NCERT Solutions for Class 6 Maths Chapter 3 playing with numbers
Here at MentorAtHome provides, NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers are helpful for exam preparation. The NCERT Class 6 Maths Solution Chapter 3 has seven exercises in this chapter. NCERT Solution for class 6 maths chapter 3 solutions are designed by a subject expert team of MentorAtHome. These are the ultimate study material for CBSE Class 6 students can to score good marks in the Maths Examination.
NCERT Solution for playing with numbers Class 6 Maths Chapter 3 discusses the concepts of multiples, divisors, factors, and how to identify factors and multiples along with HCF and LCM. Chapter 3 Playing With NumbersExercise 3.1, Solving the textual problems present in this chapter will help the students understand the various topics related to HCF and LCM in an easy way.
NCERT solutions for class 6 maths chapter 3 Exercise 3.1
NCERT Solutions for Class 6 maths chapter 3 exercise 3.1 Playing with numbers is a short exercise with 4 problems. The questions in this exercise are broadly based on finding the factors and multiples of a number. NCERT solutions class 6 maths chapter 3 exercise 3.1 helps kids identify the factors of a number. ncert solutions of class 6 maths chapter 3.
Q.1. Write all the factors of the following numbers:
(a) 24
Ans:-
Factors of 24 are:
24 = 1 x 24;
24 = 2 x 12;
24 = 3 x 8;
24 = 4 x 6
Hence, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
(b) 15
Ans:-
Factors of 15 are:
15 = 1 x 15;
15 = 3 x 5
Hence, all the factors of 15 are: 1, 3, 5 and 15.
(c) 21
Ans:-
Factors of 21 are:
21 = 1 x 21;
21 = 3 x 7
Hence, all the factors of 21 are: 1, 3, 7 and 21.
(d) 27
Ans:-
Factors of 27 are:
27 = 1 x 27;
27 = 3 x 9.
Hence, all the factors of 27 are 1, 3, 9, and 27.
(e) 12
Ans:-
Factors of 12 are:
12 = 1 x 12;
12 = 2 x 6;
12 = 3 x 4
Hence, all the factors of 12 are: 1, 2, 3, 4, 6 and 12.
(f) 20
Ans:-
Factors of 20 are:
20 = 1 x 20;
20 = 2 x 10;
20 = 4 x 5
Hence, all the factors of 20 are: 1, 2, 4, 5, 10 and 20.
(g) 18
Ans:-
Factors of 18 are:
18 = 1 x 18;
18 = 2 x 9;
18 = 3 x 6
Hence, all the factors of 18 are: 1, 2, 3, 6, 9 and 18.
(h) 23
Ans:-
Factors of 23 are:
23 = 1 x 23
Hence, all the factors of prime number 23 are 1 and 23.
(i) 36
Ans:-
Factors of 36 are:
36 = 1 x 36;
36 = 2 x 18;
36 = 3 x 12;
36 = 4 x 9;
36 = 6 x 6
Hence, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Q.2. Write first five multiples of:
(a) 5
Ans:-
First five multiples of 5 are:
5 x 1 = 5;
5 x 2 = 10;
5 x 3 = 15;
5 x 4 = 20;
5 x 5 = 25
Hence, the required multiples of 5 are: 5, 10, 15, 20 and 2
(b) 8
Ans:-
First five multiples of 8 are:
8 x 1 = 8;
8 x 2 = 16;
8 x 3 = 24;
8×4 = 32;
8 x 5 = 40
Hence, the required multiples of 8 are: 8, 16, 24, 32 and 40.
(c) 9
Ans:-
First five multiples of 9 are:
9 x 1 = 9;
9 x 2 = 18;
9 x 3 = 27;
9 x 4 = 36;
9 x 5 = 45
Hence, the required multiples of 9 are: 9,18, 27, 36 and 45.
Q.3. Match the items in column 1 with the items in column 2.
Column 1 Column 2
(i) 35 (a) Multiple of 8
(ii) 15 (b) Multiple of 7
(iii) 16 (c) Multiple of 70
(iv) 20 (d) Factor of 30
(v) 25 (e) Factor of 50
(f) Factor of 20
Ans:-
(i) ↔ (b) [∵ 7 x 5 = 35]
(ii) ↔ (d) [∵ 15 x 2 = 30]
(iii) ↔ (a) [∵ 8 x 2 = 16]
(iv) ↔ (f) [∵ 20 x 1 = 20]
(v) ↔ (e) [∵ 25 x 2 = 50]
Q.4. Find all the multiples of 9 upto 100.
Ans:-
9 x 1 = 9;
9 x 2 = 18;
9 x 3 = 27;
9 x 4 = 36;
9 x 5 = 45;
9 x 6 = 54;
9 x 7 = 63;
9 x 8 = 72;
9 x 9 = 81;
9 x 10 = 90;
9 x 11 = 99
Hence, all the multiples of 9 upto 100 are:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
NCERT solutions for class 6 maths chapter 3 Exercise 3.2
NCERT Solutions of Class 6 Maths Chapter 3 Exercise 3.2.
Q.1.What is the sum of any two:
(a) Odd numbers?
Ans:-
The sum of any two odd numbers is even.
(b) Even numbers?
Ans:-
The sum of any two even numbers is even.
Q.2. State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
Ans:-
False. (The sum of three odd numbers is odd.)
example:- 3 + 5 + 7 = 15
(b) The sum of two odd numbers and one even number is even.
Ans:-
True. (The sum of two odd numbers and one even number is even.)
example::- 3 + 5 + 4 = 12
(c) The product of three odd numbers is odd.
Ans:-
True. (The product of three odd numbers is odd.)
example:- 3 x 5 x 7 = 105
(d) If an even number is divided by 2, the quotient is always odd.
Ans:-
False. (If an even number is divided by 2, the quotient is even.)
example:- : 8 ÷ 2 = 4
(e) All prime numbers are odd.
Ans:-
False, (All prime numbers are not odd.)
example:- 2 is a prime number but it is also an even number.
(f) Prime numbers do not have any factors.
Ans:-
False [∵ 3 is a prime number having 1 and 3 as its factors]
(g) Sum of two prime numbers is always even
Ans:-
False. (The Sum of two prime numbers may also be odd number)
example:- 2 + 5 = 7
(h) 2 is the only even prime number.
Ans:-
True ( 2 is even and the lowest prime number)
(i) All even numbers are composite numbers.
Ans:-
False (2 is even but not composite number)
(j) The product of two even numbers is always even.
Ans:-
True. (The product of two even numbers is always even.)
example: 2 × 4 = 8
Q.3. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.
Ans:-
The required pair of prime numbers having same digits are:
17 and 71
37 and 73
79 and 97
Q.4. Write down separately the prime and composite numbers less than 20.
Ans:-
Prime numbers less than 20 are:
2, 3, 5, 7, 11, 13, 17 and 19
Composite numbers less than 20 are:
4, 6, 8, 9, 10, 12, 14, 15, 16 and 18
Q.5. What is the greatest prime number between 1 and 10?
Ans:-
2, 3, 5, and 7 are the prime numbers between 1 and 10. 7 is the greatest prime number among them.
Q.6 Express the following as the sum of two odd primes.
(a) 44
(b) 36
(c) 24
(d) 18
Ans:-
(a) 44 = 13 + 31
(b) 36 = 17 + 19
(c) 24 = 7 + 17
(d) 18 = 7 + 11
Q.7. Give three pairs of prime numbers whose difference is 2. [Remark: Two prime numbers whose difference is 2 are called twin primes].
Ans:-
Required pairs are: (3 and 5), (5 and 7), and (11 and 13)
8. Which of the following numbers are prime?
(a) 23
Ans:-
23 = 1 x 23]
Therefore 23 has only two factors 1 and 23. Hence, it is a prime number.
(b) 51
Ans:-
1 × 51 = 51
3 × 17 = 51
Therefore 51 has four factors 1, 3, 17, and 51. Hence, it is not a prime number, it is a composite number.
(c) 37
Ans:-
1 × 37 = 37
Therefore 37 has two factors 1 and 37. Hence, it is a prime number.
(d) 26
Ans:-
1 × 26 = 26
2 × 13 = 26
Therefore 26 has four factors 1, 2, 13, and 26. Hence, it is not a prime number, it is a composite number.
Q.9. Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Ans:-
Required seven consecutive composite numbers are:
90, 91, 92, 93, 94, 95 and 96
Q.10 Express each of the following numbers as the sum of three odd primes.
(a) 21
(b) 31
(c) 53
(d) 61
Ans:-
(a) 21 = 3 + 5 + 13
(b) 31 = 5 + 7 + 19
(c) = 13 + 17 + 23
(d) 61 = 11 + 13 + 37
Q.11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (Hint: 3 + 7 = 10)
Ans:-
Required pairs of prime numbers less than 20 are:
(i) 2 + 3 = 5
(ii) 2 + 13 = 15
(iii) 11 + 9 = 20
(iv) 17 + 3 = 20
(v) 7 + 13 = 20
Q.12 Fill in the blanks.
(a) A number which has only two factors is called a ………….
(b) A number which has more than two factors is called a ………….
(c) 1 is neither ………… nor ………….
(d) The smallest prime number is ………….
(e) The smallest composite number is ………….
(f) The smallest even number is ………….
Ans:-
(a) prime number
(b) composite number
(c) prime, composite
(d) 2
(e) 4
(f) 2
NCERT solutions for class 6 maths chapter 3 Exercise 3.3
NCERT Solutions of Class 6 Maths Chapter 3 Exercise 3.3.
1. Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes or no):
Ans:-
Q.2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:
(a) 572
Ans:-
(i) 72 are the last two digits. Since 72 is divisible by 4.
Hence, 572 is also divisible by 4
(ii) 572 are the last three digits. Since 572 is not divisible by 8.
Hence, 572 is not divisible by 8
(b) 726352
Ans:-
(i) 52 are the last two digits. Since 52 is divisible by 4.
Hence, 726352 is also divisible by 4
(ii) 352 are the last three digits. Since 352 is divisible by 8.
Hence, 726352 is divisible by 8
(c) 5500
Ans:-
(i) the last two digits 00.
Hence, 5500 is also divisible by 4
(ii) 500 are the last three digits. Since 500 is not divisible by 8.
Hence, 5500 is not divisible by 8
(d) 6000
Ans:-
(i) the last two digits 00.
Hence, 6000 is also divisible by 4
(ii) the last three digits 000.
Hence, 6000 is also divisible by 8
(e) 12159
Ans:-
(i) 59 are the last two digits. Since 59 is not divisible by 4.
Hence, 12159 is not divisible by 4
(ii) 159 are the last three digits. Since 159 is not divisible by 8.
Hence, is not divisible by 8
(f) 14560
Ans:-
(i) 60 are the last two digits. Since 60 is divisible by 4.
Hence, 14560 is also divisible by 4
(ii) 560 are the last three digits. Since 560 is divisible by 8.
Hence, 14560 is divisible by 8
(g) 21084
Ans:-
(i) 84 are the last two digits. Since 60 is divisible by 4.
Hence, 21084 is also divisible by 4
(ii) 084 are the last three digits. Since 084 is not divisible by 8.
Hence, 21084 is not divisible by 8
(h) 31795072
(i) 72 are the last two digits. Since 72 is divisible by 4.
Hence, 31795072 is also divisible by 4
(ii) 072 are the last three digits. Since 072 is divisible by 8.
Hence, 31795072 is divisible by 8
(i) 1700
(i) 00 are the last two digits. Since 00 is divisible by 4.
Hence, 1700 is also divisible by 4
(ii) 700 are the last three digits. Since 700 is not divisible by 8.
Hence, 1700 is not divisible by 8
(j) 2150
(i) 50 are the last two digits. Since 50 is divisible by 4.
Hence, 2150 is also divisible by 4
(ii) 150 are the last three digits. Since 150 is not divisible by 8.
Hence, 2150 is not divisible by 8
3. Using divisibility tests, determine which of following numbers are divisible by 6:
(a) 297144
(We know that a number is divisible by 6 if it is also divisible by both 2 and 3)
Ans:-
Since the last digit of the number is 4.
So, it is divisible by 2.
The sum of all the digits of 297144
= 2 + 9 + 7 + 1 + 4 + 4 = 27 which is divisible by 3.
Hence, the given number 297144 is divisible by 6.
(b) 1258
Ans:-
Since the last digit of the number is 8.
So, it is divisible by 2.
The sum of all digits of 1258
= l + 2 + 5 + 8 = 16 which is not divisible by 3.
Hence, the given number 1258 is not divisible by 6.
(c) 4335
Ans:-
Since the last digit of the number is 5.
So, it is not divisible by 2.
The sum of all the digits of 4335
= 4 + 3 + 3 + 5 = 15 which is divisible by 3.
Hence, the given number 4335 is not divisible by 6.
(d) 61233
Ans:-
Since the last digit of the number is 3.
So, it is not divisible by 2.
The sum of the digits of the given number 61233
= 6 + 1 + 2 + 3 + 3 = 15 which is divisible by 3.
Hence, the given number 61233 is not divisible by 6.
(e) 901352
Ans:-
Since the last digit of the number is 2.
So, it is divisible by 2.
The sum of all the digits of the given number 901352
= 9 + 0 + 1 + 3 + 5 + 2 = 20 which is not divisible by 3.
Hence, the given number 901352 is not divisible by 6.
(f) 438750
Ans:-
Since the last digit of the number is 0.
So, it is divisible by 2.
The sum of all the digits of the given number 438750
= 4 + 3 + 8 + 7 + 5 + 0 = 27 which is divisible by 3.
Hence, the given number 438750 is not divisible by 6.
(g) 1790184
Ans:-
Since the last digit of the number is 4.
So, it is divisible by 2.
sum of all the digits of the given number 1790184
= 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 which is divisible by 3.
Hence, the given number 438750 is not divisible by 6.
(h) 12583
Ans:-
The digit to ones place of the given number is odd.
So, it is not divisible by 2.
Sum of all the digits of the given number 12583
= 1 + 2 + 5 + 8 + 3 = 19 which is not divisible by 3.
Hence, the given number 12583 is not divisible by 6.
(i) 639210
Ans:-
Since the last digit of the number is 0.
So, it is divisible by 2.
The sum of all the digits of the given number 639210
= 6 + 3 + 9 + 2 + 1 + 0 = 21 which is divisible by 3.
Hence, the given number 639210 is divisible by 6.
(j) 17852
Ans:-
Since the last digit of the number is 2.
So, it is divisible by 2.
The sum of all the digits of the given number 17852
= 1 + 7 + 8 + 5 + 2 = 23 which is not divisible by 3.
Hence, the given number 17852 is not divisible by 6
Q.4. Using divisibility tests, determine which of the following numbers are divisible by 11:
(you know that a number is divisible by 11 if 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) of the number is either 0 or divisible by 11)
(a) 5445
Ans:-
Sum of the digits at odd places = 5 + 4 = 9
Sum of the digits at even places = 4 + 5 = 9
Difference = 9 – 9 = 0
Hence, the given number is divisible by 11.
(b) 10824
Ans:-
Sum of the digits at odd places = 4 + 8 + 1 = 13
Sum of the digits at even places = 2 + 0 = 2
Difference = 13 – 2 = 11
which is divisible by 11.
Hence, the given number is divisible by 11
(c) 7138965
Ans:-
Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24
Sum of the digits at even places = 6 + 8 + 1 = 15
Difference = 24 – 15 = 9
which is not divisible by 11.
Hence, the given number is not divisible by 11.
(d) 70169308
Ans:-
Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17
Sum of the digits at even places = 0 + 9 + 1 + 7 = 17
Difference = 17 – 17 = 0
Hence, the given number is not divisible by 11.
(e) 10000001
Ans:-
Sum of the digits at odd places = 1 + 0 + 0 + 0 = 1
Sum of the digits at even places = 0 + 0 + 0 + 1 = 1
Difference = 1 – 1 = 0
Hence, the given number is not divisible by 11.
(f) 901153
Ans:-
Sum of the digits at odd places = 3 + 1 + 0 = 4
Sum of the digits at even places = 5 + 1 + 9 = 15
Difference = 15 – 4 = 11
which is divisible by 11.
Hence, the given number is not divisible by 11.
Q.5. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3:
(We know that number is divisible by 3 if the sum of all the digits of the number is also divisible by 3.)
(a) __ 6724
Ans:-
Sum of the digits = 4 + 2 + 7 + 6 = 19
The smallest digit to be placed is blank space = 2
Then the sum = 19 + 2 = 21 which is divisible by 3.
The greatest digit to be placed in blank space = 8
Then, the sum = 19 + 8 = 27 which is divisible by 3
Hence, the required digits are 2 and 8.
(b) 4765 __ 2
Ans:-
Sum of digits = 2 + 5 + 6 + 7 + 4 = 24
The smallest digits to be place in blank space = 0
Then, sum = 24 + 0 = 24
which is divisible by 3.
The greatest digit to be placed in blank space = 9.
Then, the sum = 24 + 9 = 33 which is divisible by 3.
Hence, the required digits are 0 and 9.
Q.6. Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
(a) 92 __ 389
Ans:-
Sum of the digits at odd places = 9 + 3 + 2 = 14
Sum of the digits at even places = 8 + ( ) + 9 = 17
Difference = 17 + ( ) – 14 = () + 3
For the given number to be divisible by 11
( ) + 3 = 11
∴ ( ) = 11 – 3 = 8
So, the missing digit = 8
Hence, the required number is 928389.
(b) 8 __ 9484
Ans:-
Sum of the digits at odd places = 4 + 4 + ( ) = 8 + ( )
Sum of the digits at even places = 8 + 9 + 8 = 25
Difference = 25 – [8 + ( )]
= 25 – 8 – ( ) = 17 – ( )
For the given number to be divisible by 11
17 – 0 = 11
∴ 17 – 11 = 6
So, the missing digit = 6
Hence, the required number = 869484.
NCERT solutions for class 6 maths chapter 3 Exercise 3.4
NCERT Solutions of Class 6 Maths Chapter 3 Exercise 3.4.
Q.1. Find the common factors of:
(a) 20 and 28
Ans:-
Factors of 20 are 1, 2, 4, 5, 10, 20
Factors of 28 are 1, 2, 4, 7, 28
the common factors are 1, 2, and 4.
(b) 15 and 25
Ans:-
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
the common factors are 1 and 5.
(c) 35 and 50
Ans:-
Factors of 35 are: 1, 5, 7, 35
Factors of 50 are: 1, 2, 5, 10, 50
the common factors are 1 and 5.
(d) 56 and 120
Ans:-
Factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56
Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120
the common factors are 1,2, 4, and 8.
2. Find the common factors of:
(a) 4, 8 and 12
Ans:-
Factors of 4 are 1, 2, 4
Factors of 8 are 1, 2, 4, 8
Factors of 12 are 1, 2, 3, 4, 6, 12
the common factors are 1, 2, and 4.
(b) 5, 15 and 25
Ans:-
Factors of 5 are 1, 5
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
the common factors are 1 and 5.
Q.3. Find first three common multiples of:
(a) 6 and 8
Ans:-
6,12,18,24,30,36,42,48,54,60 are multiples of 6
8,16,24,32,40,48,56,64,72,80 are multiples of 8
Three common multiples are 24, 48, 72
(b) 12 and 18
Ans:-
12, 24, 36, 48, 60, 72, 84, 96, 108, 120 are multiples of 12
18, 36, 54, 72, 90, 108, 126, 144, 162, 180 are multiples of 18
Three common multiples are 36, 72 and 108
Q.4. Write all the numbers less than 100 which are common multiples of 3 and 4.
Ans:-
Multiples of 3 are 3, 6, 9, 12, 15…..and 99
Multiples of 4 are 4, 8, 12, 16, 20…..and 96
Common multiples are 12, 24, 36, 48, 60, 72, 84 and 96
Q.5. Which of the following numbers are co-prime?
(a) 18 and 35
Ans:-
Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 35 are 1, 5, 7, 35
Common factors = 1
Hence, 18 and 35 are co-prime.
(b) 15 and 37
Ans:-
Factors of 15 are 1, 3, 5, 15
Factors of 37 are 1, 37
Common factors = 1
Hence, 15 and 37 are co-prime.
(c) 30 and 415
Ans:-
Factors of 30 are 1, 2, 3, 5, 6, 15, 30
Factors of 415 are 1, 5, 83
Common factors = 1, 5
Hence, 30 and 415 are not co-prime.
(d) 17 and 68
Ans:-
Factors of 17 are 1, 17
Factors of 68 are 1, 2, 4, 17, 34, 68
Common factors = 1, 17
Hence, 17 and 68 are not co-prime.
(e) 216 and 215
Ans:-
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 54, 72, 108, 216
Factors of 215 are 1, 5, 43
Common factors = 1
Hence, 216 and 215 are co-prime.
(f) 81 and 16
Ans:-
Factors of 81 are 1, 3, 9, 27, 81
Factors of 16 are 1, 2, 4, 8, 16
Common factors = 1
Hence, 81 and 16 are co-prime.
Q.6. A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?
Ans:-
If the number is divisible by both 5 and 12 this the number will also be divisible by 5 x 12 i.e., 60
Q.7. A number is divisible by 12. By what other numbers will that number be divisible?
Ans:-
Factors of 12 are 1, 2, 3, 4, 6, 12
Hence the number which is divisible by 12, will also be divisible by its factors i.e., 1, 2, 3, 4, 6, and 12.
NCERT solutions for class 6 maths chapter 3 Exercise 3.5
NCERT Solutions Class Maths Chapter 3 Exercise 3.5
Q.1 Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
Ans:- False
(b) If a number is divisible by 9, it must be divisible by 3.
Ans:- True
(c) A number is divisible by 18 if it is divisible by both 3 and 6.
Ans:- False
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
Ans:- True
(e) If two numbers are co-primes, at least one of them must be prime.
Ans:- False
(f) All numbers which are divisible by 4 must also be divisible by 8.
Ans:- False
(g) All numbers which are divisible by 8 must also be divisible by 4.
Ans:- True
(h) If a number exactly divides two numbers separately,- it must exactly divide their sum.
Ans:- True
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Ans:- False
Q.2. Here are two different factor trees for 60. Write the missing numbers.
(a)
Ans:-
Here, 6 = 2 x missing number
∴ Missing number = 6 ÷ 2 = 3
Similarly, 10 = 5 x missing number
∴ Missing number = 10 ÷ 5 = 2
Hence, the missing numbers are 3 and 2.
(b)
Ans:-
Here, 60 = 30 x missing number
∴ Missing number = 60 ÷ 3 = 2
Similarly, 30 = 10 x missing number
∴ Missing number = 30 ÷ 10 = 3
Here, 10 = m1 x m2
⇒ m3 = 2 or 5 and m4 = 5 or 2
Hence, the missing numbers are 2, 3, 2, and 5
Q.3. Which factors are not included in the prime factorization of a composite number?
Ans:-
1 and the number itself are not included in the prime factorization of a composite number.
Q.4. Write the greatest 4-digit number and express it in terms of its prime factors.
Ans:-
The greatest 4-digit number = 9999
Hence, the prime factors of 9999 = 3 x 3 x 11 x 101.
Q.5. Write the smallest 5-digit number and express it in the form of its prime factors.
Ans:-
The smallest 5-digit number = 10000
Hence, the required prime factors: 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5.
Q.6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.
Ans:-
Given number = 1729
Hence, the prime factors of 1729 = 7 x 13 x 19.
13 – 7 = 6
19 – 13 = 6
We see that the difference between two consecutive prime factors is 6.
Q.7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Ans:-
Example:-(1)
Take three consecutive numbers 2, 3, and 4.
Here 2 and 4 are divisible by 2 and 3 is divisible by 3.
Therefore, the product 2 x 3 x 4 = 24 is divisible by 6.
Example:-(2)
Take three consecutive numbers 7, 8, and 9.
Here 8 is divisible by 2 and 9 is divisible by 3.
Therefore, the product 7 x 8 x 9 = 504 is divisible by 6.
Example:-(3)
Take three consecutive numbers 13, 14, and 15.
Here 14 is divisible by 2 and 15 is divisible by 3.
Therefore, the product 13 x 14 x 15 = 2730 is divisible by 6.
8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Ans:-
Example:-(1)
Let us take two consecutive odd numbers 13 and 15.
Sum = 13 + 15 = 28
Here, the number formed by the last two digits is 28 which is divisible by 4.
Hence, the sum of numbers 13 and 15 i.e. 196 is divisible by 4.
Example:-(2)
Let us take two consecutive odd numbers 51 and 53.
Sum = 51 + 53 = 104
Here, the number formed by the last two digits is 04 which is divisible by 4.
Hence, the sum of numbers 51 and 53 i.e. 104 is divisible by 4.
Example:-(2)
Let us take two consecutive odd numbers 105 and 107.
Sum = 105 + 107 = 212
Here, the number formed by the last two digits is 12 which is divisible by 4.
Hence, the sum of numbers 105 and 107 i.e. 212 is divisible by 4.
Q.9. In which of the following expressions, prime factorization has been done?
(a) 24 = 2 × 3 × 4
Ans:-
Here, 4 is not a prime number.
Hence, 24 = 2 x 3 x 4 is not a prime factorisation.
(b) 56 = 7 × 2 × 2 × 2
Ans:-
Here, all factors are prime numbers
Hence, 56 = 7 x 2 x 2 x 2 is a prime factorisation.
(c) 70 = 2 × 5 × 7
Ans:-
Here, all factors are prime numbers
Hence, 70 = 2 x 5 x 7 is a prime factorisation.
(d) 54 = 2 × 3 × 9
Ans:-
Here, 9 is not a prime number.
Hence, 54 = 2 x 3 x 9 is not a prime factorisation.
Q.10. Determine if 25110 is divisible by 45. [Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].
Ans:-
45 = 5 × 9
1, 5 are factors of 5
1, 3, 9 are factors of 9
Here, 5 and 9 are co-prime numbers
Test of divisibility by 5: the unit place of the given number 25110 is 0. So, it is divisible by 5.
Test of divisibility by 9:
Sum of the digits = 2 + 5 + l + l + 0 = 9 which is divisible by 9.
So, the given number is divisible by 5 and 9 both. Hence, the number 25110 is divisible by 45.
Q.11. 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.
Ans:-
No,
Example:-
since 12 and 36 are both divisible by 4 and 6. But 12 and 36 are not divisible by 24
12. I am the smallest number, having four different prime factors. Can you find me?
Ans:-
The smallest 4 prime numbers are 2, 3, 5, and 7.
Hence, the required number = 2 x 3 x 5 x 7 = 210
NCERT solutions for class 6 maths chapter 3 Exercise 3.6
NCERT Solutions Class Maths Chapter 3 Exercise 3.6.
Q.1. Find the HCF of the following numbers :
(a) 18, 48
Ans:-
Prime factorisations of 18 and 48 are
18 = (2) x 3 x (3) x 1
48 = (2) x 2 x 2 x 2 x (3) x 1
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.
(b) 30, 42
Ans:-
Prime factorisations of 30 and 42 are
30 = 2 x 3 x 5 x 1
42 = 2 x 3 x 7 x 1
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.
(c) 18, 60
Ans:-
Prime factorisations of 18 and 60 are
18 = 2 x 3 x 3 x 1
60 = 2 x 2 x 3 x 5 x 1
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.
(d) 27, 63
Ans:-
Prime factorisations of 27 and 63 are
27 = 3 x 3 x 3 x 1
63 = 3 x 3 x 7 x 1
Here, the common factors are 3 and 3.
Hence, the HCF = 3 x 3 = 9
(e) 36, 84
Ans:-
Prime factorisations of 36 and 84 are
36 = 2 x 2 x 3 x 3 x 1
84 = 2 x 2 x 3 x 7 x 1
Here, the common factors are 2, 2, and 3.
Hence, the HCF = 2 x 2 x 3 = 12
(f) 34, 102
Ans:-
Prime factorisations of 34 and 102 are
34 = 2 x 17 x 1
102 = 2 x 3 x 17 x 1
Here, the common factors are 2 and 17.
Hence, the HCF = 2 x 17 = 34
(g) 70, 105, 175
Ans:-
Prime factorisations of 70, 105 and 175 are
70 = 2 x 5 x 7 x 1
105 = 3 x 5 x 7 x 1
175 = 5 x 5 x 7 x 1
Here, the common factors are 5 and 7.
Hence, the HCF = 5 x 7 = 35
(h) 91, 112, 49
Ans:-
Prime factorisations of 91, 112 and 49 are
91 = 13 x 7 x 1
112 = 2 x 2 x 2 x 2 x 7 x 1
49 = 7 x 7 x 1
Here, the common factors are 5.
Hence, the HCF = 7 = 7
(i) 18, 54, 81
Ans:-
Prime factorisations of 18, 54 and 81 are
18 = 2 x 3 x 3 x 1
54 = 2 x 3 x 3 x 3 x 1
81 = 3 x 3 x 3 x 3 x 1
Here, the common factors 3, are 3.
Hence, the HCF = 3 x 3 = 9
(j) 12, 45, 75
Ans:-
Prime factorisations of 12, 45 and 75 are
12 = 2 x 2 x 3 x 1
45 = 3 x 3 x 5 x 1
75 = 3 x 5 x 5 x 1
Here, the common factors 3,
Hence, the HCF = 3 = 3
Q.2. What is the HCF of two consecutive
(a) numbers?
Ans:-
The HCF of two consecutive numbers is 1
Example:- The HCF of 2 and 3 is 1
(b) even numbers?
Ans:-
The HCF of two consecutive even numbers is 2
Example:- The HCF of 2 and 4 is 2
(c) odd numbers?
Ans:-
The HCF of two consecutive odd numbers is 1
Example: The HCF of 3 and 5 is 1
3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation:
4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Ans:-
No, the answer is not correct.
Reason: 0 is not the prime factor of any number.
1 is always the prime factor of co-prime number.
Hence, the correct HCF of 4 and 15 is 1.
NCERT solutions for class 6 maths chapter 3 Exercise 3.7
NCERT solution for class 6 maths chapter 3 Exercise 3.7
Q.1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser the exact number of times.
Ans:-
Given, the weight of two bags of fertiliser = 75 kg and 69 kg
Maximum weight = HCF of two bags weight i.e (75, 69)
Prime factorisations of 75 and 69 are
Prime factorisations of 75 and 69 are
75 = 3 x 5 x 5 x 1
69 = 3 x 23 x 1
Here, the common factor is 3.
∴ HCF of 75 and 69 = 3.
Hence, the required maximum value of weight = 3 kg.
Q.2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Ans:-
First boy steps measure = 63 cm
Second boy steps measure = 70 cm
Third boy steps measure = 77 cm
LCM of 63, 70, 77
∴ LCM of 63, 70 and 77 = 2 x 3 x 3 x 5 x 7 x 11 = 6930
Hence, the required minimum distance = 6930 cm.
Q.3. The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Ans:-
length of a room = 825 cm
Breadth of a room = 675 cm
Height of a room = 450 cm
Prime factorisations of 825, 675 and 450 are
825 = 3 x 5 x 5 x 11
675 = 3 x 3 x 3 x 5 x 5
450 = 2 x 3 x 3 x 5 x 5
∴ HCF of 825, 675 and 450 = 3 x 5 x 5 = 75
Hence, the required longest tape = 75 cm
Q.4. Determine the smallest 3-digit number which is exactly divisible by 6, 8, and 12.
Ans:-
The LCM of 6, 8, and 12 we have
LCM = 2 × 2 × 2 × 3 = 24
Since, all the multiples of 24 will also be divisible by 6, 8, and 12.
The smallest 3-digit number = 100
We know that 24 × 4 = 96 and 24 × 5 = 120
Hence, the required number is 120.
Q.5. Determine the greatest 3-digit number exactly divisible by 8, 10, and 12.
Ans:-
LCM of 8, 10 and 12
LCM = 2 x 2 x 2 x 3 x 5 = 120
The greatest 3-digit number = 999
We know that 120 × 8 = 960 and 120 × 9 = 1080
Hence, the required number is 960.
Q.6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds, respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
Ans:-
LCM of 48, 72, 108
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
Hence, lights will change together after every 432 seconds
Therefore, the lights will change simultaneously at 7 minutes 12 seconds.
Q.7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel, respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
Ans:-
The maximum capacity of the required measure is equal to the HCF of 403, 434 and 465.
Prime factorisations of 403, 434 and 465 are
403 = 13 × 31
434 = 2 × 7 × 31
465 = 3 × 5 × 31
HCF = 31
Hence, the maximum capacity of the required container = 31 litres.
Q.8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Ans:-
LCM of 6, 15, 18
LCM = 2 × 3 × 3 × 5 = 90
Hence, Required number = 90 + 5= 95
Q.9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Ans:-
LCM of 18, 24, 32
LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
Since, 288 is the smallest number which is exactly divisible by 18, 24 and 32.
But it is not a 4-digit numb
The smallest 4-digit number = 1000.
We find 288 × 3 = 864 and 288 × 4 = 1152
Hence, 1152 is the smallest 4-digit number which is divisible by 18, 24 and 32
Q.10. Find the LCM of the following numbers:
(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4
Observe a common property in the obtained LCM. Is LCM the product of two numbers in each case?
Ans:-
(a) To find the LCM of 9 and 4, we have
LCM = 2 × 2 × 3 × 3 = 36
(b) To find the LCM of 12 and 5, we have
LCM = 2 × 2 × 3 × 5 = 60
(c) To find the LCM of 6 and 5, we have
LCM = 2 × 3 × 5 = 30
(d) To find the LCM of 15 and 4, we have
LCM = 2 × 2 × 3 × 5 = 60
Q.11. Find the LCM of the following numbers in which one number is the factor of the other.
(a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45
What do you observe in the results obtained?
Ans:-
(a) To find the LCM of 5 and 20, we have
LCM = 2 x 2 x 5 = 20
(b) To find the LCM of 6 and 18, we have
LCM = 2 x 3 x 3 = 18
(c) To find the LCM of 12 and 48, we have
LCM = 2 x 2 x 2 x 2 x 3 = 48
(d) To find the LCM of 9 and 45, we have
LCM = 3 x 3 x 5 = 45
∴ Hence, in each case the LCM of given numbers is the larger number. When a number is a factor of other number, then their LCM will be the larger number.
| Class 6 Science NCERT Notes | Class 6 Complete Study Material |
| NCERT Solutions for Class 6 Science | Class 6 Maths Chapter 6 Solutions |
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes are based on NCERT Books. You can download ncert book for class 6 maths chapter 3.
What we learn in class 6th maths chapter 3 Solutions
In Solutions for class 6 math chapter 3, we learn these topics.
- We have discussed multiples, divisors, factors and have seen how to identify factors
and multiples. - We have discussed and discovered the following :
(a) A factor of a number is an exact divisor of that number.
(b) Every number is a factor of itself. 1 is a factor of every number.
(c) Every factor of a number is less than or equal to the given number.
(d) Every number is a multiple of each of its factors.
(e) Every multiple of a given number is greater than or equal to that number.
(f) Every number is a multiple of itself. - We have learnt that –
(a) The number other than 1, with only factors namely 1 and the number itself, is a
prime number. Numbers that have more than two factors are called composite
numbers. Number 1 is neither prime nor composite.
(b) The number 2 is the smallest prime number and is even. Every prime number other
than 2 is odd.
(c) Two numbers with only 1 as a common factor are called co-prime numbers.
(d) If a number is divisible by another number then it is divisible by each of the factors
of that number.
(e) A number divisible by two co-prime numbers is divisible by their product also. - We have discussed how we can find just by looking at a number, whether it is divisible
by small numbers 2,3,4,5,8,9 and 11. We have explored the relationship between
digits of the numbers and their divisibility by different numbers.
(a) Divisibility by 2,5 and 10 can be seen by just the last digit.
(b) Divisibility by 3 and 9 is checked by finding the sum of all digits.
(c) Divisibility by 4 and 8 is checked by the last 2 and 3 digits respectively.
(d) Divisibility of 11 is checked by comparing the sum of digits at odd and even
places. - We have discovered that if two numbers are divisible by a number then their
sum and difference are also divisible by that number. - We have learnt that –
(a) The Highest Common Factor (HCF) of two or more given numbers is the highest
of their common factors.
(b) The Lowest Common Multiple (LCM) of two or more given numbers is the lowest
of their common multiples.
FAQ for class 6 maths chapter 3 Solutions
What is a factors in math?
factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.
what is Multiple
multiple in math are the numbers you get when you multiply a certain number by an integer. For example, multiples of 5 are: 10, 15, 20, 25, 30…etc. Multiples of 7 are: 14, 21, 28, 35, 42, 49…etc. … Remember, multiples can also be negative. So, some other multiples of 5 are: -5, -10, -15…etc.
What is a factor in 5th-grade math?
A factor divides a number completely without leaving any remainder. For example 30 ÷ 6 = 5, and there is no remainder. So we can say that 5 and 6 are the factors of 30.