Published 2026-10-02
Chapter: Playing With Numbers

Playing With Numbers - Divisibility tests, prime and composite numbers, prime factorization, and finding HCF and LCM

Numbers are the foundational building blocks of mathematics. In earlier chapters, numbers were primarily used for counting, comparing, and performing basic arithmetic operations like addition, subtraction, multiplication, and division. However, numbers possess intrinsic structural properties that govern how they interact with one another.

The study of factors, multiples, prime numbers, and divisibility rules—collectively known as "Playing With Numbers"—is not merely a collection of arithmetic tricks. It forms the core foundation of number theory. Mastering these concepts develops numerical intuition and algebraic reasoning. It enables students to simplify fractions, solve complex real-world scheduling and spatial problems, and understand the fundamental properties that govern integers.


1. In-Depth Conceptual Breakdown

Factors and Multiples

To understand how numbers are built, we must first examine their basic components: factors and multiples.

Factors

A factor of a number is an exact divisor of that number. In other words, when a number is divided by its factor, the remainder is strictly 00.

Example: Consider the number 1212.

  • 12÷1=1212 \div 1 = 12 (Remainder = 00)
  • 12÷2=612 \div 2 = 6 (Remainder = 00)
  • 12÷3=412 \div 3 = 4 (Remainder = 00)
  • 12÷4=312 \div 4 = 3 (Remainder = 00)
  • 12÷6=212 \div 6 = 2 (Remainder = 00)
  • 12÷12=112 \div 12 = 1 (Remainder = 00)

Therefore, the factors of 1212 are 1,2,3,4,6,1, 2, 3, 4, 6, and 1212.

Key Properties of Factors:

  1. The Universal Factor: 11 is a factor of every number.
  2. Self-Factor: Every non-zero number is a factor of itself.
  3. Finiteness: The number of factors of a given number is finite.
  4. Bound: Every factor of a number is less than or equal to that number (f≤nf \le n).

Multiples

A multiple of a number is a number obtained by multiplying that number by any non-zero natural number (1,2,3,…1, 2, 3, \dots).

Example: The multiples of 77 are calculated as: 7×1=77 \times 1 = 7 7×2=147 \times 2 = 14 7×3=217 \times 3 = 21 7×4=28…7 \times 4 = 28 \dots

Therefore, the multiples of 77 are 7,14,21,28,35,…7, 14, 21, 28, 35, \dots

Key Properties of Multiples:

  1. Self-Multiple: Every number is a multiple of itself.
  2. Infiniteness: The number of multiples of a given number is infinite.
  3. Bound: Every multiple of a number is greater than or equal to that number (m≥nm \ge n).

Prime and Composite Numbers

Numbers can be categorized based on the count of their factors.

                   ┌───────────────────────────┐
                   │       Natural Numbers     │
                   └─────────────┬─────────────┘
                                 │
         ┌───────────────────────┼───────────────────────┐
         │                       │                       │
  ┌──────┴──────┐         ┌──────┴──────┐         ┌──────┴──────┐
  │   Unit (1)  │         │ Prime Nos.  │         │Composite Nos│
  │ (1 Factor)  │         │ (2 Factors) │         │(> 2 Factors)│
  └─────────────┘         └─────────────┘         └─────────────┘

Prime Numbers

A prime number is a natural number greater than 11 that has exactly two factors: 11 and the number itself.

  • Examples: 2,3,5,7,11,13,17,19,23,…2, 3, 5, 7, 11, 13, 17, 19, 23, \dots

Composite Numbers

A composite number is a natural number that has more than two factors.

  • Examples: 4,6,8,9,10,12,14,15,…4, 6, 8, 9, 10, 12, 14, 15, \dots

Important Note on the Number 1: The number 11 has only one factor (itself). Therefore, 11 is neither prime nor composite.

The Sieve of Eratosthenes

To find all prime numbers up to 100100, the ancient Greek mathematician Eratosthenes developed a simple method:

  1. List all natural numbers from 11 to 100100. Cross out 11.
  2. Circle 22 (the smallest prime) and cross out all its multiples (except 22 itself).
  3. Circle 33 (the next prime) and cross out all its multiples.
  4. Circle 55 and cross out all its multiples.
  5. Continue this process for 77.
  6. All remaining uncrossed numbers are prime numbers.

There are exactly 25 prime numbers between 11 and 100100.

Special Classifications of Numbers

  • Even and Odd Numbers: Any number divisible by 22 is an even number (0,2,4,6,8…0, 2, 4, 6, 8 \dots). Numbers not divisible by 22 are odd numbers (1,3,5,7,9…1, 3, 5, 7, 9 \dots).
    • 22 is the smallest prime number and the only even prime number. All other prime numbers are odd.
  • Co-prime Numbers: Two natural numbers are said to be co-prime if they have no common factor other than 11 (i.e., their HCF=1\text{HCF} = 1).
    • Note: The numbers themselves do not need to be prime. For example, 88 (composite) and 1515 (composite) are co-prime because: Factors of 8={1,2,4,8}\text{Factors of } 8 = \{1, 2, 4, 8\} Factors of 15={1,3,5,15}\text{Factors of } 15 = \{1, 3, 5, 15\} Common Factors={1}\text{Common Factors} = \{1\}
  • Twin Primes: Two prime numbers whose difference is 22 are called twin primes.
    • Examples: (3,5)(3, 5), (5,7)(5, 7), (11,13)(11, 13), (17,19)(17, 19), (29,31)(29, 31).

Summary Comparison Table

ClassificationNumber of FactorsCondition / DefinitionExamples
Prime NumberExactly 2Factors are only 11 and itself2,3,5,7,112, 3, 5, 7, 11
Composite NumberMore than 2Has factors other than 11 and itself4,6,8,9,104, 6, 8, 9, 10
Co-Prime PairN/A (Pair property)HCF(a,b)=1\text{HCF}(a, b) = 1(8,15),(9,10),(5,7)(8, 15), (9, 10), (5, 7)
Twin Prime PairN/A (Pair property)Both are prime and $a - b

Tests for Divisibility of Numbers

Divisibility tests are mental arithmetic rules that allow us to determine whether a given large integer is divisible by a smaller number without performing long division.

1. Divisibility by 2

A number is divisible by 22 if its units digit is an even number: 0,2,4,6,0, 2, 4, 6, or 88.

  • Example: 24682468 ends in 88, so it is divisible by 22.

2. Divisibility by 3

A number is divisible by 33 if the sum of its digits is a multiple of 33.

  • Example: Consider 729729. Sum of digits = 7+2+9=187 + 2 + 9 = 18. Since 1818 is divisible by 33, 729729 is divisible by 33.

3. Divisibility by 4

A number with 33 or more digits is divisible by 44 if the number formed by its last two digits (tens and units place) is divisible by 44.

  • Example: Consider 3,5243,524. The last two digits form 2424. Since 24÷4=624 \div 4 = 6, 3,5243,524 is divisible by 44.

4. Divisibility by 5

A number is divisible by 55 if its units digit is either 00 or 55.

  • Example: 1,0551,055 and 3,4203,420 are both divisible by 55.

5. Divisibility by 6

A number is divisible by 66 if it is divisible by both 22 and 33.

  • Example: Consider 438438.
    • Units digit is 88 (even), so it is divisible by 22.
    • Sum of digits = 4+3+8=154 + 3 + 8 = 15, which is divisible by 33.
    • Hence, 438438 is divisible by 66.

6. Divisibility by 8

A number with 44 or more digits is divisible by 88 if the number formed by its last three digits is divisible by 88.

  • Example: Consider 7,1207,120. The last three digits form 120120. Since 120÷8=15120 \div 8 = 15, 7,1207,120 is divisible by 88.

7. Divisibility by 9

A number is divisible by 99 if the sum of its digits is a multiple of 99.

  • Example: Consider 4,6894,689. Sum of digits = 4+6+8+9=274 + 6 + 8 + 9 = 27. Since 2727 is divisible by 99, 4,6894,689 is divisible by 99.

8. Divisibility by 10

A number is divisible by 1010 if its units digit is 00.

  • Example: 5,6305,630 ends in 00, so it is divisible by 1010.

9. Divisibility by 11

Find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right). If the difference is either 00 or divisible by 1111, then the number is divisible by 1111.

Example: Consider 61,80961,809.

  • Digits at odd places (1st, 3rd, 5th from right): 9,8,6  ⟹  Sum=9+8+6=239, 8, 6 \implies \text{Sum} = 9 + 8 + 6 = 23
  • Digits at even places (2nd, 4th from right): 0,1  ⟹  Sum=0+1=10, 1 \implies \text{Sum} = 0 + 1 = 1
  • Difference = 23−1=2223 - 1 = 22
  • Since 2222 is divisible by 1111, 61,80961,809 is divisible by 1111.

Prime Factorization

Prime Factorization is the process of expressing a composite number strictly as a product of prime numbers. According to the Fundamental Theorem of Arithmetic, every composite number can be factored uniquely into prime numbers (disregarding the order of factors).

Method 1: Factor Tree Method

We break down the composite number into pairs of factors until every branch ends in a prime number.

       36
      /  \
     2    18
         /  \
        2    9
            / \
           3   3

Prime factorization of 36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2.

Method 2: Continuous Division Method

We divide the given number continuously by the smallest possible prime numbers until the quotient becomes 11.

2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$ Prime factorization of $90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$. --- ### Highest Common Factor (HCF) The **Highest Common Factor (HCF)**, also known as the **Greatest Common Divisor (GCD)**, of two or more given numbers is the highest (or largest) of their common factors. #### Method 1: Prime Factorization Method for HCF 1. Express each number as a product of prime factors. 2. Identify the common prime factors. 3. Multiply the lowest powers of these common prime factors. *Example*: Find HCF of $24$ and $36$. $$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$$ $$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$$ Common prime factors = $2 \times 2 \times 3 = 12$. Hence, $\text{HCF}(24, 36) = 12$. #### Method 2: Continued Division Method for HCF 1. Divide the larger number by the smaller number. 2. Take the remainder as the new divisor and the previous divisor as the new dividend. 3. Repeat this process until the remainder becomes $0$. The last divisor is the required HCF. --- ### Lowest Common Multiple (LCM) The **Lowest Common Multiple (LCM)** of two or more given numbers is the smallest (non-zero) number that is a multiple of each of the given numbers. #### Method 1: Prime Factorization Method for LCM 1. Express each number as a product of prime factors. 2. Find the product of the highest powers of all prime factors involved in the numbers. *Example*: Find LCM of $24$ and $36$. $$24 = 2^3 \times 3^1$$ $$36 = 2^2 \times 3^2$$ $\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72$. #### Method 2: Common Division Method for LCM 1. Write the numbers in a row separated by commas. 2. Divide by a prime number that divides at least one of the numbers. 3. Carry down any number that is not completely divisible. 4. Repeat until the last row consists only of $1$s. 5. Multiply all the divisors to get the LCM. $$\begin{array}{r|l} 2 & 12, 18, 20 \\ \hline 2 & 6, 9, 10 \\ \hline 3 & 3, 9, 5 \\ \hline 3 & 1, 3, 5 \\ \hline 5 & 1, 1, 5 \\ \hline & 1, 1, 1 \end{array}$$ $$\text{LCM} = 2 \times 2 \times 3 \times 3 \times 5 = 180$$ --- ### Relationship Between HCF and LCM For **any two given positive integers $a$ and $b$**, the product of their HCF and LCM is strictly equal to the product of the two numbers. $$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$ > **Crucial Warning:** This formula holds true **only for two numbers**. It is not directly valid for three or more numbers! --- ## 2. Real-World Applications ### Application 1: Spatial Tiling and Cutting (HCF) Suppose a mason needs to lay square tiles on a rectangular floor measuring $12\text{ meters}$ by $8\text{ meters}$. What is the size of the largest possible square tile that can fit perfectly without cutting any tile? * **Why HCF?** The edge length of the square tile must divide both the length ($12\text{m}$) and breadth ($8\text{m}$) completely without leaving any fractional space. The *largest* size requires finding the *Highest Common Factor*. * $\text{HCF}(12, 8) = 4\text{ meters}$. The largest square tile is $4\text{m} \times 4\text{m}$. ### Application 2: Periodic Synchronization (LCM) Three traffic lights at different road crossings change after every $40\text{ seconds}$, $60\text{ seconds}$, and $90\text{ seconds}$ respectively. If they change simultaneously at $8:00\text{ AM}$, at what time will they change together again? * **Why LCM?** Each light repeats its sequence at multiples of its interval ($40, 80, 120\dots$; $60, 120, 180\dots$). The lights will flash together at a time value that is a *common multiple* of all three intervals. The *next* occurrence requires the *Lowest Common Multiple*. * $\text{LCM}(40, 60, 90) = 360\text{ seconds} = 6\text{ minutes}$. * They will change together again at $8:06\text{ AM}$. ### Application 3: Equal Grouping and Distribution (HCF/LCM) A school teacher wants to arrange $48$ boys and $36$ girls in equal rows for a drill exercise such that every row contains students of only one gender and all rows have the exact same number of students. To minimize the number of rows, she needs the maximum number of students per row, which corresponds to $\text{HCF}(48, 36) = 12\text{ students per row}$. --- ## 3. Step-by-Step Solved Textbook Examples ### Example 1: Testing Divisibility **Question:** Test whether the number $10,824$ is divisible by $6$ and $11$. **Solution:** *Part A: Test for Divisibility by 6* 1. Check divisibility by $2$: * The units digit of $10,824$ is $4$, which is an even number. * Therefore, $10,824$ is divisible by $2$. 2. Check divisibility by $3$: * Sum of digits $= 1 + 0 + 8 + 2 + 4 = 15$. * Since $15$ is divisible by $3$ ($15 \div 3 = 5$), $10,824$ is divisible by $3$. 3. Conclusion for 6: * Since $10,824$ is divisible by both $2$ and $3$, **it is divisible by $6$**. *Part B: Test for Divisibility by 11* 1. Identify digits from right to left: * Position 1 (Odd): $4$ * Position 2 (Even): $2$ * Position 3 (Odd): $8$ * Position 4 (Even): $0$ * Position 5 (Odd): $1$ 2. Sum of digits at odd places $= 4 + 8 + 1 = 13$. 3. Sum of digits at even places $= 2 + 0 = 2$. 4. Difference $= 13 - 2 = 11$. 5. Since $11$ is divisible by $11$, **$10,824$ is divisible by $11$**. --- ### Example 2: Prime Factorization **Question:** Find the prime factorization of $980$ using the Division Method. **Solution:** Step 1: Divide $980$ by the smallest prime factor, $2$. $$980 \div 2 = 490$$ Step 2: Divide $490$ by $2$. $$490 \div 2 = 245$$ Step 3: $245$ is not divisible by $2$ or $3$ (sum of digits = $11$). Divide $245$ by $5$. $$245 \div 5 = 49$$ Step 4: Divide $49$ by $7$. $$49 \div 7 = 7$$ Step 5: Divide $7$ by $7$. $$7 \div 7 = 1$$ $$\begin{array}{r|l} 2 & 980 \\ \hline 2 & 490 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$ **Final Answer:** The prime factorization of $980$ is **$2 \times 2 \times 5 \times 7 \times 7$** or **$2^2 \times 5 \times 7^2$**. --- ### Example 3: Finding HCF by Prime Factorization **Question:** Find the HCF of $144$, $180$, and $192$. **Solution:** Step 1: Express each number as a product of prime factors. $$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$$ $$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$$ $$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$$ Step 2: Identify common prime bases present in all three numbers. * The common prime bases are $2$ and $3$. Step 3: Select the minimum exponent for each common prime factor: * For base $2$: minimum power is $\min(4, 2, 6) = 2 \implies 2^2$ * For base $3$: minimum power is $\min(2, 2, 1) = 1 \implies 3^1$ Step 4: Multiply these lowest powers: $$\text{HCF} = 2^2 \times 3^1 = 4 \times 3 = 12$$ **Final Answer:** The HCF of $144, 180,$ and $192$ is **$12$**. --- ### Example 4: Real-world Word Problem (HCF vs. LCM) **Question:** Two tankers contain $850\text{ liters}$ and $680\text{ liters}$ of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both tankers when used an exact number of times. **Solution:** Step 1: Understand the mathematical requirement. * We need a container capacity that divides $850$ and $680$ completely without leaving any remainder. * Since we require the *maximum capacity*, we must calculate the **Highest Common Factor (HCF)** of $850$ and $680$. Step 2: Find prime factorizations. $$850 = 2 \times 5 \times 5 \times 17 = 2^1 \times 5^2 \times 17^1$$ $$680 = 2 \times 2 \times 2 \times 5 \times 17 = 2^3 \times 5^1 \times 17^1$$ Step 3: Calculate HCF by taking product of minimum powers of common factors: $$\text{Common factors} = 2^1 \times 5^1 \times 17^1$$ $$\text{HCF} = 2 \times 5 \times 17 = 170$$ **Final Answer:** The maximum capacity of the required container is **$170\text{ liters}$**. --- ## 4. Common Student Mistakes to Avoid ### Mistake 1: Misinterpreting $1$ as a Prime Number * **Incorrect Assumption:** Students often list $1$ as the smallest prime number. * **Correction:** A prime number must have **exactly two distinct positive factors** ($1$ and itself). The number $1$ has only one factor ($1$). Hence, $1$ is neither prime nor composite. The smallest prime number is $2$. ### Mistake 2: Mixing Up HCF and LCM Applications in Word Problems * **Incorrect Strategy:** Using LCM when a problem asks to divide or group things into smaller equal parts, or using HCF when finding repeating cyclical time intervals. * **Correction:** * If the problem requires **dividing/cutting/arranging into maximum equal sizes**, calculate **HCF**. * If the problem involves **repeating events, loops, or finding a future common time/distance**, calculate **LCM**. ### Mistake 3: Errors in the Divisibility Test for 11 * **Incorrect Method:** Counting digit positions from left to right instead of right to left, or taking the difference between the sums of odd/even *digits* instead of digits at odd/even *places*. * **Correction:** Place positions must always be counted from the **units digit (right side)**: $$\text{Number: } \quad \underset{\text{5th}}{5} \quad \underset{\text{4th}}{4} \quad \underset{\text{3rd}}{3} \quad \underset{\text{2nd}}{2} \quad \underset{\text{1st}}{1}$$ Calculate $(\text{Sum of digits at 1st, 3rd, 5th places}) - (\text{Sum of digits at 2nd, 4th places})$. ### Mistake 4: Believing Co-Prime Numbers Must Be Prime * **Incorrect Assumption:** Assuming that numbers in a co-prime pair must individual prime numbers. * **Correction:** "Co-prime" refers to a **relationship** between two numbers having no common factor other than $1$. For example, $8$ and $9$ are both composite, yet $\text{HCF}(8, 9) = 1$, making them co-prime. --- ## 5. Practice Questions for Self-Assessment ### Question 1 Find the smallest digit $K$ such that the four-digit number $5,3K2$ is completely divisible by $6$. ### Question 2 Determine the HCF and LCM of $360$ and $450$. Verify that: $$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$ ### Question 3 Three runners run along a circular track. They complete one round in $20\text{ seconds}$, $30\text{ seconds}$, and $45\text{ seconds}$ respectively. If they start together from the same point at $10:00\text{ AM}$, at what exact time will they meet next at the starting point? ### Question 4 Find the greatest $3$-digit number which is exactly divisible by $8, 10,$ and $12$. --- ### Step-by-Step Solutions #### Solution to Question 1: 1. For $5,3K2$ to be divisible by $6$, it must be divisible by both $2$ and $3$. 2. **Divisibility by 2:** The units digit is $2$ (even), so $5,3K2$ is divisible by $2$ for any digit value of $K$. 3. **Divisibility by 3:** Sum of digits must be a multiple of $3$. $$\text{Sum} = 5 + 3 + K + 2 = 10 + K$$ 4. Possible single-digit values for $K$ ($0$ to $9$) that make $(10 + K)$ divisible by $3$: * If $K = 2 \implies 10 + 2 = 12$ (Divisible by 3) * If $K = 5 \implies 10 + 5 = 15$ (Divisible by 3) * If $K = 8 \implies 10 + 8 = 18$ (Divisible by 3) 5. The *smallest* digit value required is $K = 2$. **Final Answer:** **$K = 2$** (The number is $5,322$). --- #### Solution to Question 2: 1. **Prime Factorization:** $$360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5^1$$ $$450 = 2 \times 3 \times 3 \times 5 \times 5 = 2^1 \times 3^2 \times 5^2$$ 2. **Calculate HCF:** $$\text{HCF} = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90$$ 3. **Calculate LCM:** $$\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1,800$$ 4. **Verification:** $$\text{LHS} = \text{HCF} \times \text{LCM} = 90 \times 1,800 = 162,000$$ $$\text{RHS} = \text{Product of numbers} = 360 \times 450 = 162,000$$ $$\text{LHS} = \text{RHS}$$ **Final Answer:** $\text{HCF} = \mathbf{90}$, $\text{LCM} = \mathbf{1,800}$. Relation verified successfully. --- #### Solution to Question 3: 1. The runners will meet at a time duration that is a common multiple of $20, 30,$ and $45$. 2. To find the *first* time they meet, calculate $\text{LCM}(20, 30, 45)$. $$\begin{array}{r|l} 2 & 20, 30, 45 \\ \hline 2 & 10, 15, 45 \\ \hline 3 & 5, 15, 45 \\ \hline 3 & 5, 5, 15 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1, 1 \end{array}$$ $$\text{LCM} = 2 \times 2 \times 3 \times 3 \times 5 = 180\text{ seconds}$$ 3. Convert seconds to minutes: $$180\text{ seconds} = \frac{180}{60} = 3\text{ minutes}$$ 4. Add $3\text{ minutes}$ to the start time ($10:00\text{ AM}$). **Final Answer:** They will meet next at **$10:03\text{ AM}$**. --- #### Solution to Question 4: 1. Find the smallest number divisible by $8, 10,$ and $12$ by taking their LCM. $$8 = 2^3$$ $$10 = 2 \times 5$$ $$12 = 2^2 \times 3$$ $$\text{LCM}(8, 10, 12) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$$ 2. Any number divisible by $8, 10,$ and $12$ must be a multiple of $120$. 3. The largest $3$-digit number is $999$. Divide $999$ by $120$ to find the remainder: $$999 \div 120 = 8 \text{ with a remainder of } 39$$ 4. Subtract the remainder from $999$: $$\text{Required Number} = 999 - 39 = 960$$ **Final Answer:** The greatest 3-digit number is **$960$**. --- ## 6. Exam Revision & FAQs ### FAQ 1: Can two numbers have $15$ as their HCF and $110$ as their LCM? Explain. **Answer:** No. The **HCF of two numbers must always completely divide their LCM**. Let us divide the given LCM ($110$) by the given HCF ($15$): $$110 \div 15 = 7 \text{ with a remainder of } 5$$ Since $110$ is not perfectly divisible by $15$, it is mathematically impossible for two natural numbers to have an HCF of $15$ and an LCM of $110$. --- ### FAQ 2: What is the HCF of: (a) two consecutive numbers, (b) two consecutive even numbers, and (c) two consecutive odd numbers? **Answer:** * **(a) Two consecutive numbers:** Always **$1$**. * *Reason:* Any two consecutive numbers (e.g., $14, 15$) have no common factor other than $1$; they are always co-prime. * **(b) Two consecutive even numbers:** Always **$2$**. * *Reason:* Two consecutive even numbers (e.g., $18, 20$) are both divisible by $2$, and their difference is $2$. * **(c) Two consecutive odd numbers:** Always **$1$**. * *Reason:* Two consecutive odd numbers (e.g., $21, 23$) share no common factors other than $1$. --- ### FAQ 3: State the main differences between Twin Primes and Co-Primes. **Answer:** 1. **Individual Primality Requirement:** * **Twin Primes:** Both numbers in the pair **must be prime numbers** (e.g., $5$ and $7$). * **Co-Primes:** Neither number is required to be prime. Composite numbers can be co-prime to each other (e.g., $8$ and $9$). 2. **Numerical Difference Rule:** * **Twin Primes:** The absolute numerical difference between the two prime numbers must be exactly $2$ ($|a - b| = 2$). * **Co-Primes:** There is no fixed difference required between the numbers (e.g., $4$ and $35$ are co-prime). 3. **Core Definition:** * **Twin Primes:** A set of prime numbers that differ by $2$. * **Co-Primes:** Any pair of numbers whose $\text{HCF} = 1$.

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