Q.1 Write all the factors of the following numbers:
(a) 24

(b) 15

(c) 21

(d) 27

(e) 12

(f) 20

(g) 18

(h) 23

(i) 36

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 24 \(\rightarrow \) 1, 2, 3, 4, 6, 8, 12 and 24.

(b) 15 \(\rightarrow \) 1,3,5,and 15

(c) 21 \(\rightarrow \) 1, 3 , 7 and 21

(d) 27 \(\rightarrow \) 1,3,9 and 27

(e) 12 \(\rightarrow \) 1,2,3,4, 6 and 12

(f) 20 \(\rightarrow \) 1,2,4,5,10 and 20

(g) 18 \(\rightarrow \) 1,2,3,6,9 and 18

(h) 23 \(\rightarrow \)1 and 23

(i) 36 \(\rightarrow \) 1,2, 3, 4, 6, 9, 12, 18, and 36

Q.3 Match the items in column 1 with the items in column 2.

Column 1

(i) 35

(ii) 15

(iii) 16

(iv) 20

(v) 25

Column 2

(a) Multiple of 8

(b) Multiple of 7

(c) Multiple of 70

(d) Factor of 30

(e) Factor of 50

(f) Factor of 20

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

Column 1

(i) 35\(\rightarrow \). multiple of 7

(ii) 15\(\rightarrow \). Factor of 30

(iii) 16\(\rightarrow \). Multiple of 8

(iv) 20\(\rightarrow \). Factor of 20

(v) 25\(\rightarrow \). Factor of 50

Q.2 State whether the following statements are True or False:

(a) The sum of three odd numbers is even.

(b) The sum of two odd numbers and one even number is even.

(c) The product of three odd numbers is odd.

(d) If an even number is divided by 2, the quotient is always odd.

(e) All prime numbers are odd.

(f) Prime numbers do not have any factors.

(g) Sum of two prime numbers is always even.

(h) 2 is the only even prime number.

(i) All even numbers are composite numbers.

(j) The product of two even numbers is always even.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(?) False

(b) True

(c) True

(d) False

(e) False

(f) False

(g) False

(h) True

(i) False

(j) True

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 ______.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) A number which has only two factors is called a prime number.

(b) A number which has more than two factors is called a composite number.

(c) 1 is neither prime number nor composite number.

(d) The smallest prime number is 2

(e) The smallest composite number is 4

(f) The smallest even number is 2.

Q.2 Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

(a) 572

(b) 726352

(c) 5500

(d) 6000

(e) 12159

(f) 14560

(g) 21084

(h) 31795072

(i) 1700

(j) 2150

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 72 are the last two digits. Since, 72 is divisible by 4. Hence, 572 is also divisible by 4

572 are the last three digits. Since, 572 is not divisible by 8.

Hence, 572 is not divisible by 8

(b) 726352

52 are the last two digits. Since, 52 is divisible by 4. Hence, 726352 is divisible by 4

352 are the last three digits. Since 352 is divisible by 8.

Hence, 726352 is divisible by 8

(c) 5500

Since, last two digits are 00. Hence 5500 is divisible by 4

500 are the last three digits. Since, 500 is not divisible by 8.

Hence, 5500 is not divisible by 8

(d) 6000

Since, last two digits are 00. Hence 6000 is divisible by 4

Since, last three digits are 000.

Hence, 6000 is divisible by 8

(e) 12159

59 are the last two digits. Since, 59 is not divisible by 4. Hence, 12159 is not divisible by 4

159 are the last three digits. Since, 159 is not divisible by 8.

Hence, 12159 is not divisible by 8

(f) 14560

60 are the last two digits. Since 60 is divisible by 4. Hence, 14560 is divisible by 4

560 are the last three digits. Since, 560 is divisible by 8. Hence, 14560 is divisible by 8

(g) 21084

84 are the last two digits. Since, 84 is divisible by 4. Hence, 21084 is divisible by 4

084 are the last three digits. Since, 084 is not divisible by 8. Hence, 21084 is not divisible by 8

(h) 31795072

72 are the last two digits. Since, 72 is divisible by 4. Hence, 31795072 is divisible by 4

072 are the last three digits. Since, 072 is divisible by 8. Hence, 31795072 is divisible by 8

(i) 1700

Since, the last two digits are 00. Hence, 1700 is divisible by 4

700 are the last three digits. Since, 700 is not divisible by 8. Hence, 1700 is not divisible by 8

(j) 2150

50 are the last two digits. Since, 50 is not divisible by 4. Hence, 2150 is not divisible by 4

150 are the last three digits. Since, 150 is not divisible by 8. Hence, 2150 is not divisible by 8

Q.3 Using divisibility tests, determine which of following numbers are divisible by 6:

(a) 297144

(b) 1258

(c) 4335

(d) 61233

(e) 901352

(f) 438750

(g) 1790184

(h) 12583

(i) 639210

(j) 17852

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 297144 – it is divisible by 2 because it’s unit digit is 4, it is also divisible by 3 because the sum of all digit is divisible by 3 so it is divisible by 6.

(b) 1258 – it is divisible by 2 because it’s unit digit is 8 but it is not divisible by 3 so it is not divisible by 6

(c) 4335 – it is not divisible by 2 because it’s unit digit is 5, so it is not divisible by 6

(d) 61233 - it is not divisible by 2 because it’s unit digit is 3, so it is not divisible by 6

(e) 901352- it is divisible by 2 because it’s unit digit is 2 but it is not divisible by 3 so it is not divisible by 6

(f) 438750 - it is divisible by 2 because it’s unit digit is 0, it is also divisible by 3 because the sum of all digit is divisible by 3 so it is divisible by 6.

(g) 1790184 - it is divisible by 2 because it’s unit digit is 4, it is also divisible by 3 because the sum of all digit is divisible by 3 so it is divisible by 6.

(h) 12583 - it is not divisible by 2 because it’s unit digit is 3, so it is not divisible by 6

(i) 639210 - it is divisible by 2 because it’s unit digit is 0, it is also divisible by 3 because the sum of all digit is divisible by 3 so it is divisible by 6.

(j) 17852 - it is divisible by 2 because it’s unit digit is 2, but it is not divisible by 3 so it is not divisible by 6

4. Using divisibility tests, determine which of the following numbers are divisible by 11:

(a) 5445

(b) 10824

(c) 7138965

(d) 70169308

(e) 10000001

(f) 901153

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) Given number = 5445

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) Given number = 10824

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) Given number = 7138965

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) Given number = 70169308

Sum of all the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of all the digits at even places = 0 + 9 + 1 + 7 = 17

Difference = 17-17 = 0

Hence, the given number is divisible by 11.

(e) Given number = 10000001

Sum of all the digits at odd places = 1 + 0 + 0 + 0 = 1

Sum of all the digits at even places = 0 + 0 + 0 + 1 = 1

Difference = 1 – 1 = 0

Hence, the given number is 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:

(a) __ 6724

(b) 4765 __ 2

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) ___ 6724

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.

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

(b) 8 __ 9484

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 92 ___ 389

Let, at the missing digit be x

Sum of the digits at odd places = 9 + 3 + 2 = 14

Sum of the digits at even places = 8 + x + 9 = 17

Difference = 17 + x – 14 = x + 3

For the given number to be divisible by 11

x + 3 = 11

x = 11 – 3 = 8

So, the missing digit = 8

Hence, the required number is 928389.

(b) 8 ___ 9484

Let the missing digit be x

Sum of the digits at odd places = 4 + 4 + x = 8 + x

Sum of the digits at even places = 8 + 9 + 8 = 25

? Difference = 25 – (8 + x)

= 25 – 8 – x = 17 – x

For the given number to be divisible by 11

17 – 0 = 11

\(\therefore \)17 – 11 = 6

So, the missing digit = 6

Hence, the required number = 869484.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 20 and 28

Factors of 20 are 1, 2, 4, 5, 10, 20

Factors of 28 \(\rightarrow \) 1, 2, 4, 7, 28

Hence, the common factors \(\rightarrow \) 1, 2 and 4.

(b) 15 and 25

Factors of 15 \(\rightarrow \) 1, 3, 5, 15

Factors of 25 \(\rightarrow \)1, 5, 25

Hence, the common factors are 1 and 5.

(c) 35 and 50

Factors of 35 \(\rightarrow \) 1, 5, 7, 35

Factors of 50 \(\rightarrow \) 1, 2, 5, 10, 50

Hence, the common factors are 1 and 5.

(d) 56 and 120

Factors of 56 \(\rightarrow \) 1, 2, 4, 7, 8, 14, 28, 56

Factors of 120 \(\rightarrow\) 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120

Hence, the common factors are 1,2, 4, and 8.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 4, 8 and 12

Factors of 4 = 1, 2, 4

Factors of 8 = 1, 2, 4, 8

Factors of 12 = 1, 2, 3, 4, 6, 12

Hence, the common factors are 1, 2 and 4.
(b) 5, 15 and 25

Factors of 5 = 1, 5

Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25

Hence, the common factors are 1 and 5.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 6 and 8

multiples of 6 = 6,12,18,24,30....

multiples of 8 = 8,16,24,32......

Common multiples of 6 and 8 = 24, 48, 72

(c) 12 and 18.

multiples of 12 = 12,24,36,48,60.....

multiples of 18 = 18,36,54,72....

Common multiples of 12 and 18 = 36,72,108

5. Which of the following numbers are co-prime?

(a) 18 and 35

(b) 15 and 37

(c) 30 and 415

(d) 17 and 68

(e) 216 and 215

(f) 81 and 16

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 18 and 35

Factors of 18.= 1, 2, 3, 6, 9, 18

Factors of 35 = 1, 5, 7, 35

Common factor = 1

Since, their common factor is 1. Hence, the given two numbers are co-prime

(b) 15 and 37

Factors of 15 = 1, 3, 5, 15

Factors of 37 = 1, 37

Common factors = 1

Since, their common factor is 1. Hence, the given two numbers are co-prime

(c) 30 and 415

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Factors of 415 = 1, 5, 83, 415

Common factors = 1, 5

Since, their common factor is other than 1. Hence, the given two numbers are not co-prime

(d) 17 and 68

Factors of 17 = 1, 17

Factors of 68 = 1, 2, 4, 17, 34, 68

Common factors = 1, 17

Since, their common factor is other than 1. Hence, the given two numbers are not co-prime

(e) 216 and 215

Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216

Factors of 215 = 1, 5, 43, 215

Common factors = 1

Since, their common factor is 1. Hence, the given two numbers are co-prime

(f) 81 and 16

Factors of 81 = 1, 3, 9, 27, 81

Factors of 16 = 1, 2, 4, 8, 16

Common factors = 1

Since, their common factor is 1. Hence, the given two numbers are co-prime

Q.6. A number is divisible by both 5 and 12. By which other number will that number be always divisible?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

Multiples of 5 = 5 , 10, 15, 20,............60, 65,.........

Multiple of 12 = 12, 24, 36, 48, 60,..............

Common multiples of 5 and 12 = 60, 120,......

If the number is divisible by both 5 and 12 this the number will also be divisible by common multiple of 5 and 12 which is 60, 120 , 180,........

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

Factors of 12 = 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.

Q. 1. Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.

(b) If a number is divisible by 9, it must be divisible by 3.

(c) A number is divisible by 18, if it is divisible by both 3 and 6.

(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.

(e) If two numbers are co-primes, at least one of them must be prime.

(f) All numbers which are divisible by 4 must also be divisible by 8.

(g) All numbers which are divisible by 8 must also be divisible by 4.

(h) If a number exactly divides two numbers separately, it must exactly divide their sum.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) False

(b) True

(c) False

(d) True

(e) False

(f) False

(g) True

(h) True

(i) False

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The greatest 4-digit number = 9999

$$
\begin{array}{|l}
\llap{3~~~~} 9999 \\ \hline
\llap{3~~~~} 3333 \\ \hline
\llap{11~~~~} 1111 \\ \hline
\llap{101~~~~} 101 \\ \hline
1
\end{array}
$$
Hence, the prime factors of 9999 = 3 x 3 x 11 x 101.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The smallest 5-digit number = 10000

$$
\begin{array}{|l}
\llap{2~~~~} 10000 \\ \hline
\llap{2~~~~} 5000 \\ \hline
\llap{2~~~~} 2500 \\ \hline
\llap{2~~~~} 1250 \\ \hline
\llap{5~~~~} 625 \\ \hline
\llap{5~~~~} 125 \\ \hline
\llap{5~~~~} 25 \\ \hline
5
\end{array}
$$
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.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

$$
\begin{array}{|l}
\llap{7~~~~} 1729 \\ \hline
\llap{13~~~~} 247 \\ \hline
\llap{19~~~~} 19 \\ \hline
1
\end{array}
$$
Hence, the prime factors of 1729 = 7 x 13 x 19.

Here, 13 – 7 = 6 and 19 – 13 = 6

Hence the difference between two consecutive prime factors is 6.

Q.8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(i) 1 + 3 = 4 which is divisible by 4

(ii) 9 + 11 = 20 which is divisible by 4

(iii) 13 + 15 = 28 which is divisible by 4

Q.9. In which of the following expressions, prime factorisation has been done?

(a) 24 = 2 × 3 × 4

(b) 56 = 7 × 2 × 2 × 2

(c) 70 = 2 × 5 × 7

(d) 54 = 2 × 3 × 9

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 24 = 2 x 3 x 4

Here, 4 is not a prime number.

Hence, it is not a prime factorisation.

(b) 56 = 7 x 2 x 2 x 2

Here, all factors are prime numbers

Hence, it is a prime factorisation.

(c) 70 = 2 x 5 x 7

Here, all factors are prime numbers.

Hence, it is a prime factorisation.

(d) 54 = 2 x 3 x 9

Here, 9 is not a prime number.

Hence, it is not a prime factorisation.

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].

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

45 = 5 x 9

Here, 5 and 9 are co-prime numbers.

unit place of the given number is 0. So, it is divisible by 5.

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.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

No, since, 12 and 36 are both divisible by 4 and 6. But 12 and 36 are not divisible by 24

1. Find the HCF of the following numbers :

(a) 18, 48

(b) 30, 42

(c) 18, 60

(d) 27, 63

(e) 36, 84

(f) 34, 102

(g) 70, 105, 175

(h) 91, 112, 49

(i) 18, 54, 81

(j) 12, 45, 75

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 18, 48
$$
\begin{array}{|l}
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 48 \\ \hline
\llap{2~~~~} 24 \\ \hline
\llap{2~~~~} 12 \\ \hline
\llap{2~~~~} 6 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

HCF = 2 × 3 = 6

(b) 30, 42
$$
\begin{array}{|l}
\llap{2~~~~} 30 \\ \hline
\llap{3~~~~} 15 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 42 \\ \hline
\llap{3~~~~} 21 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
30 = 2 × 3 × 5

42 = 2 × 3 × 7

HCF = 2 × 3 = 6

(c) 18, 60

$$
\begin{array}{|l}
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 60 \\ \hline
\llap{2~~~~} 30 \\ \hline
\llap{3~~~~} 15 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

HCF = 2 × 3 = 6

(d) 27, 63

$$
\begin{array}{|l}
\llap{3~~~~} 27 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 63 \\ \hline
\llap{7~~~~} 21 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
27 = 3 × 3 × 3

63 = 3 × 3 × 7

HCF = 3 × 3 = 9

(e) 36, 84
$$
\begin{array}{|l}
\llap{2~~~~} 36 \\ \hline
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 84 \\ \hline
\llap{2~~~~} 42 \\ \hline
\llap{3~~~~} 21 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 36 \\ \hline
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 84 \\ \hline
\llap{2~~~~} 42 \\ \hline
\llap{3~~~~} 21 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

HCF = 2 × 2 × 3 = 12

(f) 34, 102
$$
\begin{array}{|l}
\llap{2~~~~} 34 \\ \hline
\llap{17~~~~} 17 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 102 \\ \hline
\llap{3~~~~} 51 \\ \hline
\llap{17~~~~} 17 \\ \hline
1
\end{array}
$$
34 = 2 × 17

102 = 2 × 3 × 17

HCF = 2 × 17 = 34

(g) 70, 105, 175
$$
\begin{array}{|l}
\llap{2~~~~} 70 \\ \hline
\llap{5~~~~} 35 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 105 \\ \hline
\llap{5~~~~} 35 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{5~~~~} 175 \\ \hline
\llap{5~~~~} 35 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
70 = 2 × 5 × 7

105 = 3 × 5 × 7

175 = 5 × 5 × 7

HCF = 5 × 7 = 35

(h) 91, 112, 49
$$
\begin{array}{|l}
\llap{7~~~~} 91 \\ \hline
\llap{13~~~~} 13 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 112 \\ \hline
\llap{2~~~~} 56 \\ \hline
\llap{2~~~~} 28 \\ \hline
\llap{2~~~~} 14 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{7~~~~} 49 \\ \hline
\llap{7~~~~} 7 \\ \hline
1
\end{array}
$$
91 = 7 × 13

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

HCF = 7

(I) 18, 54 , 81
$$
\begin{array}{|l}
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 54 \\ \hline
\llap{3~~~~} 27 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 81 \\ \hline
\llap{3~~~~} 27 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 18 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

HCF = 3 × 3 = 9

(i) 12, 45, 75
$$
\begin{array}{|l}
\llap{2~~~~} 12 \\ \hline
\llap{2~~~~} 6 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 45 \\ \hline
\llap{3~~~~} 15 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
12 = 2 × 2 × 3

45 = 3 × 3 × 5

75 = 3 × 5 × 5

HCF = 3

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) The common factor of two consecutive numbers is always 1.
Hence, their HCF = 1.

(b) The common factors of two consecutive even numbers are 1 and 2.
Hence, their HCF = 1 x 2 = 2.

(c) The common factor of two consecutive odd numbers is 1.
Hence, their HCF = 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?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

No, answer is not correct.

It’s correct answer is 1

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 exact number of times.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

Maximum value of weight which can measure the given weight exact number of time = HCF of 75 g and 69 kg
Prime factorisations of 75 and 69 are

$$
\begin{array}{|l}
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 69 \\ \hline
\llap{23~~~~} 23 \\ \hline
1
\end{array}
$$
75 = 3×5×5

69 = 3× 23

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?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The minimum distance that each boy should walk must be the least common multiple (LCM) of the measure of their steps.

$$
\begin{array}{|l}
\llap{7~~~~} 63, 70, 77 \\ \hline
9, 10, 11
\end{array}
$$
LCM = 7×9 ×10× 11 = 6930 cm

Hence the minimum distance is 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.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The longest tape required to measure the three dimensions of the room = HCF of 825, 675 and 450
Prime factorisations of 825, 675 and 450 are

$$
\begin{array}{|l}
\llap{3~~~~} 825 \\ \hline
\llap{5~~~~} 275 \\ \hline
\llap{5~~~~} 55 \\ \hline
\llap{11~~~~} 11 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 675 \\ \hline
\llap{3~~~~} 225 \\ \hline
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
3
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 450 \\ \hline
\llap{3~~~~} 225 \\ \hline
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
3
\end{array}
$$
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 = 3× 5× 5 = 75
Hence the required tape = 75cm

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The smallest 3-digit number = 100

Since LCM of 6, 8, 12

$$
\begin{array}{|l}
\llap{2~~~~} 6 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 8 \\ \hline
\llap{2~~~~} 4 \\ \hline
\llap{2~~~~} 2 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 12 \\ \hline
\llap{2~~~~} 6 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
6= 2× 3

8 = 2× 2 × 2

12 = 2 × 2 × 3

Then LCM = 2 × 2 × 3 × 2 = 24

Now we need to find the smallest 3-digit multiple of 24

\(\therefore \) 24 × 4 = 96 and 24 × 5 = 120

Hence, 120 is the smallest 3-digit number which is exactly divisible by 6, 8 and 12

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The greatest 3-digit number = 999
$$
\begin{array}{|l}
\llap{2~~~~} 8, 10, 12 \\ \hline
\llap{2~~~~} 4, 5, 6 \\ \hline
2, 5, 3
\end{array}
$$
LCM = 2× 2× 2 × 5 × 3 = 120

Now we need to find the greatest 3-digit multiple of 120

\(\therefore \) 120 × 8 = 960 and 120 × 9 = 1080

Hence, 960 is the greatest 3-digit number exactly divisible by 8, 10 and 12

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?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

LCM = 2× 2× 2×3×3×2×3 = 432

432 sec = 7 min 12 sec

Hence, lights will change together after every 432 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.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

Maximum capacity of the required measure is equal to the HCF of 403, 434 and 465.

$$
\begin{array}{|l}
\llap{13~~~~} 403 \\ \hline
\llap{31~~~~} 31 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{2~~~~} 434 \\ \hline
\llap{7~~~~} 217 \\ \hline
\llap{31~~~~} 31 \\ \hline
1
\end{array}
$$
$$
\begin{array}{|l}
\llap{3~~~~} 465 \\ \hline
\llap{5~~~~} 155 \\ \hline
\llap{31~~~~} 31 \\ \hline
1
\end{array}
$$
403=13× 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

Hence, the maximum capacity of the required container = 31 litres.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

$$
\begin{array}{|l}
\llap{2~~~~} 6, 15, 18 \\ \hline
\llap{3~~~~} 3, 15, 9 \\ \hline
1, 5, 3
\end{array}
$$
LCM = 2×3×5×3 = 90

Here, 90 is the least number exactly divisible by 6, 15 and 18.

To get a remainder 5, the least number will be 90 + 5 = 95.

Hence, 95 is the required number.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

The smallest 4-digit number = 1000.

$$
\begin{array}{|l}
\llap{2~~~~} 18, 24, 32 \\ \hline
\llap{2~~~~} 9, 12, 16 \\ \hline
\llap{2~~~~} 9, 6, 8 \\ \hline
\llap{3~~~~} 9, 3, 4 \\ \hline
3, 1, 4
\end{array}
$$
LCM = 2 x 2 x 2 x 3 x 3 x 4 = 288

Here, we need to find the smallest 4-digit multiple of 288

\(\therefore \) 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 ’ LCMs. Is LCM the product of two numbers in each case?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 9 and 4

It is co prime number then LCM = product of the number

LCM = 9 × 4 = 36

(b) 12 and 5

LCM = 12 × 5 = 60

(c) 6 and 5

LCM = 6 × 5 = 30

(d) 15 and 4

LCM = 15 × 4 = 60

Yes in each case the LCM of given numbers is the product of these numbers.

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?

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Answer :

(a) 5, 20
$$
\begin{array}{|l}
\llap{5~~~~} 5, 20 \\ \hline
1, 4
\end{array}
$$
LCM = 5×4 = 20

(b) 6, 18

$$
\begin{array}{|l}
\llap{2~~~~} 6, 18 \\ \hline
\llap{3~~~~} 3, 9 \\ \hline
1, 3
\end{array}
$$
LCM = 2 × 3 × 3 = 18

(c) 12, 48

$$
\begin{array}{|l}
\llap{2~~~~} 12, 48 \\ \hline
\llap{2~~~~} 6, 24 \\ \hline
\llap{3~~~~} 3, 12 \\ \hline
1, 4
\end{array}
$$
LCM = 2 × 2 × 3 × 4 = 48

(d) 9, 45

$$
\begin{array}{|l}
\llap{3~~~~} 9, 45 \\ \hline
\llap{3~~~~} 3, 15 \\ \hline
1, 5
\end{array}
$$
LCM = 3 × 3 × 5 = 45

There are total 55 questions present in ncert solutions for class 6 maths chapter 3 playing with numbers

There are total 5 long question/answers in ncert solutions for class 6 maths chapter 3 playing with numbers

There are total 7 exercise present in ncert solutions for class 6 maths chapter 3 playing with numbers