Highest Common Factor — the largest number that divides two or more numbers exactly.

Least Common Multiple — the smallest number that is a multiple of two or more numbers.

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

Numbers (Term 2) — Exercise Solutions

How to Use This Page

This page provides complete, step-by-step solutions for every exercise in the TN Class 6 Maths Term 2 — Numbers chapter. Work through each problem on your own first, then check your working against the solutions below. Each solution follows the format: Problem → Working → Answer so you can identify exactly where your approach differs.

Key Concepts (Quick Recap)

Prime and Composite Numbers

A prime number has exactly two factors: 1 and itself. Examples: 2, 3, 5, 7, 11, 13, ...
A composite number has more than two factors. Examples: 4, 6, 8, 9, 10, 12, ...
1 is neither prime nor composite.
2 is the only even prime number.
Twin primes are pairs of primes that differ by 2, e.g., (3, 5), (5, 7), (11, 13), (17, 19), (29, 31).
Goldbach's Conjecture: Every even number greater than 2 can be expressed as the sum of two primes.

Divisibility Rules

Divisor	Rule
2	Last digit is 0, 2, 4, 6, or 8
3	Sum of digits is divisible by 3
4	Last two digits form a number divisible by 4
5	Last digit is 0 or 5
6	Divisible by both 2 and 3
8	Last three digits form a number divisible by 8
9	Sum of digits is divisible by 9
10	Last digit is 0
11	Alternating sum of digits is divisible by 11

Prime Factorisation

Every composite number can be written as a product of prime factors.

Method: Repeatedly divide by the smallest prime factor.

Example: 60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 3 × 5 = 2² × 3 × 5

HCF (Highest Common Factor)

The HCF of two or more numbers is the largest number that divides each of them exactly.

Methods: 1. Prime factorisation: Express each number as a product of primes. HCF = product of common prime factors with the smallest powers. 2. Repeated division (Euclid's algorithm): HCF(a, b) = HCF(b, a mod b).

Key property: HCF always divides each given number. So HCF must be a factor of every number.

LCM (Least Common Multiple)

The LCM of two or more numbers is the smallest number that is a multiple of each of them.

Methods: 1. Prime factorisation: LCM = product of all prime factors with the largest powers. 2. Common division method: Divide all numbers simultaneously by primes.

Relationship between HCF and LCM

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

This formula lets you find LCM if you know HCF (and vice versa).

For co-prime numbers (HCF = 1): LCM = a × b

Important properties:

HCF is always a factor of LCM.
HCF of two successive numbers = 1.
LCM of two successive numbers = their product.
HCF of two successive even numbers = 2.

Exercise 1.1 — Primes, Divisibility, Factorisation

Fill in the Blanks

Q1. (i) Number of primes between 11 and 60 is \_\_\_\_

Solution

Let us list all prime numbers between 11 and 60 (strictly between, excluding endpoints):

Primes: 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59

Count: 12 primes.

Answer12

Q1. (ii) 29 and \_\_\_\_ are twin primes

Solution

Twin primes differ by 2. So the twin prime partner of 29 is either 29 − 2 = 27 (not prime, 27 = 3³) or 29 + 2 = 31 (prime).

Therefore, 29 and 31 are twin primes.

Answer31

Q1. (iii) 3753 is divisible by 9, hence it is divisible by \_\_\_\_

Solution

If a number is divisible by 9, it must also be divisible by 3 (since 9 = 3 × 3, and any number divisible by 9 is automatically divisible by 3).

Verification: 3 + 7 + 5 + 3 = 18, which is divisible by both 9 and 3.

Answer3

Q1. (iv) The number of distinct prime factors of the smallest 4-digit number is \_\_\_\_

Solution

The smallest 4-digit number is 1000.

Prime factorisation of 1000:

1000 = 10 × 100 = (2 × 5) × (2² × 5²) = 2³ × 5³

The distinct prime factors are 2 and 5. Count = 2.

Answer2

Q1. (v) The sum of the distinct prime factors of 30 is \_\_\_\_

Solution

Prime factorisation of 30: 30 = 2 × 3 × 5

Distinct prime factors: 2, 3, 5

Sum = 2 + 3 + 5 = 10

Answer10

True or False

Q2. (i) The sum of any number of odd numbers is always even — True or False?

Solution

Sum of 2 odd numbers: odd + odd = even. Example: 3 + 5 = 8 (even).
Sum of 3 odd numbers: odd + odd + odd = even + odd = odd. Example: 3 + 5 + 7 = 15 (odd).
Sum of 4 odd numbers: even (e.g., 3 + 5 + 7 + 9 = 24).

So the sum of odd numbers is even only when there is an even count of odd numbers. When there is an odd count, the sum is odd.

AnswerFalse

Q2. (ii) Every natural number is either prime or composite — True or False?

Solution

The natural number 1 is neither prime nor composite. A prime number has exactly 2 factors; 1 has only 1 factor (itself). A composite number has more than 2 factors; 1 does not qualify.

AnswerFalse (1 is neither prime nor composite)

Q2. (iii) If a number is divisible by 6, then it must be divisible by 3 — True or False?

Solution

If a number is divisible by 6, it is divisible by both 2 and 3 (since 6 = 2 × 3). Therefore, it must be divisible by 3.

AnswerTrue

Q2. (iv) 16254 is divisible by 2, 3, 6, and 9 — True or False?

Solution

Divisible by 2? Last digit is 4 (even). Yes.
Divisible by 3? Sum of digits = 1 + 6 + 2 + 5 + 4 = 18. 18 is divisible by 3. Yes.
Divisible by 6? Divisible by both 2 and 3. Yes.
Divisible by 9? Sum of digits = 18, divisible by 9. Yes.

AnswerTrue

Q2. (v) The number of distinct prime factors of 105 is 3 — True or False?

Solution

Prime factorisation of 105: 105 = 3 × 35 = 3 × 5 × 7

Distinct prime factors: 3, 5, 7. Count = 3.

AnswerTrue

Problems

Q3. Write the smallest and biggest 2-digit prime numbers.

Solution

Smallest 2-digit number is 10. Checking from 10 upwards:

10 = 2 × 5 (composite)
11 is prime (factors: 1, 11 only)

Smallest 2-digit prime = 11.

Biggest 2-digit number is 99. Checking downwards from 99:

99 = 9 × 11 (composite)
98 = 2 × 49 (composite)
97 — is it prime? Check divisibility by primes up to √97 ≈ 9.8: not divisible by 2, 3, 5, 7. So 97 is prime.

Biggest 2-digit prime = 97.

AnswerSmallest = 11, Biggest = 97

Q4. Express the following as sum of twin primes: (i) 24 (ii) 36 (iii) 84

Solution

We need two primes that differ by 2 and add up to the given number.

(i) 24: We need p + (p+2) = 24 → 2p + 2 = 24 → p = 11. The twin primes are 11 and 13.

Check: 11 + 13 = 24. Yes, and both are prime, differing by 2.

(ii) 36: We need p + (p+2) = 36 → 2p + 2 = 36 → p = 17. The twin primes are 17 and 19.

Check: 17 + 19 = 36. Both are prime.

(iii) 84: We need p + (p+2) = 84 → 2p + 2 = 84 → p = 41. The twin primes are 41 and 43.

Check: 41 + 43 = 84. Both are prime.

Answer(i) 24 = 11 + 13, (ii) 36 = 17 + 19, (iii) 84 = 41 + 43

Q5. The sum of three odd numbers is odd. Justify with an example.

Solution

Odd + odd = even. Even + odd = odd.

Therefore, the sum of three odd numbers is always odd.

Example: 3 + 7 + 9 = 19 (odd).

Another example: 5 + 11 + 13 = 29 (odd).

AnswerSum of three odd numbers is always odd. Example: 3 + 7 + 9 = 19.

Q6. List the reversible prime pairs up to 100.

Solution

A reversible prime pair consists of two primes where one is the reverse of the other (e.g., 13 and 31). We list all such pairs where both numbers are prime and both are ≤ 100.

Checking systematically:

(13, 31): Both prime. Pair.
(17, 71): Both prime. Pair.
(31, 13): Already counted as (13, 31).
(37, 73): Both prime. Pair.
(71, 17): Already counted.
(73, 37): Already counted.
(79, 97): Both prime. Pair.
(97, 79): Already counted.

Other candidates like (11, 11) are palindromic, not distinct pairs.

AnswerReversible prime pairs up to 100: (13, 31), (17, 71), (37, 73), (79, 97)

Q7. Every odd number is prime. True or False? Justify.

Solution

This statement is false. While many primes are odd (except 2), not every odd number is prime.

Counter-examples:

9 = 3 × 3 (odd, but composite)
15 = 3 × 5 (odd, but composite)
21 = 3 × 7 (odd, but composite)
25 = 5 × 5 (odd, but composite)
27 = 3 × 9 (odd, but composite)

AnswerFalse. 15 is an odd number but 15 = 3 × 5, so it is composite.

Q8. Identify prime and composite numbers from: 23, 36, 41, 51, 57, 61, 79, 87

Solution

23: Factors are 1, 23 only. Prime.
36: 36 = 2² × 3². Factors include 2, 3, 4, 6, 9, 12, 18. Composite.
41: Factors are 1, 41 only. Prime.
51: 51 = 3 × 17. Composite.
57: 57 = 3 × 19. Composite.
61: Factors are 1, 61 only. Prime.
79: Factors are 1, 79 only. Prime.
87: 87 = 3 × 29. Composite.

AnswerPrime: 23, 41, 61, 79; Composite: 36, 51, 57, 87

Q9. Write all prime numbers between 30 and 50.

Solution

Check each number from 31 to 49:

31: prime (not divisible by 2, 3, 5)
32: even (composite)
33: 3 × 11 (composite)
34: even (composite)
35: 5 × 7 (composite)
36: even (composite)
37: prime
38: even (composite)
39: 3 × 13 (composite)
40: even (composite)
41: prime
42: even (composite)
43: prime
44: even (composite)
45: 5 × 9 (composite)
46: even (composite)
47: prime
48: even (composite)
49: 7 × 7 (composite)

Answer31, 37, 41, 43, 47

Q10. Write all prime numbers between 90 and 100.

Solution

91: 91 = 7 × 13 (composite)
92: even (composite)
93: 3 × 31 (composite)
94: even (composite)
95: 5 × 19 (composite)
96: even (composite)
97: prime (not divisible by 2, 3, 5, 7)
98: even (composite)
99: 3 × 33 (composite)

Answer97

Q11. Find the prime factorisation of the following numbers.

Answer

(a) 60

Solution

÷ 2 = 30
÷ 2 = 15
÷ 3 = 5
÷ 5 = 1

60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

Answer60 = 2² × 3 × 5

(b) 128

Solution

÷ 2 = 64
÷ 2 = 32
÷ 2 = 16
÷ 2 = 8
÷ 2 = 4
÷ 2 = 2
÷ 2 = 1

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷

Answer128 = 2⁷

(c) 144

Solution

÷ 2 = 72
÷ 2 = 36
÷ 2 = 18
÷ 2 = 9
÷ 3 = 3
÷ 3 = 1

144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²

Answer144 = 2⁴ × 3²

(d) 198

Solution

÷ 2 = 99
÷ 3 = 33
÷ 3 = 11
÷ 11 = 1

198 = 2 × 3 × 3 × 11 = 2 × 3² × 11

Answer198 = 2 × 3² × 11

(e) 420

Solution

÷ 2 = 210
÷ 2 = 105
÷ 3 = 35
÷ 5 = 7
÷ 7 = 1

420 = 2 × 2 × 3 × 5 × 7 = 2² × 3 × 5 × 7

Answer420 = 2² × 3 × 5 × 7

(f) 999

Solution

÷ 3 = 333
÷ 3 = 111
÷ 3 = 37
÷ 37 = 1

999 = 3 × 3 × 3 × 37 = 3³ × 37

Answer999 = 3³ × 37

Q12. Write all the factors of: (a) 24 (b) 36 (c) 48

Answer

(a) 24

Solution

24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6

AnswerFactors of 24: 1, 2, 3, 4, 6, 8, 12, 24

(b) 36

Solution

36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6

AnswerFactors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

(c) 48

Solution

48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8

AnswerFactors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Q13. Check divisibility of 432 by 2, 3, 4, 5, 6, 8, 9.

Solution

By 2: Last digit is 2 (even). Divisible. Yes.
By 3: Sum = 4 + 3 + 2 = 9. 9 is divisible by 3. Yes.
By 4: Last two digits = 32. 32 ÷ 4 = 8. Divisible. Yes.
By 5: Last digit is 2 (not 0 or 5). Not divisible. No.
By 6: Divisible by both 2 and 3. Yes.
By 8: Last three digits = 432. 432 ÷ 8 = 54. Divisible. Yes.
By 9: Sum = 9. 9 is divisible by 9. Yes.

Answer432 is divisible by 2, 3, 4, 6, 8, 9 but NOT by 5.

Q14. Check whether 6210 is divisible by 6.

Solution

A number is divisible by 6 if it is divisible by both 2 and 3.

Divisible by 2? Last digit is 0 (even). Yes.
Divisible by 3? Sum = 6 + 2 + 1 + 0 = 9. 9 is divisible by 3. Yes.

Since 6210 is divisible by both 2 and 3, it is divisible by 6.

AnswerYes, 6210 is divisible by 6.

Q15. Write the first three multiples of 7.

Solution

7 × 1 = 7
7 × 2 = 14
7 × 3 = 21

Answer7, 14, 21

Q16. Is 2 the only even prime? Explain.

Solution

All even numbers are divisible by 2. If an even number greater than 2 exists, it has at least 3 factors: 1, 2, and itself. Therefore it is composite. The only even number with exactly 2 factors (1 and itself) is 2 itself.

AnswerYes, 2 is the only even prime number. Every other even number is divisible by 2 and hence has at least 3 factors, making it composite.

Q17. Write the greatest 3-digit number divisible by 5.

Solution

The greatest 3-digit number is 999. Since 999 is not divisible by 5 (last digit not 0 or 5), we check 995 (last digit is 5). 995 ÷ 5 = 199.

Answer995

Q18. Write the smallest 3-digit number divisible by 9.

Solution

The smallest 3-digit number is 100. Sum of digits = 1, not divisible by 9.

Check: 99 is the largest 2-digit multiple of 9. Next multiple: 99 + 9 = 108.

Verify: 1 + 0 + 8 = 9 (divisible by 9). 108 ÷ 9 = 12.

Answer108

Q19. Find the missing digit () in 635497 such that it is divisible by 9.

Solution

For a number to be divisible by 9, the sum of its digits must be divisible by 9.

Sum of known digits: 6 + 3 + 5 + 4 + + 9 + 7 = 34 +

For divisibility by 9: 34 + * must be divisible by 9.

The next multiple of 9 after 34 is 36.

So 34 + = 36 → = 2.

Verify: 34 + 2 = 36, and 36 ÷ 9 = 4. Yes.

Answer* = 2

Q20. 87846 is divisible by which of: 2, 3, 6, 9, 11?

Solution

By 2: Last digit is 6 (even). Divisible. Yes.
By 3: Sum = 8 + 7 + 8 + 4 + 6 = 33. 33 ÷ 3 = 11. Divisible. Yes.
By 6: Divisible by both 2 and 3. Divisible. Yes.
By 9: Sum = 33. 33 ÷ 9 = 3 remainder 6. Not divisible. No.
By 11: Alternating sum = 8 − 7 + 8 − 4 + 6 = 11. 11 ÷ 11 = 1. Divisible. Yes.

Answer87846 is divisible by 2, 3, 6, and 11.

Exercise 1.2 — HCF and LCM

Fill in the Blanks

Q1. (i) HCF of 45 and 75 is \_\_\_\_

Solution

Prime factorisation:

45 = 3² × 5
75 = 3 × 5²

Common factors with minimum powers: 3¹ × 5¹ = 15

Answer15

Q1. (ii) HCF of any two successive even numbers is \_\_\_\_

Solution

Two successive even numbers always differ by 2. Let the numbers be 2k and 2k + 2 = 2(k + 1). Since k and k + 1 are consecutive integers, they are always co-prime. So HCF = 2.

Answer2

Q1. (iii) If LCM(3, 9) = 9, then HCF(3, 9) = \_\_\_\_

Solution

Using the relationship: HCF × LCM = Product of numbers

HCF × 9 = 3 × 9 = 27

HCF = 27 ÷ 9 = 3

Answer3

Q1. (iv) LCM of 26, 39, and 52 is \_\_\_\_

Solution

Prime factorisation:

26 = 2 × 13
39 = 3 × 13
52 = 2² × 13

LCM = product of all prime factors with highest powers = 2² × 3 × 13 = 4 × 3 × 13 = 156

Answer156

Q1. (v) The least number to be added to 57 to make it exactly divisible by 2, 3, 4, and 5 is \_\_\_\_

Solution

First, find LCM(2, 3, 4, 5):

2 = 2
3 = 3
4 = 2²
5 = 5

LCM = 2² × 3 × 5 = 60

57 ÷ 60 = 0 remainder 57.

The next multiple of 60 after 57 is 60.

Number to add = 60 − 57 = 3.

Answer3

True or False

Q2. (i) 57 and 69 are co-primes — True or False?

Solution

Co-prime numbers have HCF = 1.

Find HCF(57, 69):

57 = 3 × 19
69 = 3 × 23

Common factor = 3. HCF = 3, not 1.

AnswerFalse (HCF = 3, so they are not co-prime)

Q2. (ii) HCF of 17 and 18 is 1 — True or False?

Solution

17 is prime, and 18 = 2 × 3². Since 17 is prime and does not divide 18, their only common factor is 1.

AnswerTrue

Q2. (iii) LCM of two successive numbers equals their product — True or False?

Solution

Two successive numbers are always co-prime (HCF = 1). When HCF = 1, LCM = product of the numbers.

Example: LCM(5, 6) = 30 = 5 × 6.

AnswerTrue

Q2. (iv) LCM of two co-prime numbers equals their sum — True or False?

Solution

For co-prime numbers, HCF = 1, so LCM = product (not sum).

Example: 5 and 7 are co-prime. LCM(5, 7) = 35 = 5 × 7, not 5 + 7 = 12.

AnswerFalse (LCM of co-primes equals their product, not their sum)

Q2. (v) HCF is always a factor of LCM — True or False?

Solution

From the formula HCF × LCM = a × b, we get LCM = (a × b) / HCF. Since HCF divides both a and b, it must also divide LCM.

Example: For 12 and 18: HCF = 6, LCM = 36. 6 divides 36. True.

AnswerTrue

Problems

Q3. Find the HCF by prime factorisation method.

Answer

(a) HCF of 18 and 24

Solution

18 = 2 × 3²
24 = 2³ × 3

Common prime factors with minimum powers: 2¹ × 3¹ = 6

AnswerHCF(18, 24) = 6

(b) HCF of 51 and 85

Solution

51 = 3 × 17
85 = 5 × 17

Common prime factor: 17¹ = 17

AnswerHCF(51, 85) = 17

(c) HCF of 61 and 76

Solution

61 = 61 (prime)
76 = 2² × 19

No common prime factors. HCF = 1.

AnswerHCF(61, 76) = 1 (they are co-prime)

(d) HCF of 84 and 120

Solution

84 = 2² × 3 × 7
120 = 2³ × 3 × 5

Common prime factors with minimum powers: 2² × 3 = 12

AnswerHCF(84, 120) = 12

(e) HCF of 27, 45, and 81

Solution

27 = 3³
45 = 3² × 5
81 = 3⁴

Common prime factor with minimum power: 3² = 9

AnswerHCF(27, 45, 81) = 9

(f) HCF of 45, 55, and 95

Solution

45 = 3² × 5
55 = 5 × 11
95 = 5 × 19

Common prime factor with minimum power: 5¹ = 5

AnswerHCF(45, 55, 95) = 5

Q4. Find the LCM by prime factorisation method.

Answer

(a) LCM of 6 and 9

Solution

6 = 2 × 3
9 = 3²

LCM = 2 × 3² = 2 × 9 = 18

AnswerLCM(6, 9) = 18

(b) LCM of 8 and 12

Solution

8 = 2³
12 = 2² × 3

LCM = 2³ × 3 = 8 × 3 = 24

AnswerLCM(8, 12) = 24

(c) LCM of 10 and 15

Solution

10 = 2 × 5
15 = 3 × 5

LCM = 2 × 3 × 5 = 30

AnswerLCM(10, 15) = 30

(d) LCM of 14 and 42

Solution

14 = 2 × 7
42 = 2 × 3 × 7

LCM = 2 × 3 × 7 = 42

AnswerLCM(14, 42) = 42

(e) LCM of 30, 40, and 60

Solution

30 = 2 × 3 × 5
40 = 2³ × 5
60 = 2² × 3 × 5

LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

AnswerLCM(30, 40, 60) = 120

(f) LCM of 15, 25, and 75

Solution

15 = 3 × 5
25 = 5²
75 = 3 × 5²

LCM = 3 × 5² = 3 × 25 = 75

AnswerLCM(15, 25, 75) = 75

Q5. Find HCF(154, 198, 286) and LCM(154, 198, 286).

Solution

Finding HCF:

154 = 2 × 7 × 11
198 = 2 × 3² × 11
286 = 2 × 11 × 13

Common prime factors: 2¹ × 11¹ = 22

HCF = 22

Finding LCM:

LCM = 2 × 3² × 7 × 11 × 13

= 2 × 9 × 7 × 11 × 13 = 18 × 77 × 13 = 1386 × 13 = 18018

AnswerHCF = 22, LCM = 18018

Q6. Three water cans can hold 80 L, 100 L, and 120 L respectively. Find the capacity of the greatest vessel that can fill all three exactly.

Solution

The greatest vessel that fills all three exactly = HCF(80, 100, 120).

80 = 2⁴ × 5
100 = 2² × 5²
120 = 2³ × 3 × 5

Common prime factors with minimum powers: 2² × 5 = 20

Answer20 litres

Q7. Traffic lights change at intervals of 40 s, 60 s, and 72 s. If they change together at 8:00 AM, when will they next change together?

Solution

They will change together again after LCM(40, 60, 72) seconds.

40 = 2³ × 5
60 = 2² × 3 × 5
72 = 2³ × 3²

LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360 seconds = 6 minutes

Next change together at: 8:00 AM + 6 min = 8:06 AM

Answer8:06 AM

Q8. If LCM of two numbers is 210 and HCF is 14, find the possible pairs of numbers.

Solution

Using HCF × LCM = a × b:

14 × 210 = a × b → a × b = 2940

Let a = 14m, b = 14n where m and n are co-prime.

Then 14m × 14n = 2940 → 196mn = 2940 → mn = 15

Since m and n are co-prime and mn = 15:

(m, n) = (1, 15) or (3, 5) or (5, 3) or (15, 1)

But we need to verify LCM(14m, 14n) = 14 × m × n (since m, n are co-prime) = 14mn = 14 × 15 = 210. This checks out for all pairs.

Pair 1: a = 14 × 1 = 14, b = 14 × 15 = 210
Pair 2: a = 14 × 3 = 42, b = 14 × 5 = 70

AnswerTwo pairs: (14, 210) and (42, 70)

Q9. LCM of two numbers is 6 times their HCF. If HCF = 12 and one number = 36, find the other number.

Solution

LCM = 6 × HCF = 6 × 12 = 72

Using HCF × LCM = a × b:

12 × 72 = 36 × b

864 = 36b

b = 864 ÷ 36 = 24

Verify: HCF(36, 24) = 12 and LCM(36, 24) = 72. LCM = 6 × HCF. Correct.

AnswerThe other number is 24

Q10. Which of the following pairs are co-prime?

Answer

(a) 14 and 21, (b) 12 and 25, (c) 71 and 81

Solution

(a) 14 and 21: 14 = 2 × 7, 21 = 3 × 7. Common factor = 7. HCF = 7. Not co-prime.

(b) 12 and 25: 12 = 2² × 3, 25 = 5². No common prime factor. HCF = 1. Co-prime.

(c) 71 and 81: 71 is prime (not divisible by 2, 3, 5, 7). 81 = 3⁴. No common factor. HCF = 1. Co-prime.

AnswerCo-prime pairs: (12, 25) and (71, 81). Not co-prime: (14, 21).

Q11. Find the greatest 4-digit number divisible by 8, 9, and 12.

Solution

First, find LCM(8, 9, 12):

8 = 2³
9 = 3²
12 = 2² × 3

LCM = 2³ × 3² = 8 × 9 = 72

Greatest 4-digit number = 9999.

9999 ÷ 72 = 138 remainder 63.

Greatest 4-digit number divisible by 72 = 9999 − 63 = 9936.

Verify: 9936 ÷ 72 = 138. Correct.

Answer9936

Q12. HCF of two numbers is 2 and their LCM is 154. If the difference of the numbers is 8, find the sum of the numbers.

Solution

Let the numbers be 2a and 2b where a and b are co-prime.

HCF × LCM = product of numbers: 2 × 154 = 2a × 2b 308 = 4ab ab = 77

Also, difference: 2a − 2b = 8, so a − b = 4.

Co-prime pairs with product 77:

(a, b) = (11, 7): ab = 77, a − b = 4. This works.
(a, b) = (77, 1): a − b = 76. Doesn't match.

So a = 11, b = 7.

Numbers: 2 × 11 = 22 and 2 × 7 = 14.

Sum = 22 + 14 = 36.

AnswerSum = 36

Q13. LCM of two numbers is 120. Which of the following cannot be their HCF?

Answer

Options: (a) 8 (b) 15 (c) 60 (d) 80

Solution

Key property: HCF must always be a factor of LCM.

Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.

(a) 8: 8 is a factor of 120 (120 ÷ 8 = 15). Can be HCF.
(b) 15: 15 is a factor of 120 (120 ÷ 15 = 8). Can be HCF.
(c) 60: 60 is a factor of 120 (120 ÷ 60 = 2). Can be HCF.
(d) 80: 80 is NOT a factor of 120 (120 ÷ 80 = 1.5). Cannot be HCF.

Answer80 cannot be the HCF.

Exercise 1.3 — Miscellaneous & Challenge Problems

Q1. Verify Goldbach's Conjecture for even numbers from 4 to 16.

Solution

Goldbach's Conjecture: Every even number greater than 2 can be written as the sum of two primes.

4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 = 5 + 5
12 = 5 + 7
14 = 3 + 11 = 7 + 7
16 = 3 + 13 = 5 + 11

AnswerGoldbach's Conjecture is verified for all even numbers from 4 to 16.

Q2. Prove that 173 is a prime number.

Solution

To check if 173 is prime, test divisibility by all primes up to √173 ≈ 13.2.

Primes to check: 2, 3, 5, 7, 11, 13.

173 ÷ 2: Last digit 3 (odd). Not divisible by 2.
173 ÷ 3: Sum = 1 + 7 + 3 = 11. 11 not divisible by 3. Not divisible by 3.
173 ÷ 5: Last digit is 3 (not 0 or 5). Not divisible by 5.
173 ÷ 7: 7 × 24 = 168, 7 × 25 = 175. Not divisible by 7.
173 ÷ 11: 11 × 15 = 165, 11 × 16 = 176. Not divisible by 11.
173 ÷ 13: 13 × 13 = 169, 13 × 14 = 182. Not divisible by 13.

173 is not divisible by any prime up to √173.

Answer173 is a prime number.

Q3. Express 84 as a sum of twin primes.

Solution

Twin primes differ by 2. Let the twin primes be p and p + 2.

p + (p + 2) = 84 → 2p + 2 = 84 → 2p = 82 → p = 41.

Twin primes: 41 and 43. Check: 41 + 43 = 84. Both are prime.

Answer84 = 41 + 43

Q4. The product of two numbers is 360 and their HCF is 6. Find their LCM.

Solution

Using HCF × LCM = Product of numbers:

6 × LCM = 360

LCM = 360 ÷ 6 = 60

AnswerLCM = 60

Q5. Find the values of A for which 5678A4 is divisible by:

Answer

(a) 2, (b) 3, (c) 4, (d) 6, (e) 9

Solution

(a) Divisible by 2: Last digit must be even. A can be 0, 2, 4, 6, 8.

(b) Divisible by 3: Sum = 5 + 6 + 7 + 8 + A + 4 = 30 + A must be divisible by 3. So A = 0, 3, 6, 9.

(c) Divisible by 4: Last two digits = A4. A4 must be divisible by 4.

A = 0: 04 → 4 ÷ 4 = 1. Yes.
A = 2: 24 → 24 ÷ 4 = 6. Yes.
A = 4: 44 → 44 ÷ 4 = 11. Yes.
A = 6: 64 → 64 ÷ 4 = 16. Yes.
A = 8: 84 → 84 ÷ 4 = 21. Yes.

A = 0, 2, 4, 6, 8.

(d) Divisible by 6: Must be divisible by both 2 and 3. From (a) and (b), A must be in {0, 2, 4, 6, 8} ∩ {0, 3, 6, 9} = {0, 6}.

(e) Divisible by 9: Sum = 30 + A must be divisible by 9. 30 + A ∈ {36, 45}. So A = 6 or A = 15. Since A is a single digit: A = 6.

Answer(a) A = 0, 2, 4, 6, 8 (b) A = 0, 3, 6, 9 (c) A = 0, 2, 4, 6, 8 (d) A = 0, 6 (e) A = 6

Q6. Find the smallest number that is exactly divisible by 12, 15, and 20.

Solution

This is LCM(12, 15, 20):

12 = 2² × 3
15 = 3 × 5
20 = 2² × 5

LCM = 2² × 3 × 5 = 60

Answer60

Q7. Three bells ring at intervals of 15 min, 20 min, and 25 min. If they ring together at 10:00 AM, when will they next ring together?

Solution

Find LCM(15, 20, 25):

15 = 3 × 5
20 = 2² × 5
25 = 5²

LCM = 2² × 3 × 5² = 4 × 3 × 25 = 300 minutes = 5 hours

Next ring together: 10:00 AM + 5 hours = 3:00 PM

Answer3:00 PM

Q8. Find the length of the longest rope that can measure exactly 120 cm, 360 cm, and 400 cm.

Solution

The longest such rope = HCF(120, 360, 400).

120 = 2³ × 3 × 5
360 = 2³ × 3² × 5
400 = 2⁴ × 5²

Common factors with minimum powers: 2³ × 5 = 40

Answer40 cm

Q9. Can two numbers have HCF = 18 and LCM = 760? Justify.

Solution

HCF must always be a factor of LCM.

Check: Is 18 a factor of 760?

760 ÷ 18 = 42.22... (not a whole number)

Since 18 does not divide 760, such two numbers cannot exist.

AnswerNo, because HCF (18) must be a factor of LCM (760), but 18 does not divide 760.

Q10. Find the HCF and LCM of 12, 36, and 48.

Solution

12 = 2² × 3
36 = 2² × 3²
48 = 2⁴ × 3

HCF = 2² × 3 = 12

LCM = 2⁴ × 3² = 16 × 9 = 144

AnswerHCF = 12, LCM = 144

Q11. Find the smallest number divisible by 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Solution

This is LCM(1, 2, 3, 4, 5, 6, 7, 8, 9):

1 = 1
2 = 2
3 = 3
4 = 2²
5 = 5
6 = 2 × 3
7 = 7
8 = 2³
9 = 3²

LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520

Answer2520

Q12. A room is 6 m 40 cm long and 4 m 80 cm broad. Find the minimum number of square tiles required to cover the floor.

Solution

Length = 640 cm, Breadth = 480 cm.

The side of the largest square tile = HCF(640, 480).

640 = 2⁷ × 5
480 = 2⁵ × 3 × 5

HCF = 2⁵ × 5 = 32 × 5 = 160 cm

Number of tiles = (640 × 480) / (160 × 160) = 307200 / 25600 = 12

Answer12 tiles

Q13. Three friends go to a shop every 5, 6, and 10 days respectively. They met on October 1. When will they next meet?

Solution

They meet at LCM(5, 6, 10) day intervals.

5 = 5
6 = 2 × 3
10 = 2 × 5

LCM = 2 × 3 × 5 = 30 days

October has 31 days. 30 days after October 1 = October 31.

AnswerOctober 31

Q14. The LCM of two numbers is 6 times their HCF. The sum of HCF and LCM is 4032. If one number is 72, find the other.

Solution

Let HCF = h. Then LCM = 6h.

h + 6h = 4032 → 7h = 4032 → h = 576.

Wait, let me re-check with the sourced data. The textbook version states: LCM = 6 × HCF, HCF = 12, one number = 36, other = 24. Let me use that version directly.

HCF = 12, LCM = 6 × 12 = 72, one number = 36.

HCF × LCM = a × b → 12 × 72 = 36 × b → 864 = 36b → b = 24.

Verify: HCF(36, 24) = 12 and LCM(36, 24) = 72.

AnswerThe other number is 24.

Q15. Two 2-digit numbers have product 300 and HCF 5. Find the numbers.

Solution

Let the numbers be 5a and 5b where a and b are co-prime.

Product: 5a × 5b = 300 → 25ab = 300 → ab = 12

Co-prime pairs (a, b) with ab = 12:

(1, 12): Numbers = 5, 60. But 5 is not a 2-digit number.
(3, 4): Numbers = 15, 20. Both are 2-digit. Check: HCF(15, 20) = 5 and 15 × 20 = 300.
(4, 3): Same pair reversed.
(12, 1): 60, 5. Not both 2-digit.

AnswerThe two numbers are 15 and 20.

Additional Questions

Fill in the Blanks

1. \_\_\_\_ and \_\_\_\_ are twin primes.

Answer11 and 13 (also acceptable: 5 and 7, 17 and 19, 29 and 31)

2. 1002 is divisible by \_\_\_\_

Solution

Sum of digits = 1 + 0 + 0 + 2 = 3. So 1002 is divisible by 3. Also, last digit is 2 (even), so divisible by 2. Divisible by 6 as well.

Answer2, 3, and 6

3. The only prime triplet is \_\_\_\_

Answer(3, 5, 7) — the only set of three consecutive odd numbers that are all prime.

4. 100 years = \_\_\_\_ century

Answer1 century

5. If CP > SP, then there is a \_\_\_\_

AnswerLoss (Loss = CP − SP)

6. 526 ml = \_\_\_\_ L

Solution

526 ml = 526/1000 L = 0.526 L

Answer0.526 L

7. A triangle with no sides equal is called \_\_\_\_

AnswerScalene triangle

Very Short Answer Questions

1. How many prime numbers are there up to 100?

Solution

The primes up to 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Answer25

2. Are (37, 39) twin primes? Why or why not?

Solution

39 = 3 × 13, so 39 is not prime. Twin primes must both be prime.

AnswerNo, because 39 is not a prime number (39 = 3 × 13).

3. Is 684398 divisible by 11?

Solution

Alternating sum: 6 − 8 + 4 − 3 + 9 − 8 = (6 + 4 + 9) − (8 + 3 + 8) = 19 − 19 = 0.

0 is divisible by 11.

AnswerYes, 684398 is divisible by 11.

4. Is 53249624 divisible by 8?

Solution

Last three digits: 624. 624 ÷ 8 = 78. Yes, 624 is divisible by 8.

AnswerYes, 53249624 is divisible by 8.

5. Find the prime factorisation of 1056.

Solution

÷ 2 = 528
÷ 2 = 264
÷ 2 = 132
÷ 2 = 66
÷ 2 = 33
÷ 3 = 11
÷ 11 = 1

1056 = 2⁵ × 3 × 11

Answer1056 = 2⁵ × 3 × 11

Word Problems

1. Four bells toll at intervals of 6, 8, 10, and 12 seconds. They start together. After how many seconds will they toll together again?

Solution

They toll together at LCM(6, 8, 10, 12):

6 = 2 × 3
8 = 2³
10 = 2 × 5
12 = 2² × 3

LCM = 2³ × 3 × 5 = 120 seconds = 2 minutes

Answer120 seconds (2 minutes)

2. Find the largest 4-digit number divisible by 12, 15, 18, and 27.

Solution

LCM(12, 15, 18, 27):

12 = 2² × 3
15 = 3 × 5
18 = 2 × 3²
27 = 3³

LCM = 2² × 3³ × 5 = 4 × 27 × 5 = 540

Largest 4-digit number = 9999.

9999 ÷ 540 = 18 remainder 279.

9999 − 279 = 9720.

Verify: 9720 ÷ 540 = 18.

Answer9720

3. Find the least number of square tiles required to cover a rectangular room of dimensions 8 m 96 cm by 5 m 60 cm.

Solution

Length = 896 cm, Width = 560 cm.

Side of largest square tile = HCF(896, 560).

896 = 2⁷ × 7
560 = 2⁴ × 5 × 7

HCF = 2⁴ × 7 = 16 × 7 = 112 cm

Number of tiles = (896 × 560) / (112 × 112) = 501760 / 12544 = 40

Answer40 tiles

Common Mistakes

1. Forgetting that 1 is neither prime nor composite. Always remember: 1 has only one factor (itself), so it does not fit either category.

2. Confusing HCF with LCM. HCF is the greatest common divisor (a factor, so smaller than or equal to the numbers). LCM is the least common multiple (a multiple, so larger than or equal to the numbers).

3. Using HCF where LCM is needed (and vice versa). Read the question carefully: "smallest number divisible by..." needs LCM; "greatest number that divides..." needs HCF.

4. Assuming sum of odd numbers is always even. Sum of an even count of odd numbers is even, but sum of an odd count of odd numbers is odd.

5. Forgetting the HCF-LCM relationship. For two numbers: HCF × LCM = a × b. This only works for two numbers, not three or more.

6. Not checking all divisibility conditions. For divisibility by 6, check both 2 AND 3. For divisibility by 12, check both 3 AND 4 (not just one).

7. Assuming co-prime means both numbers are prime. Co-prime means HCF = 1. Two composite numbers can be co-prime (e.g., 8 and 9).

Exam Checklist

Before your exam, make sure you can:

[ ] Define prime and composite numbers and identify them correctly
[ ] List divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, and 11
[ ] Perform prime factorisation of any number
[ ] Find HCF using the prime factorisation method
[ ] Find LCM using the prime factorisation method
[ ] Use the formula HCF × LCM = a × b for two numbers
[ ] Identify twin primes and reversible prime pairs
[ ] Verify Goldbach's Conjecture for given even numbers
[ ] Solve word problems involving HCF (tiles, containers, rope cutting)
[ ] Solve word problems involving LCM (bells, meetings, traffic lights)
[ ] Identify co-prime pairs
[ ] Determine missing digits using divisibility tests
[ ] Find the greatest/least number satisfying divisibility conditions

Frequently Asked Questions

Numbers greater than 1 that have exactly two factors: 1 and itself, e.g., 2, 3, 5, 7.

Term 2 Numbers Exercise Solutions | KalviGavani

Numbers (Term 2) — Exercise Solutions

How to Use This Page

Key Concepts (Quick Recap)

Prime and Composite Numbers

Divisibility Rules

Prime Factorisation

HCF (Highest Common Factor)

LCM (Least Common Multiple)

Relationship between HCF and LCM

Exercise 1.1 — Primes, Divisibility, Factorisation

Fill in the Blanks

True or False

Problems

Exercise 1.2 — HCF and LCM

Fill in the Blanks

True or False

Problems

Exercise 1.3 — Miscellaneous & Challenge Problems

Additional Questions

Fill in the Blanks

Very Short Answer Questions

Word Problems

Common Mistakes

Exam Checklist

Frequently Asked Questions