NCERT Solutions Class 6 Mathematics Chapter 3 Playing with Numbers

Playing with Numbers

 

EXERCISE 3.1

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
Sol.1) The factor of a number is an exact divisor of that number.
a) 24
1 × 24 = 24; 2 × 12 = 24; 3 × 8 = 24 ; 4 × 6 = 24
∴ factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
b) 15
1 × 15 = 15 ; 3 × 5 = 15
∴ Factors of 15 are 1, 3, 5 and 15
c) 21
1 × 21 = 21 ; 3 × 7 = 21
∴ Factors of 21 are 1, 3, 7 and 21.
d) 27
1 × 27 = 27 ; 3 × 9 = 27
∴ Factors of 27 are 1, 3, 9 and 27
e) 12
1 × 12 = 12 ; 2 × 6 = 12 ; 3 × 4 = 12
∴ Factors of 12 are 1, 2, 3, 4, 6 and 12
f) 20
1 × 20 = 20 ; 2 × 10 = 20 ; 4 × 5 = 20
∴ Factors of 20 are 1, 2, 4, 5, 10, 20
g) 18
1 × 18 = 18 ; 2 × 9 = 18 ; 3 × 6 = 18
∴ Factors of 18 are 1, 2, 3, 6, 9, 18
h) 23
1 × 23 = 23
∴ Factors of 23 are 1 and 23. [23 is a prime number]
i) 36
1 × 36 = 36; 2 × 18 = 36 ; 3 × 12 = 36; 4 × 9 = 36 ; 6 × 6 = 36
∴ Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36

Q.2) Write first five multiples of:
(a) 5 (b) 8 (c) 9
Sol.2) a) 5
5 × 1 = 5 ; 5 × 2 = 10; 5 × 3 = 15; 5 × 4 = 20 ; 5 × 5 = 25
The first five multiples of 5 are 5, 10, 15, 20 and 25
b) 8
8 × 1 = 8 ; 8 × 2 = 16; 8 × 3 = 24; 8 × 4 = 32; 8 × 5 = 40
The first five multiples of 8 are 8, 16, 24, 32 and 40
c) 9
9 × 1 = 9 ; 9 × 2 = 10 ; 9 × 3 = 27; 9 × 4 = 36 ; 9 × 5 = 45
The first five 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
Sol.3) 35 is the multiple of 7 because 7 × 5 = 35. Therefore, (i) matches with (b).
15 is the factor of 30 because 30 = 2 × 15. Therefore, (ii) matches with (d).
16 is the multiple of 8 because 8 × 2 = 16. Therefore, (iii) matches with (a).
20 is the factor of 20 because 20 × 1 = 20. Therefore, (iv) matches with (f).
25 is the factor of 50 because 50 = 2 × 25. Therefore, (v) matches with (e).

Q.4) Find all the multiples of 9 up to 100.
Sol.4) 9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 × 4 = 36, 9 × 5 = 45 , 9 × 6 = 54
9 × 7 = 63, 9 × 8 = 72 , 9 × 9 = 81, 9 × 10 = 90, 9 × 11 = 99, 9 × 12 = 108
Therefore, all the multiples of 9 up to 100 are:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99

Exercise 3.2

Q.1) What is the sum of any two?
(a) Odd numbers (b) Even numbers
Sol.1) Consider any two odd numbers say 7 and 13.
7 + 13 = 20.
Similarly, 23 + 21 = 44
20 and 44 are both even numbers.
Thus, the sum of any two odd numbers is an even number.
Now, consider even numbers 12 and 14.
12 + 14 = 28.
Similarly, 24 + 64 = 88
28 and 88 are even numbers.
Thus, the sum of any two even numbers is an even number.

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.
Sol.2) (a) This is false. The sum of three odd numbers is even..
5 + 3 + 5 = 13, 13 is an odd number.
(b) This is true. We know that sum of two odd numbers is always even. Adding one more even number will keep the result an even number. [𝑒𝑥: 7 + 3 + 4 = 14. ]
(c) True. The product of three odd numbers is odd.
[𝐸𝑥: 7 × 3 × 9 = 21 × 9 = 189]
(d) If an even number is divided by 2, the quotient is always odd.
This is false. Example, 1002 = 50 𝑎𝑛𝑑 50 is an even number.
(e) All prime numbers are odd.
False. 2 is the only prime number which is even.
(f) Prime numbers do not have any factors.
False. Factors of given prime number is 1 and number itself. Example, factors of prime number 23 are 1 and 23.
(g) Sum of two prime numbers is always even.
False. 2 and 3 are both prime numbers. Sum of 2 and 3 is 5 which is odd.
(h) 2 is the only even prime number.
True.
(i) All even numbers are composite numbers.
False. 2 is not a composite number and it is even.
(j) The product of two even numbers is always even.
True. examples like 4 × 2 = 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.`
Sol.3) 17 and 71 are prime numbers and they both use 1 and 7 as digits.
Similarly, 37 and 73 are prime numbers and they both use 3 and 7 as digits.
And, 79 and 97 are prime numbers and they both use 7 and 9 as digits.

Q.4) Write down separately the prime and composite numbers less than 20.
Sol.4) Prime numbers less than 20:
2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20:
4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Q.5) What is the greatest prime number between 1 and 10?
Sol.5) Prime numbers between 1 and 10 are: 2, 3, 5, 7.
Greatest prime number between 1 and 10 is 7.

Q.6) Express the following as the sum of two odd primes.
(a) 44 (b) 36 (c) 24 (d) 18
Sol.6) There are many ways of writing a number as the sum of two odd primes. The following s
show a few of them
(a) 44
= 41 + 3 𝑂𝑅 37 + 7
(b) 36
= 31 + 5 𝑂𝑅 29 + 7
(c) 24
= 17 + 7 𝑂𝑅 19 + 5
(d) 18
= 11 + 7 𝑂𝑅 13 + 5

Q.7) Give three pairs of prime numbers whose difference is 2.
Sol.7) (5, 7), (11, 13) and (17, 19) are three pairs of prime numbers whose difference is 2.
Two prime numbers whose difference is 2 are called twin primes

Q.8) Which of the following numbers are prime?
(a) 23 (b) 51 (c) 37 (d) 26
Sol.8) The factors of 23 are 1 and 23.
The factors of 51 are 1, 51, 13 and 7.
The factors of 37 are 1 and 37
The factors of 26 are 1, 26, 13, 2
Prime numbers are numbers whose only factors are 1 and itself.
So, a) and c) are prime numbers.

Q.9) Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Sol.9) 90, 91, 92, 93, 94, 95 and 96 are 7 consecutive composite numbers less than 100.

Q.10) Express each of the following numbers as the sum of three odd primes:
(a) 21 (b) 31 (c) 53 (d) 61
Sol.10) a) 21 = 7 + 9 + 5
b) 31 = 23 + 5 + 3
c) 53 = 23 + 17 + 1
d) 61 = 47 + 11 + 3

Q.11) Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
Sol.11) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.
Let us write five pairs of prime numbers whose sum is divisible by 5.
2 + 3 = 5
2 + 13 = 15
3 + 17 = 20
7 + 13 = 20
11 + 19 = 30

Q.12) Fill in the blanks:
A number which has only two factors is called a ______.
A number which has more than two factors is called a ______.
1 is neither ______ nor ______.
The smallest prime number is ______.
The smallest composite number is _____.
The smallest even number is ______.
Sol.12) A number which has only two factors is called a prime number.
A number which has more than two factors is called a composite number.
1 is neither composite number nor prime number.
The smallest prime number is 2.
The smallest composite number is 4.
The smallest even number is 2.

EXERCISE 3.3

Q.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):

Sol.1) 990
Divisible by 2 as it has 0 in its one’s place.
Divisible by 3 as the sum of the digits, 18, is a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 90, is not divisible by 4.
Divisible by 5 as the number has 0 in its one’s place.
Divisible by 6 as the number is divisible by both 2 and 3.
Not divisible by 8.
Divisible by 9 as the sum of the digits of the number, 9 + 9 = 18, is divisible by 9.
Divisible by 10 as the number has 0 in its unit’s place.
The number is divisible by 11.
The sum of the digits in the odd places from the right is 9 + 0 = 9. The sum of the digits in the even places from the right is 9. The difference is 0. If the difference is 0 or a divisible by 11, then the number is divisible by 11.
1586
Divisible by 2 as the number in the unit’s place is 2.
Not divisible by 3 as the sum of the digits, 20, is not a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 86, is not divisible by 4.
Not divisible by 5 as the number does not have 0 or 5 in its one’s place.
Not divisible by 6 as the number is divisible by 2 but not by 3.
Not divisible by 8 as the number formed by the last three digits, 586, is not divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 20, is not divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 6 + 5 = 11. The sum of the digits in the even places from the right is 8 + 1 = 9. The difference is 2. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.
275
Not divisible by 2 as the number does not have 0, 2, 4, 6 or 8 in its unit’s place.
Not divisible by 3 as the sum of the digits, 14, is not a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 75, is not divisible by 4.
Divisible by 5 as the number has 5 in its one’s place.
Not divisible by 6 as the number is neither divisible by 2 nor by 3.
Not divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 14, is not divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is divisible by 11.
The sum of the digits in the odd places from the right is 5 + 2 = 7. The sum of the digits in the even places from the right is 7. The difference is 0. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is divisible by 11.
6686
Divisible by 2 as the number has 6 in its unit’s place.
Not divisible by 3 as the sum of the digits, 26, is not a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 86, is not divisible by 4.
Not divisible by 5 as the number does not have 0 or 5 in its one’s place.
Not divisible by 6 as the number is divisible by 2 but not by 3.
Not divisible by 8 as the number formed by the last three digits 686 is not divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 86, is not divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 6 + 6 = 12. The sum of the digits in the even places from the right is 14. The difference is 2. If the difference is 0 or a number divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.
639210
Divisible by 2 as the number has 0 in its unit’s place.
Divisible by 3 as the sum of the digits, 21, is a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 10, is not divisible by 4.
Divisible by 5 as the number has 0 in its one’s place.
Divisible by 6 as the number is divisible by 2 and by 3.
Not divisible by 8 as the number formed by the last three digits, 210, is not divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 21, is not divisible by 9.
Divisible by 10 as the number has 0 in its unit’s place.
The number is divisible by 11.
The sum of the digits in the odd places from the right is 5. The sum of the digits in the even places from the right is 16. The difference is 11. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is divisible by 11.
429714
Divisible by 2 as the number has 4 in its unit’s place.
Divisible by 3 as the sum of the digits, 27, is a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 14, is not divisible by 4.
Not divisible by 5 as the number does not have 0 or 5 in its one’s place.
Divisible by 6 as the number is divisible by both 2 and by 3.
Not divisible by 8 as the number formed by the last three digits, 714, is not divisible by 8.
Divisible by 9 as the sum of the digits of the number, 27, is divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 13. The sum of the digits in the even places from the right is 14. The difference is 1. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.
2856
Divisible by 2 as the number has 6 in its unit’s place.
Divisible by 3 as the sum of the digits, 21, is a multiple of 3.
Divisible by 4 as the number formed by its last two digits, 56, is divisible by 4.
Not divisible by 5 as the number does not have 0 or 5 in its one’s place.
Divisible by 6 as the number is divisible by both 2 and by 3.
Divisible by 8 as the number formed by the last three digits, 856, is divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 21, is not divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 14. The sum of the digits in the even places from the right is 7. The difference is 7. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.
3060
Divisible by 2 as the number has 0 in its unit’s place.
Divisible by 3 as the sum of the digits, 9, is a multiple of 3.
Divisible by 4 as the number formed by its last two digits, 60, is divisible by 4.
Divisible by 5 as the number has 0 in its one’s place.
Divisible by 6 as the number is divisible by both 2 and by 3.
Not divisible by 8 as the number formed by the last three digits, 60, is not divisible by 8.
Divisible by 9 as the sum of the digits of the number, 9, is divisible by 9.
Divisible by 10 as the number has 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 0. The sum of the digits in the even places from the right is 9. The difference is 9. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.
406839
Not divisible by 2 as the number does not have 0, 2, 4, 6 or 8 in its unit’s place.
Divisible by 3 as the sum of the digits, 30, is a multiple of 3.
Not divisible by 4 as the number formed by its last two digits, 39, is not divisible by 4.
Not divisible by 5 as the number does not have 0 or 5 in its one’s place.
Not divisible by 6 as the number is not divisible by 2.
Not divisible by 8 as the number formed by the last three digits, 839, is not divisible by 8.
Not divisible by 9 as the sum of the digits of the number, 30, is not divisible by 9.
Not divisible by 10 as the number does not have 0 in its unit’s place.
The number is not divisible by 11.
The sum of the digits in the odd places from the right is 17. The sum of the digits in the even places from the right is 13. The difference is 4. If the difference is 0 or a divisible by 11, then the number is divisible by 11. So, the number is not divisible by 11.

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
Sol.2) A number with 3 or more digits is divisible by 4 if the number formed by its last two digits is divisible by 4.
A number with 3 or more digits is divisible by 8 if the number formed by its last three digits is divisible by 8.

a) 572
72 is divisible by 4, hence the number is divisible by 4.
The number is not divisible by 8.

b) 726352
52 is divisible by 4, hence the number is divisible by 4.
352 is divisible by 8, hence the number is divisible by 8.

c) 5500
0 is divisible by 4, hence the number is divisible by 4.
500 is not divisible by 8, hence the number is not divisible by 8.

d) 6000
0 is divisible by 4 and 8. Hence, the number is divisible by 4 and 8.

e) 12159
59 is not divisible by 4. Hence, the number is not divisible by 4.
159 is not divisible by 8, hence the number is not divisible by 8.

f) 14560
60 is divisible by 4, hence the number is divisible by 4.
560 is divisible by 8, hence the number is divisible by 8.

g) 21084
84 is divisible by 4, hence the number is divisible by 4.
84 is not divisible by 8, hence the number is not divisible by 8.

h) 31795072
72 is divisible by 4 and 8. Hence the number is divisible by 4 and 8.

i) 1700
The number is divisible by 4.
700 is not divisible by 8, hence the number is not divisible by 8.

j) 2150
50 is not divisible by 4, hence the number is not divisible by 4.
150 is not divisible by 8, hence the number 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
Sol.3) A number is divisible by 6 if it is divisible by both 2 and 3.

a) 297144
The number is divisible by 2, as the number has 4 in its one’s place.
The number is divisible by 3 as the sum of the digits, 27, is divisible by 3.
Hence the number is divisible by 6.

b) 1258
The number is not divisible by 3 as the sum of the digits, 16, is not divisible by 3.
Hence, the number is not divisible by 6.

c) 4335
The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its
unit’s place. Hence the number is not divisible by 6.

d) 61233
The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its
unit’s place. Hence the number is not divisible by 6.

e) 901352
The number is not divisible by 3 as the sum of the digits, 20, is not divisible by 3.
Hence, the number is not divisible by 6.

f) 438750
The number is divisible by 2, as the number has 0 in its one’s place.
The number is divisible by 3 as the sum of the digits, 27, is divisible by 3.
Hence the number is divisible by 6.

g) 1790184
The number is divisible by 2, as the number has 4 in its one’s place.
The number is divisible by 3 as the sum of the digits, 30, is divisible by 3.
Hence the number is divisible by 6.

h) 12583
The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its
unit’s place. Hence the number is not divisible by 6.

i) 639210
The number is divisible by 2, as the number has 4 in its one’s place.
The number is divisible by 3 as the sum of the digits, 21, is divisible by 3.
Hence the number is divisible by 6.

j) 17852
The number is not divisible by 3 as the sum of the digits, 23, is not divisible by 3.
Hence, the number is not divisible by 6.
From the discussion above, we can say that (a), (f), (g) and (i) are the numbers which are
divisible by 6.

Q.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
Sol.4) (a) 5445
Sum of odd digits = 5 + 4 = 9
Sum of even digits = 4 + 5 = 9
Difference = 0
Difference is divisible by 11. Therefore, 5445 is divisible by 11.

(b) 10824
Sum of odd digits = 1 + 8 + 4 = 13
Sum of even digits = 0 + 2 = 2
Difference = 13 − 2 = 11
Difference is divisible by 11. Therefore, 10824 is divisible by 11.

(c) 7138965
Sum of odd digits = 7 + 3 + 9 + 5 = 24
Sum of even digits = 1 + 8 + 6 = 15
Difference = 24 − 15 = 9
Difference is not divisible by 11. Therefore, 7138965 is not divisible by 11.

(d) 70169308
Sum of odd digits counting from right = 8 + 3 + 6 + 0 = 17
Sum of even digits counting from right = 0 + 9 + 1 + 7 = 17
Difference = 17 − 17 = 0
Difference is divisible by 11. Therefore, 70169308 is divisible by 11.

(e) 10000001
Sum of odd digits = 1 + 0 + 0 + 0 = 1
Sum of even digits = 0 + 0 + 0 + 1 = 1
Difference = 1 − 1 = 0
Difference is divisible by 11. Therefore, 10000001 is divisible by 11.

(f) 901153
Sum of odd digits = 9 + 1 + 5 = 15
Sum of even digits = 0 + 1 + 3 = 4
Difference = 15 − 4 = 11
Difference is divisible by 11. Therefore, 901153 is divisible by 11.
From the discussion above, we can say that numbers (a), (b), (d), (e) and (f) are 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
Sol.5) Number is divisible by 3 if the sum of digits is multiple of 3.
a) A number is divisible by 3, if the sum of the digits is a multiple of 3.
The sum of the digits of the given number is 6 + 7 + 2 + 4 + __ = 19.
The greatest digit that could be added is 9.
But, 19 + 9 = 28, which is not a multiple of 3.
So, find 19 + 8 = 27, which is a multiple of 3.
The greatest digit in the blank space should be 8.
Similarly, the smallest digit which can be added to 19 is 0. But 19 is not a multiple of 3.
19 + 1 = 20, not a multiple of 3.
19 + 2 = 21, is a multiple of 3.
Hence, the smallest digit that should be in the blank space is 2.
b) The Sum of the digits is 4 + 7 + 6 + 5 + 2 = 24
The greatest digit that could be added is 9.
24 + 9 = 33; which is a multiple of 3.
The smallest digit which can be added to 24 is 0, as 24 is a multiple of 3.
The greatest digit to be added is 9 and the smallest digit to be added is 0.

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
Sol.6) (a) 92 __ 389
Let p be the missing number in the blank space.
Find the sum of the digits at the odd places from the right.
9 + 3 + 2 = 14
Find the sum of the digits at the even places from the right.
8 + 𝑝 + 9 = 17 + 𝑝
If the difference is either 0 or 11, then the number is divisible by 11.
That is, 17 + 𝑝 – 14 = 0 OR 17 + 𝑝 – 14 = 11
3 + 𝑝 = 0 OR 3 + 𝑝 = 11
𝑝 = −3 OR 𝑝 = 8
But, 𝑝 cannot be negative.
Hence, 𝑝 = 8.
The missing digit in the blank space is 8.
(b) 8 __ 9484
Let 𝑝 be the missing number in the blank space.
Find the sum of the digits at the odd places from the right.
4 + 4 + 𝑝 = 8 + 𝑝
Find the sum of the digits at the even places from the right.
8 + 9 + 8 = 25
If the difference is either 0 or 11, then the number is divisible by 11.
That is, 25 – (8 + 𝑝) = 0 OR 25 – (8 + 𝑝) = 11
17 – 𝑝 = 0 OR 17 – 𝑝 = 11
𝑝 = 17 OR 𝑝 = 6
But, 𝑝 = 17 is not possible.
Hence, p = 6.
The missing digit in the blank space is 6.

EXERCISE 3.4

Q.1) Find the common factors of:
(a) 20 and 28 (b) 15 and 25
(c) 35 and 50 (d) 56 and 120
Sol.1) a) 20 and 28
Factors of 20 are 1, 2, 4, 5, 10, 20
Factors of 28 are 1, 2, 4, 7, 14, 28
Common factors are: 1, 2, 4

b) 15 and 25
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
Common factors are 1, 5

c) 35 and 50
Factors of 35 are 1, 5, 7, 35
Factors of 50 are 1, 2, 5, 10, 25, 50
Common factors are 1, 5,

d) 56 and 120
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, 24, 30, 40, 60, 120
Common factors are 1, 2, 4, 8

Q.2) Find the common factors of:
(a) 4, 8 and 12 (b) 5, 15 and 25
Sol.2) a) 4, 8 and 12
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
Common factors are 1, 2, 4

b) 5, 15 and 25
Factors of 5 are 1, 5
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
Common factors are 1, 5

Q.3) Find first three common multiples of:
(a) 6 and 8 (b) 12 and 18
Sol.3) a) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72…
Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96…
∴ First three common multiples are 24, 48, 72
b) Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 108, 120, 132, 144,…
Multiples of 18 are 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198…
∴ First three common multiples are 36, 72, 108

Q.4) Write all the numbers less than 100 which are common multiples of 3 and 4.
Sol.4) Multiples of 3 less than 100:
3, 6, 9, 12, 15, 18, 21, 24,…93, 96, 99
Multiples of 4 less than 100:
4, 8, 12, 16, …96
∴ Common multiples are: 12, 24, 36, 48, 60, 72, 84, 96

Q.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
Sol.5) Two numbers having only 1 as a common factor are called co-prime numbers.
a) 18 and 35
Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 35 are 1, 5, 7, 35
Common factor is 1.
So, 18 and 35 are Co-prime numbers

b) 15 and 37
Factors of 15 are 1, 3, 5, 15
Factors of 37 are 1, 37
Common factor is 1.
15 and 37 are Co-prime numbers.

c) 30 and 415
Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 are 1, 5, 83, 415
Common factors are 1 and 5.
The numbers 30 and 415 are not Co-prime

d) 17 and 68
Factors of 17 are 1, 17
Factors of 68 are 1, 2, 4, 17, 34, 68
Common factors are 1, 17
The numbers are not co-prime.

e) 216 and 215
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216,
Factors of 215 are 1, 5, 43, 215
Common factor is 1.
The numbers are co-prime.

f) 81 and 16
Factors of 81 are 1, 3, 9, 27, 81
Factors of 16 are 1, 2, 4, 8, 16
Common factor 1
The 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?
Sol.6) The number is divisible by 5 and 12 and as 5 and 12 are co-prime numbers, the number must be divisible by the product 5 × 12 = 60.
So, the given number will always be divisible by 60.

Q.7) A number is divisible by 12. By what other numbers will that number be divisible?
Sol.7) The number is divisible by 12.
So, the number will be divisible by all the factors of 12.
Factors of 12 are 1, 2, 3, 4, 6, 12
The number will also be divisible by the numbers 2, 3, 4 and 6.

Exercise 3.5

Q.1) Which of the following statements are true:-
Sol.1) (a) If a number is divisible by 3, it must be divisible by 9.
False. We can prove it with the help of an example. 12 is divisible by 3 but it is not divisible by 9.

(b) If a number is divisible by 9, it must be divisible by 3.
True. If a given number is divisible by some number, then it is also divisible by its factor.

(c) A number is divisible by 18, if it is divisible by both 3 and 6.
False. 12 is divisible by both 3 and 6 but it is not divisible by 18.

(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
True. This is because if a number is divisible by two given co-prime numbers then it is also divisible by their product. We know that 9 and 10 are co-primes and their product is equal to 90.

(e) If two numbers are co-primes, at least one of them must be prime.
False. Take an example, 8 and 9 are co-prime numbers and none of them is a prime number.

(f) All numbers which are divisible by 4 must also be divisible by 8.
False. 12 is divisible by 4 but it is not divisible by 8.

(g) All numbers which are divisible by 8 must also be divisible by 4.
True. If a number is divisible by another number, then it is divisible by each of the factors of that number.

(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
True. Take different examples and check this property.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
False. Let's take an example, 9 and 3 are not divisible by 4 but their sum is divisible by 4.

Q.2) Here are two different factor trees for 60. Write the missing numbers?

Sol.2) a) 6 × 10 = 60
Now, 2 × 3 = 6 and 5 × 2 = 10
So, the missing number are 3 and 2.
b) 60 = 30 × 2
30 = 10 × 3
10 = 2 × 5
The missing numbers are 2, 3 and 5

Q.3) Which factors are not included in the prime factorisation of a composite number?
Sol.3) 1. The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers.
2. Numbers having more than two factors are called Composite numbers.1 and the number itself
3. 1 is neither a prime nor a composite number.

Q.4) Write the greatest 4-digit number and express it in terms of its prime factors
Sol.4) The greatest 4-digit number is 9999 & its factor can be found as described follows:

Hence, greatest 4-digit number 9999 can be expressed in the form of its prime factors as
3 × 3 × 11 × 101

Q.5) Write the smallest 5-digit number and express it in the form of its prime factors.
Sol.5) The smallest 5-digit number is 10000 & its factors can be found as described below:

Hence, the smallest 5-digit number 10000 can be expressed in the form of its prime factors as 2 × 2 × 2 × 2 × 5 × 5 × 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.
Sol.6) 1729 = 7 × 13 × 19
Prime factors of 1729 in ascending order are 7, 13 and 19.
Clearly 13 − 7 = 6 and 19 − 13 = 6.
Here, difference of 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.
Sol.7) Case 1:
Let us take the three consecutive numbers 6, 7, 8
Product of these numbers: 6 × 7 × 8 = 336. Also 336/6 = 56
Case 2:
Let us take the three consecutive numbers 4, 5, 6
Product of these numbers: 4 × 5 × 6 = 120. Also 120/6 = 20
Case 3:
Let us take one more set of three consecutive numbers 9, 10, 11
Product of these numbers: 9 × 10 × 11 = 990. Also 990/6 = 165
From the above examples it is clear that the product of three consecutive numbers is always divisible by 6

Q.8) The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Sol.8) Example 1:
Let us take the two consecutive odd numbers 3 and 5
Sum of these numbers: 3 + 5 = 8. Also 8/4 = 2
Example 2:
Let us take the two consecutive odd numbers 11 and 3
Sum of these numbers: 11 + 13 = 24. Also 24/4 = 6
Example 3:
Let us take others set of two consecutive odd numbers 21 and 23
Sum of these numbers: 21 + 23 = 44. Also 44/11 = 4
From the above examples it is clear that the sum of two consecutive odd numbers is always 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
Sol.9) (a) 24 = 2 × 3 × 4

9 is not a prime factor.
∴ the answer is (b) and (c).

Q.10) Determine if 25110 is divisible by 45.
Sol.10) 45 = 5 × 9
As 5 and 9 are co-primes, a number divisible by 5 and 9 will be divisible by 45.
Consider divisibility of 25110 by 5.
As the unit’s digit is 0, the number is divisible by 5.
Consider divisibility of 25110 by 9.
The number is divisible by 9, as the sum of the digits of the number is 9.
The number is divisible by both 5 and 9, which are co-prime numbers. Hence, the number is divisible by the product, 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.
Sol.11) No. Number 12 is divisible by both 4 and 6; but 12 is not divisible by 4 × 6 = 24.
Similarly Number 36 is divisible by both 4 and 6; but 36 is not divisible by 4 × 6 = 24.
Also Number 60 is divisible by both 4 and 6; but 60 is not divisible by 4 × 6 = 24.

Q.12) I am the smallest number, having four different prime factors. Can you find me?
Sol.12) We know that starting from the beginning we have 4 different prime numbers as 2, 3, 5, 7 and their product will be the required number i.e. 2 × 3 × 5 × 7 = 210

Exercise 3.6

Q.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
Sol.1) The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.

12 = 2 × 2 × 3
45 = 5 × 3 × 3
75 = 5 × 3 × 5
Common factor is 3. HCF = 3

Q.2) What is the HCF of two consecutive
(a) numbers? (b) even numbers? (c) odd numbers?
Sol.2) a) The HCF of two consecutive numbers is 1. For example, HCF of 4 and 5 is 1.
b) The HCF of two consecutive even numbers is 2. For example, HCF of 8 and 10 is 2.
c) The HCF of two consecutive odd numbers is 1. For example, HCF of 13 and 15 is 1.

Q.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?
Sol.3) No, the answer is not correct because 1 is the correct answer. Confusion arises because 1 is not shown in the prime factorization of any number.

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 exact number of times.
Sol.1) The required value of weight should measure the weight of the fertiliser exact number of times. That is, this number should be an exact divisor of the weights of the two bags. Also, this value should be the maximum. Thus, the maximum value of weight can be obtained by finding the HCF of the two weights, 75 kg and 69 kg.

75 = 5 × 5 × 3
69 = 23 × 3
HCF is 3.
Hence, the maximum value of weight that can measure the weight of the fertiliser exact number of times is 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?
Sol.2) The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps.
To find minimum distance = 𝐿. 𝐶. 𝑀 𝑜𝑓 63, 70, 77

∴ The LCM of the three numbers is 3 × 3 × 7 × 2 × 5 × 11 = 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?
Sol.3) The measurement of longest tape = 𝐻. 𝐶. 𝐹. 𝑜𝑓 825 𝑐𝑚, 675 𝑐𝑚, 450 𝑐𝑚

825 = 5 × 5 × 3 × 11
675 = 5 × 5 × 3 × 3 × 3
450 = 5 × 5 × 3 × 3 × 2
HCF of the numbers is 5 × 5 × 3 = 75 𝑐𝑚
So, the longest tape that can measure the dimensions of the room exactly is 75 𝑐𝑚.

Q.4) Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Sol.4) Smallest number = LCM of 6, 8, 12

LCM = 2 × 2 × 2 × 3 = 24
We have to find the smallest 3-digit multiple of 24.
It can be seen that 24 × 4 = 96 and 24 × 5 = 120.
Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

Q.5) Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
Sol.5) Let's find L.C.M of 8, 10 and 12. 

∴ L.C.M of 8, 10 and 12 = 2 × 2 × 2 × 5 × 3 = 120.
∴ 120 is the smallest number which is exactly divisible by 8, 10 and 12.
We need to find the greatest 3-digit number which is divisible by 8, 10 and 12. This number would be the multiple of 120. So, let's find multiples of 120.
120 × 1 = 120
120 × 2 = 240
120 × 3 = 360
120 × 4 = 480
120 × 5 = 600
120 × 8 = 960
The greatest number of 3-digits 999

The greatest 3-digits number divisible by 120
999 − 39 = 960
Therefore, 960 is the greatest 3-digit multiple of 120.
It means that 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?
Sol.6) Given that traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds respectively.
L.C.M. of 48, 72, 108 is:

48 = 2 × 2 × 2 × 2 × 3
72 = 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
Hence, LCM of 48, 72 and 108 is (2 × 2 × 2 × 2 × 3 × 3 × 3) = 432
That is after 432 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 they will change simultaneously
432/60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 = 7 𝑚𝑖𝑛 12 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
So, the traffic lights at three different road crossings change every 432 seconds, that is,
7: 00 𝑎. 𝑚 + 7 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 12 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 = 07: 07: 12 𝑎𝑚
Thus the traffic lights change simultaneously at 7: 7: 12 𝑎𝑚

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.
Sol.7) To find the maximum capacity of the container that can measure the diesel of three tankers we have to take their HCF

Factors of 403 = 1, 13, 31, 403
Factors of 434 = 1, 2, 7, 14, 31, 62, 217, 434
Factors of 465 = 1, 3, 5, 15, 31, 93, 155, 465
Highest common factor of 403, 434 and 465 is 31.
∴ 31 𝑙𝑖𝑡𝑟𝑒𝑠 is the maximum capacity of a container that can measure the diesel of the three containers exact number of times

Q.8) Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case?
Sol.8) To find the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case, we have to find the L.C.M. of 6, 15 and 18.
L.C.M. of 6, 5 and 18

6 = 2 × 3
15 = 3 × 5
18 = 2 × 3 × 3
L.C.M. of 6, 15 and 18 = 3 × 2 × 3 × 5 = 90
But, we need the least number that leaves remainder 5 in each case then add 5 in that number
Now,
5 + 90 = 95
Hence, 95 is the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case.
Let us check our answer.
1) 95/6
Quotient = 15
Remainder = 5
2) 95/15
Quotient = 6
Remainder = 5
3) 95/18
Quotient = 5
Remainder = 5
So, the required least number is 95.

Q.9) Find the smallest 4-digit number which is divisible by 18, 24 and 32
Sol.9) Given numbers are 18, 24 & 32

L.C.M of 18, 24 and 32 = 2 × 3 × 3 × 4 × 4 = 288 
We need to find smallest 4-digit number which is divisible by 18, 24 and 32. This number would be factor of 288.
So, let's find the factors of 288.
288 × 1 = 288
288 × 2 = 576
288 × 3 = 864
288 × 4 = 1152
We require a 4-digit number. So, 1152 is the smallest 4-digit multiple of 288.
∴ 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?
Sol.10) (a) 9 and 4
Factors of 9 & 4:

Prime factor 3 is occurring most number of times in the prime factorization of 9.
Prime factor 2 is occurring most number of times in the prime factorization of 4.
Therefore, L.CM = 3 × 3 × 2 × 2 = 36
(b) 12 and 5

Prime factor 2 is occurring most number of times in the prime factorization of 12.
Prime factor 3 is occurring most number of times in the prime factorization of 12.
Prime factor 5 is occurring most number of times in the prime factorization of 5.
Therefore, L.C.M = 2 × 2 × 3 × 5 = 60
(c) 6 and 5
Factors of 6 & 5:

2 is coming most number of times in the prime factorization of 6.
3 is coming most number of times in the prime factorization of 6.
5 is coming most number of times in the prime factorization of 5.
Therefore, L.C.M = 2 × 3 × 5 = 30

is coming most number of times in the prime factorization of 15.
5 is coming most number of times in the prime factorization of 15.
2 is coming most number of times in the prime factorization of 4.
Therefore, L.C.M = 3 × 5 × 2 × 2 = 60
L.C.M is equal to the product of given numbers. This happens when given numbers are co-prime.

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?
Sol.11) a) 5 and 20