2 \times 5, 75 = 3 \times 5^2 \implies \text{LCM} = 2 \times 3^2 \times 5^2 = 2 \times 9 \times 25 = \mathbf{450}$
• (v) $49, 70, 105, 84$: $49 = 7^2, 70 = 2 \times 5 \times 7, 105 = 3 \times 5 \times 7, 84 = 2^2 \times 3 \times 7 \implies \text{LCM} = 2^2 \times 3 \times 5 \times 7^2 = 4 \times 3 \times 5 \times 49 = \mathbf{2940}$
• (vi) $480, 560, 720$: $480 = 2^5 \times 3 \times 5, 560 = 2^4 \times 5 \times 7, 720 = 2^4 \times 3^2 \times 5 \implies \text{LCM} = 2^5 \times 3^2 \times 5 \times 7 = 32 \times 9 \times 5 \times 7 = \mathbf{10080}$
Q3. Find the LCM by division method:
• (ii) $60, 96, 108$: $\text{LCM} = 2^5 \times 3^3 \times 5 = 32 \times 27 \times 5 = \mathbf{4320}$
• (iii) $70, 210, 126$: $\text{LCM} = 2 \times 3^2 \times 5 \times 7 = \mathbf{630}$
• (iv) $105, 140, 280$: $\text{LCM} = 2^3 \times 3 \times 5 \times 7 = 8 \times 15 \times 7 = \mathbf{840}$
• (v) $120, 256, 288$: $\text{LCM} = 2^8 \times 3^2 \times 5 = 256 \times 9 \times 5 = \mathbf{11520}$
Q4. Find the least number which, when divided by $14, 21$ and $18$, gives no remainder:
Q5. Find the length of the shortest piece of string that can be cut into equal lengths of $16\text{ cm}, 20\text{ cm}, 24\text{ cm}$:
Q6. On a light house a red light flashes after $10\text{ minutes}$, a green light after $12\text{ minutes}$ and a yellow light after $15\text{ minutes}$. If they flash together at $9\text{ pm}$, at what time will they flash together again?
Time $= 9\text{ pm} + 1\text{ hour} = \mathbf{10\text{ pm}}$.
Q7. Find the shortest length of a pipe that can be measured exactly with $5\text{m}, 15\text{m}$ and $25\text{m}$ long measuring tapes respectively:
Q8. Three bells ring at interval of $8\text{ minutes}, 15\text{ minutes}$ and $24\text{ minutes}$ respectively. If they ring together at $3\text{ pm}$, at what time will they next ring together again?
Time $= 3\text{ pm} + 2\text{ hours} = \mathbf{5\text{ pm}}$.
Exercise 1.6
Q1. The product of two numbers is $432$. Find their LCM if HCF is $6$:
Q2. The product of two numbers is $6048$. Find their HCF if LCM is $504$:
Q3. The HCF and LCM of two numbers are $15$ and $180$ respectively. Find product of the numbers:
Q4. The LCM and HCF of two numbers are $408$ and $34$ respectively. Find the second number if the first number is $102$:
Q5. The LCM of two co-prime numbers is $555$. Find the second number if the first number is $15$:
$$\text{Second Number} = \dfrac{\text{LCM}}{\text{First Number}} = \dfrac{555}{15} = \mathbf{37}$$
Review Exercise 1
Q1. Encircle the correct option:
(ii) $93$ is a multiple of: (a) $3$ ($3 \times 31 = 93$)
(iii) Number of multiples of $7$ between $21$ and $50$ is: (b) $4$ ($28, 35, 42, 49$)
(iv) $7521$ is divisible by: (b) $3$ ($7+5+2+1=15$)
(v) A number is divisible by $6$ if it is divisible by: (c) $2$ and $3$
(vi) $990$ is divisible by: (a) $2, 3, 5, 11$
(vii) $5^4 =$: (c) $5 \times 5 \times 5 \times 5$
(viii) There are $\dots\dots$ methods for finding HCF: (a) $2$ (Prime Factorization & Long Division)
(ix) HCF of $16$ and $48$ is: (a) $16$
(x) LCM of $4$ and $8$ is: (b) $8$
Q2. Write the factors of $84$ and $96$. What are the common factors of $84$ and $96$?
• Factors of $96$: $1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96$
• Common Factors: $1, 2, 3, 4, 6, 12$
Q3. From the set: $\{5, 9, 12, 14, 15, 18, 20, 21, 24, 27, 28, 30, 33, 36, 39, 41, 42, 44, 45, 49, 50\}$, which are multiples of:
(ii) Multiples of $3$: $9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45$
(iii) Multiples of $4$: $12, 20, 24, 28, 36, 44$
(iv) Multiples of $5$: $5, 15, 20, 30, 45, 50$
(v) Multiples of $6$: $12, 18, 24, 30, 36, 42$
(vi) Multiples of $7$: $14, 21, 28, 42, 49$
(vii) Multiples of $8$: $24$
(viii) Multiples of $9$: $9, 18, 27, 36, 45$
Q4. One of the numbers $2, 3, 4, 5, 6, 8, 9, 11$ does not divide $6480$. Find it:
Answer: $11$
Q5. Express as a product of prime factors in index notation:
(ii) $6912 = 2^8 \times 3^3 = \mathbf{2^8 \times 3^3}$
Q6. Find the greatest mass that can be taken an exact number of times from $360\text{g}, 504\text{g}$ and $672\text{g}$:
Q7. A rectangular field measures $308\text{m}$ by $228\text{m}$. Fencing posts are placed along its sides at equal distances apart. If the posts are as far apart as possible, what is the distance between them?
$308 = 2^2 \times 7 \times 11$, $228 = 2^2 \times 3 \times 19 \implies \text{HCF} = 2^2 = \mathbf{4\text{m}}$.
Q8. Find the smallest mass that can be measured out in equal amount of $6\text{kg}, 9\text{kg}$ and $12\text{kg}$:
Q9. Find the least length of a rope which can be cut into pieces of lengths $30\text{m}, 36\text{m}$ and $54\text{m}$:
Q10. Find the smallest sum of money that is an exact multiple of $\text{Rs. }72, \text{Rs. }80$ and $\text{Rs. }96$:
$\text{LCM} = 2^5 \times 3^2 \times 5 = 32 \times 9 \times 5 = \mathbf{\text{Rs. }1440}$.
🎯 Unit 1 Synthesis & Common Misconceptions
| Common Misconception | Mathematical Reality | Quick Memory Tip |
|---|---|---|
| Thinking $1$ is a prime number. | $1$ is neither prime nor composite because it has only one factor ($1$ itself). A prime number must have exactly two distinct factors. | Primes have 2 distinct doors; $1$ only has 1 door. |
| Confusing HCF and LCM. | $\text{HCF} \le$ the smallest number in the set; $\text{LCM} \ge$ the largest number in the set. | HCF divides; LCM is divided into. |
| Thinking all odd numbers are prime. | Numbers like $9, 15, 21, 25, 27$ are odd but composite ($3 \times 3 = 9$, $3 \times 5 = 15$). | Check digit sum and last digit before declaring prime! |
More Chapter Notes for Class 6 (FBISE)
MathematicsTest Your Knowledge on Chapter 1: Class 6 Mathematics - Ch 1: Mastery Guide: Factors, Multiples, Divisibility Rules, Prime Factorization, HCF & LCM with Real-Life Applications (FBISE)
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Class 6 Mathematics - Ch 1: Factors and Multiples Chapter Mock Test
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