Samacheer Class 6 Maths Numbers Book Back Answers & Solutions

Q.1Fill in the blanks. (i) The number of prime numbers between 11 and 60 is _______ (ii) The numbers 29 and _______ are twin primes. (iii) 3753 is divisible by 9 and hence divisible by _______ (iv) The number of distinct prime factors of the smallest 4 digit number is______ (v) The sum of distinct prime factors of 30 is ________v

Solution

(i) 12
(ii) 31
(iii) 3
(iv) 2
(v) 10
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

(i) 12
(ii) 31
(iii) 3
(iv) 2
(v) 10
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.2Say True or False. (i) The sum of any number of odd numbers is always even. (ii) Every natural number is either prime or composite. (iii) If a number is divisible by 6, then it must be divisible by 3. (iv) 16254 is divisible by 2, 3, 6, and 9. (v) The number of distinct prime factors of 105 is 3.v

Solution

(i) False
(ii) False
(iii) True
(iv) True
(v) True

Answer:

(i) False
(ii) False
(iii) True
(iv) True
(v) True

Q.3Write the smallest and the biggest two-digit prime number.v

Solution

Smallest two-digit prime number – 11
Biggest two-digit prime number – 97.

Answer:

Smallest two-digit prime number – 11
Biggest two-digit prime number – 97.

Q.4Write the smallest and the biggest three-digit composite number.v

Solution

Smallest three-digit composite number – 100
Biggest three-digit composite number – 999
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

Smallest three-digit composite number – 100
Biggest three-digit composite number – 999
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.5The sum of any three odd natural number is odd. Justify this statement with an example.v

Solution

True.
1 + 3 + 5 = 9 (Odd)

Answer:

True.
1 + 3 + 5 = 9 (Odd)

Q.6The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pair exists up to 100.v

Solution

(17, 71), (37, 73) and (79, 97)

Answer:

(17, 71), (37, 73) and (79, 97)

Q.7Your friend says that every odd number is prime. Give an example to prove him/her wrong.v

Solution

False. 15 is an odd number. But not prime.

Answer:

False. 15 is an odd number. But not prime.

Q.8Each of the composite numbers has at least three factors. Justify this statement with an example.v

Solution

Factors of 4 are 1, 2, 4
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

Factors of 4 are 1, 2, 4
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.9Find the dates of any month in a calendar which are divisible by both 2 and 3.v

Solution

Every month the dates 6, 12, 18, 24, and 30 (excluding February) are divisible by both 2 and 3.

Answer:

Every month the dates 6, 12, 18, 24, and 30 (excluding February) are divisible by both 2 and 3.

Q.10I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. Who am I?v

Solution

Two-digit prime numbers with a sum of digits 10 are 19, 37, 73.
of these factors of 57 is 19. the number is 19.

Answer:

Two-digit prime numbers with a sum of digits 10 are 19, 37, 73.
of these factors of 57 is 19. the number is 19.

Q.11Find the prime factorisation of each number by factor tree method and division method. (i) 60 (ii) 128 (iii) 144 (iv) 198 (v) 420 (vi) 999v

Solution

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 1
60 = 2 × 30 = 2 × 2 × 15
= 2 × 2 × 3 × 5

128 = 2 × 2 × 2 × 2 × 2 ×2 × 2
128 = 2 × 64 = 2 × 2 × 32
= 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8
128 = 2 × 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2 × 2

144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 2 × 72 = 2 × 2 × 36
= 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 4
198 = 2 × 3 × 3 × 11
198 = 2 × 99 = 2 × 3 × 33 = 2 × 3 × 3 × 11

420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7

999 = 3 × 333 = 3 × 3 × 111
= 3 × 3 × 3 × 37

Answer:

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 1
60 = 2 × 30 = 2 × 2 × 15
= 2 × 2 × 3 × 5

128 = 2 × 2 × 2 × 2 × 2 ×2 × 2
128 = 2 × 64 = 2 × 2 × 32
= 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8
128 = 2 × 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2 × 2

144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 2 × 72 = 2 × 2 × 36
= 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 4
198 = 2 × 3 × 3 × 11
198 = 2 × 99 = 2 × 3 × 33 = 2 × 3 × 3 × 11

420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7

999 = 3 × 333 = 3 × 3 × 111
= 3 × 3 × 3 × 37

Q.12If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.v

Solution

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 7
143 = 11 × 13 (11, 13) or (13, 11)
143 = 11 × 13

Answer:

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1 7
143 = 11 × 13 (11, 13) or (13, 11)
143 = 11 × 13

Q.13The difference between two successive odd numbers isv

A. 1
B. 2
C. 3
D. 0

Solution

(b) 2
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

(b) 2
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.14The only even prime number isv

A. 4
B. 6
C. 2
D. 0

Solution

(c) 2

Answer:

(c) 2

Q.15Which of the following numbers is not a prime?v

A. 53
B. 92
C. 97
D. 71

Solution

(b) 92

Answer:

(b) 92

Q.16The sum of the factors of 27 isv

A. 28
B. 37
C. 40
D. 31

Solution

(c) 40
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

(c) 40
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.17The factors of the number are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. What is the number?v

A. 80
B. 100
C. 128
D. 160

Solution

(a) 80

Answer:

(a) 80

Q.18The prime factorisation of 60 is 2 × 2 × 3 × 5. Any other number which has the same prime factorisation as 60 isv

A. 30
B. 120
C. 90
D. Impossible

Solution

(d) Impossible

Answer:

(d) Impossible

Q.19If the number 6354*97 is divisible by 9 then the value of * isv

A. 2
B. 4
C. 6
D. 7

Solution

(a) 2
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Answer:

(a) 2
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.1

Q.20The number 87846 is divisible byv

A. 2 only
B. 3 only
C. 11 only
D. all of these

Solution

(d) all of these
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi

Answer:

(d) all of these
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi

Q.1Fill in the blanks (i) The HCF of 45 and 75 is ……. (ii) The HCF of two successive even numbers is ……. (iii) If the LCM of 3 and 9 is 9, then their HCF is ……… (iv) The LCM of 26, 39 and 52 is …….. (v) The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ………v

Solution

(i) 15
(ii) 2
(iii) 3
(iv) 156
(v) 3
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
The Least common multiple (LCM) of 15 and 20 is 60.

Answer:

(i) 15
(ii) 2
(iii) 3
(iv) 156
(v) 3
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
The Least common multiple (LCM) of 15 and 20 is 60.

Q.2Say True or False (i) The numbers 57 and 69 are co-primes. (ii) The HCF of 17 and 18 is 1. (iii) The LCM of two successive numbers is the product of the numbers. (iv) The LCM of two co-primes is the sum of the numbers. (v) The HCF of two numbers is always a factor of their LCM.v

Solution

(i) False
(ii) True
(iii) True
(iv) False
(v) True
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
So, now the prime factorization of 84 with the upside-down division method is 2 × 2 × 3 × 7.

Answer:

(i) False
(ii) True
(iii) True
(iv) False
(v) True
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
So, now the prime factorization of 84 with the upside-down division method is 2 × 2 × 3 × 7.

Q.3Find the HCF of each set of numbers using the prime factorisation method. (i) 18, 24 (ii) 51, 85 (iii) 61, 76 (iv) 84, 120 (v) 27, 45, 81 (vi) 45, 55, 95v

Solution

(i) 18, 24
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 1
HCF = 2 × 3 = 6
(ii) 51, 85

HCF = 17

(iii) 61, 76
61 = 1 × 61
For 61, 76, the common factor is 1
HCF = 1
(iv) 84, 120

HCF = 2 × 2 × 3 = 12
(v) 27, 45, 81

HCF = 3 × 3 = 9
(vi) 45, 55, 95

Factors of 55 = 5 × 11
Factors of 45 = 3 × 3 × 5
Factors of 95 = 5 × 19
HCF = 5
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Answer:

(i) 18, 24
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 1
HCF = 2 × 3 = 6
(ii) 51, 85

HCF = 17

(iii) 61, 76
61 = 1 × 61
For 61, 76, the common factor is 1
HCF = 1
(iv) 84, 120

HCF = 2 × 2 × 3 = 12
(v) 27, 45, 81

HCF = 3 × 3 = 9
(vi) 45, 55, 95

Factors of 55 = 5 × 11
Factors of 45 = 3 × 3 × 5
Factors of 95 = 5 × 19
HCF = 5
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Q.4Find the LCM of each set of numbers using the prime factorisation method. (i) 6, 9 (ii) 8, 12 (iii) 10, 15 (iv) 14, 42 (v) 30, 40, 60 (vi) 15, 25, 75v

Solution

(i) 6, 9
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 6
Factors of 6 = 2 × 3
Factors of 9 = 3 × 3
LCM of 6 and 9 = 3 × 2 × 3 = 18
(ii) 8, 12

(iii) 10, 15

Factors of 10 = 2 × 5
Factors of 15 = 3 × 5
LCM of 10 and 15 = 5 × 2 × 3 = 30

(iv) 14, 42

Factors of 14 = 2 × 7
Factors of 42 = 2 × 3 × 7
LCM of 14 and 42 = 7 × 2 × 3 = 42
(v) 30, 40, 60
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 10
Factors of 30 = 2 × 3 × 5
Factors of 40 = 2 × 2 × 2 × 5
Factors of 60 = 2 × 2 × 3 × 5
LCM of 30, 40, 60 = 2 × 5 × 3 × 2 × 2 = 120
(vi) 15, 25, 75

Factors of 15 = 3 × 5
Factors of 25 =5 × 5
Factors of 75 = 3 × 5 × 5
LCM of 15, 25, 75 = 5 × 3 × 5 = 75
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Answer:

(i) 6, 9
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 6
Factors of 6 = 2 × 3
Factors of 9 = 3 × 3
LCM of 6 and 9 = 3 × 2 × 3 = 18
(ii) 8, 12

(iii) 10, 15

Factors of 10 = 2 × 5
Factors of 15 = 3 × 5
LCM of 10 and 15 = 5 × 2 × 3 = 30

(iv) 14, 42

Factors of 14 = 2 × 7
Factors of 42 = 2 × 3 × 7
LCM of 14 and 42 = 7 × 2 × 3 = 42
(v) 30, 40, 60
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 10
Factors of 30 = 2 × 3 × 5
Factors of 40 = 2 × 2 × 2 × 5
Factors of 60 = 2 × 2 × 3 × 5
LCM of 30, 40, 60 = 2 × 5 × 3 × 2 × 2 = 120
(vi) 15, 25, 75

Factors of 15 = 3 × 5
Factors of 25 =5 × 5
Factors of 75 = 3 × 5 × 5
LCM of 15, 25, 75 = 5 × 3 × 5 = 75
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Q.5Find the HCF and the LCM of the numbers 154, 198, 286v

Solution

HCF
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 12
Factors of 154 = 2 × 7 × 11
Factors of 198 = 2 × 3 × 3 × 11
Factors of 286 = 2 × 13 × 11
HCF of 154, 198, 286 = 11 × 2 = 22
LCM

LCM = 2 × 11 × 7 × 9 × 13
= 22 × 63 × 13
= 18018

Answer:

HCF
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 12
Factors of 154 = 2 × 7 × 11
Factors of 198 = 2 × 3 × 3 × 11
Factors of 286 = 2 × 13 × 11
HCF of 154, 198, 286 = 11 × 2 = 22
LCM

LCM = 2 × 11 × 7 × 9 × 13
= 22 × 63 × 13
= 18018

Q.6What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?v

Solution

This is an HCF related problem. So, we need to find the HCF of 80, 100, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 14
80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

Answer:

This is an HCF related problem. So, we need to find the HCF of 80, 100, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 14
80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

Q.7The traffic lights at three different road junctions change after every 40 seconds, 60 seconds, and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?v

Solution

This is an LCM related problem. So, we need to find the LCM of 40, 60, 72
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 15
LCM of 40, 60 and 72
= 2 × 2 × 3 × 5 × 2 × 1 × 1 × 3
= 360 seconds
= 6 minutes
The traffic lights will change simultane¬ously again at 8 : 06 am.

Answer:

This is an LCM related problem. So, we need to find the LCM of 40, 60, 72
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 15
LCM of 40, 60 and 72
= 2 × 2 × 3 × 5 × 2 × 1 × 1 × 3
= 360 seconds
= 6 minutes
The traffic lights will change simultane¬ously again at 8 : 06 am.

Q.8The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible.v

Solution

Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs are possible.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 16

Answer:

Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs are possible.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 16

Q.9The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.v

Solution

36x = 6 × 12 × 12
⇒ \(x=\frac{6 \times 12 \times 12}{36}=24\)
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
Objective Type Questions

Answer:

36x = 6 × 12 × 12
⇒ \(x=\frac{6 \times 12 \times 12}{36}=24\)
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2
Objective Type Questions

Q.10Which of the following pairs is co-prime?v

A. 51, 63
B. 52, 91
C. 71, 81
D. 81, 99

Solution

(c) 71, 81

Answer:

(c) 71, 81

Q.11The greatest four-digit number which is exactly divisible by 8, 9, and 12 isv

A. 9999
B. 9996
C. 9696
D. 9936

Solution

(d) 9936
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Answer:

(d) 9936
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Q.12The HCF of two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum isv

A. 26
B. 36
C. 46
D. 56

Solution

(b) 36

Answer:

(b) 36

Q.13Which of the following cannot be the HCF of two numbers whose LCM is 120?v

A. 60
B. 40
C. 80
D. 30

Solution

(c) 80
Posted in Class 6 on January 5, 2025 January 6, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi

Answer:

(c) 80
Posted in Class 6 on January 5, 2025 January 6, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi

Q.1Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number upto 16.v

Solution

Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.2Is 173 a prime? Why?v

Solution

Yes, 173 is a prime. Because it has only 1 and itself as factors.

Answer:

Yes, 173 is a prime. Because it has only 1 and itself as factors.

Q.3For which of the numbers, from n = 2 to 8. Is 2n – 1 a prime?v

Solution

n = 2 ⇒ 2n – 1 = 2 × 2 – 1
= 4 – 1
= 3 (prime)
n = 3 ⇒ 2n – 1 = 2 × 3 – 1
= 6 – 1
= 5 (prime)
n = 4 ⇒ 2n – 1 = 2 × 4 – 1
= 8 – 1
= 7 (prime)
n = 5 ⇒ 2n – 1 = 2 × 5 – 1
= 10 – 1
= 9 (Not prime)
n = 6 ⇒ 2n – 1 = 2 × 6 – 1
= 12 – 1
= 11 (prime)
n = 7 ⇒ 2n – 1 = 2 × 7 – 1
= 14 – 1
= 13
n = 8 ⇒ 2n – 1 = 2 × 8 – 1
= 16 – 1
= 15 (Not prime)
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

n = 2 ⇒ 2n – 1 = 2 × 2 – 1
= 4 – 1
= 3 (prime)
n = 3 ⇒ 2n – 1 = 2 × 3 – 1
= 6 – 1
= 5 (prime)
n = 4 ⇒ 2n – 1 = 2 × 4 – 1
= 8 – 1
= 7 (prime)
n = 5 ⇒ 2n – 1 = 2 × 5 – 1
= 10 – 1
= 9 (Not prime)
n = 6 ⇒ 2n – 1 = 2 × 6 – 1
= 12 – 1
= 11 (prime)
n = 7 ⇒ 2n – 1 = 2 × 7 – 1
= 14 – 1
= 13
n = 8 ⇒ 2n – 1 = 2 × 8 – 1
= 16 – 1
= 15 (Not prime)
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.4Explain your answer with the reason for the following statements.v

A. A number is divisible by 9 if it is divisible by 3.
B. A number is divisible by 6 if it is divisible by 12.

Solution

(i) False, 42 is divisible by 3 but it is not divisible by 9
(ii) True, 36 is divisible by 12. Also divisible by 6.

Answer:

(i) False, 42 is divisible by 3 but it is not divisible by 9
(ii) True, 36 is divisible by 12. Also divisible by 6.

Q.5Find A as required (i) The greatest 2 digit number 9 A is divisible by 2. (ii) The least number 567A is divisible by 3. (iii) The greatest 3 digit number 9A6 is divisible by 6. (iv) The number A08 is divisible by 4 and 9. (v) The number 225A85 is divisible by 11.v

Solution

(i) 98 A = 8
(ii) 5670 A = 0
(iii) 996 A = 9
(iv) 108 A = 1
(v) 225885 A = 8
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

(i) 98 A = 8
(ii) 5670 A = 0
(iii) 996 A = 9
(iv) 108 A = 1
(v) 225885 A = 8
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.6Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.v

Solution

False 12 is divisible by both 4 and 6. But not divisible by 24

Answer:

False 12 is divisible by both 4 and 6. But not divisible by 24

Q.7The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.v

Solution

True 3 + 5 = 8 is divisible by 4.

Answer:

True 3 + 5 = 8 is divisible by 4.

Q.8Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20cm, 3m 60 cm and 4 m.v

Solution

1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem. So, we need to find the HCF of 120,360 and 400.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 1
120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cm

Challenge Problems

Answer:

1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem. So, we need to find the HCF of 120,360 and 400.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 1
120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cm

Challenge Problems

Q.9The sum of three prime numbers is 80. The difference between the two of them is 4. Find the numbers.v

Solution

Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 [∵ one number must be even]
Also their difference = 4 (given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41

Answer:

Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 [∵ one number must be even]
Also their difference = 4 (given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41

Q.10Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.v

Solution

Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Answer:

Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Q.11Find the smallest number which is exactly divisible by all the numbers from 1 to 9.v

Solution

2520
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

2520
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.12The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.v

Solution

Yes. Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7

Answer:

Yes. Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7

Q.13Malarvizhi, Karthiga, and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively, Assuming that they met each other on the 1st of October, when will all the three meet again?v

Solution

This is an LCM related problem. So, we need to find the LCM of 5, 6, and 10.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 2
LCM = 5 × 2 × 1 × 3 × 1
= 30
All the three will meet again once in 30 days.

Answer:

This is an LCM related problem. So, we need to find the LCM of 5, 6, and 10.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 2
LCM = 5 × 2 × 1 × 3 × 1
= 30
All the three will meet again once in 30 days.

Q.14In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?v

Solution

LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90, and 105.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90, and 105.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.15The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?v

Solution

15 × 20 = 300
HCF of 15 and 20 is 5
The numbers are 15 and 20

Answer:

15 × 20 = 300
HCF of 15 and 20 is 5
The numbers are 15 and 20

Q.16Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help)v

Solution

564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Answer:

564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3

Q.17Wilson, Mathan, and Guna can complete one round of a circular track in 10, 15, and 20 minutes respectively. If they start together from at the starting point of 7 am, at what time will they meet together again at the same starting point?v

Solution

This is an LCM related problem. So, we need to find the LCM of 10, 15, and 20.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 3
LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7.am + 60 minutes = 8 am.
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi

Answer:

This is an LCM related problem. So, we need to find the LCM of 10, 15, and 20.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.3 3
LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7.am + 60 minutes = 8 am.
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi