fbpx

 CH 3  –  Playing with Numbers

Exercise 3.1

  1. Write all the factors of the following numbers:

The factor of a number is an exact divisor of that number.

  1. a)24

Solution

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

Thus, the factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

  1. b)15

Solution

1 × 15 = 15      3 × 5 = 15

Factors of 15 are 1, 3, 5 and 15

  1. c)21

Solution

1 × 21= 21       3 × 7 = 21

Factors of 21 are 1, 3, 7 and 21.

  1. d)27

Solution

1 × 27 = 27      3 × 9 = 27

Factors of 27 are 1, 3, 9 and 27

  1. e)12

Solution

1 × 12 = 12      2 × 6 = 12        3 × 4 = 12

Factors of 12 are 1, 2, 3, 4, 6 and 12

  1. f)20

1 × 20 = 20      2 × 10 = 20      4 × 5 = 20

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

  1. g)18

1 × 18 = 18      2 × 9 = 18        3 × 6 = 18

Factors of 18 are 1, 2, 3, 6, 9, 18

  1. h)23

1 × 23 = 23

Factors of 23 are 1 and 23. [23 is a prime number]

  1. 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

  1. Write first five multiples of :
  2. a)5

Solution

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

  1. 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

  1. 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

  1. Match the items in column 1 with the items in column 2.
Column 1Column 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

Solution

  1.             35 is a multiple of 7. (b)
  2.             15 is a factor of 30 (d)

iii.            16 is a multiple of 8 (a)

  1.             20 is a factor of 20 (f)
  2.             25 is a factor of 50 (e)
  1. Find all the multiples of 9 up to 100.

Solution

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

Multiples of 9 up to 100 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99

 

Exercise 3.2

  1. What is the sum of any two (a) Odd numbers? (b) Even numbers?

Solution

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.

  1. State whether the following statements are True or False:

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

False. [Ex: 7 + 3 + 9 = 19.]

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

True. [ex: 7 + 3 + 4 = 14.]

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

True. [Ex: 7 × 3 × 9 = 21 × 9 = 189]

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

False. [Ex: 32 ÷ 4 = 8]

(e)    All prime numbers are odd.

False, every prime number except 2 is odd.

(f)    Prime numbers do not have any factors.

False, prime numbers have two factors, the number 1 and itself.

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

False. [3 + 2 = 5]

(h)   2 is the only even prime number.

True

(i)     All even numbers are composite numbers.

False, 2 is an even prime number.

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

True. [Ex: 12 × 4 = 36]

  1. 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.

Solution

17 and 71

37 and 73

79 and 97

  1. Write down separately the prime and composite numbers less than 20.

Solution

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

  1. What is the greatest prime number between 1 and 10?

Solution

Prime numbers between 1 and 10 are: 2, 3, 5, 7.

Greatest prime number between 1 and 10 is 7.

  1. Express the following as the sum of two odd primes.

[There are many ways of writing a number as the sum of two odd primes. The following solutions show a few of them]

(a)    44

Solution

44 = 41 + 3 OR 37 + 7

(b)   36

Solution

36 = 31 + 5 OR 29 + 7

(c)    24

Solution

24 = 17 + 7 OR 19 + 5

(d)   18

Solution

18 = 11 + 7 OR 13 + 5

  1. Give three pairs of prime numbers whose difference is 2. [Remark: Two prime numbers whose difference is 2 are called twin primes].

Solution

3 and 5

5 and 7

11and 13

  1. Which of the following numbers are prime?
  2. a)23
  3. b)51
  4. c)37
  5. d)26

Solution

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.

  1. Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Solution

90, 91, 92, 93, 94, 95, 96

  1. Express each of the following numbers as the sum of three odd primes:

(a)    21

Solution

21 = 7 + 9 + 5

(b)   31

Solution

31 = 23 + 5 + 3

(c)    53

Solution

53 = 23 + 17 + 13

(d)   61

Solution

61 = 47 + 11 + 3

  1. Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (Hint : 3+7 = 10)

Solution

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

  1. Fill in the blanks :

(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 ______ nor ______.

Prime, composite

(d)   The smallest prime number is ______.

2

(e)    The smallest composite number is _____.

4

(f)    The smallest even number is ______.

2

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):

Solution

990

Divisible by 2 as it has 0 in its ones 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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 units 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 ones 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 units 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.

NumberDivisible by
23456891011
128yesnoyesNonoyesnonono
990yesyesnoYesyesnoyesyesYes
1586yesnonoNononononoNo
275NononoYesnonononoYes
6686YesNoNoNoNoNoNoNoNo
639210YesYesNoYesYesNoNoYesYes
429714YesYesNoNoYesNoyesNoNo
2856YesYesYesNoYesYesNoNoNo
3060YesYesYesYesYesNoYesYesNo
406839NoYesNoNoNoNoNoNoNo
  1. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

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.

  1. a)572

72 is divisible by 4, hence the number is divisible by 4.

The number is not divisible by 8.

  1. 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.

  1. 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.

  1. d)6000

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

  1. 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.

  1. 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.

  1. 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.

  1. h)31795072

72 is divisible by 4 and 8. Hence the number is divisible by 4 and 8.

  1. i)1700

The number is divisible by 4.

700 is not divisible by 8, hence the number is not divisible by 8.

  1. 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.

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

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

  1. a)297144

The number is divisible by 2, as the number has 4 in its ones 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.

  1. 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.

  1. c)4335

The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the number is not divisible by 6.

  1. d)61233

The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the number is not divisible by 6.

  1. 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.

  1. f)438750

The number is divisible by 2, as the number has 0 in its ones 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.

  1. g)1790184

The number is divisible by 2, as the number has 4 in its ones 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.

  1. h)12583

The number is not divisible by 2 as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the number is not divisible by 6.

  1. i)639210

The number is divisible by 2, as the number has 4 in its ones 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.

  1. 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.

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

Find 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. If the difference is either 0 or divisible by 11, then the number is divisible by 11.

  1. a)5445

The sum of the digits in the odd places from the right is 9. The sum of the digits in the even places from the right is 9. The difference is 0. The number is not divisible by 11.

  1. b)10824

Sum of the digits in the odd places from the right is 13.

Sum of the digits in the even places from the right is 2.

Difference is 11.

The number is divisible by 11.

  1. c)7138965

Sum of the digits in the odd places from the right is 24.

Sum of the digits in the even places from the right is 15.

Difference is 9.

The number is not divisible by 11.

  1. d)70169308

Sum of the digits in the odd places from the right is 17.

Sum of the digits in the even places from the right is 17.

Difference is 0.

The number is divisible by 11.

  1. e)10000001

Sum of the digits in the odd places from the right is 1.

Sum of the digits in the even places from the right is 1.

Difference is 0.

The number is divisible by 11.

  1. f)901153

Sum of the digits in the odd places from the right is 4.

Sum of the digits in the even places from the right is 15.

Difference is 11.

The number is divisible by 11.

  1. 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:
  1. a)__ 6724

Solution

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.

  1. b)4765 __ 2

Solution

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.

  1. 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

Solution

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

Solution

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

  1. Find the common factors of:
  2. a)20 and 28

Solution

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

  1. b)15 and 25

Solution

Factors of 15 are 1, 3, 5, 15

Factors of 25 are 1, 5, 25

Common factors are 1, 5

  1. 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,

  1. 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

  1. Find the common factors of :
  2. a)4, 8 and 12

Solution

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

  1. b)5, 15 and 25

Solution

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

  1. Find first three common multiples of :
  2. a)6 and 8

Solution

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

  1. b)12 and 18

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

  1. Write all the numbers less than 100 which are common multiples of 3 and 4.

Solution

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

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

Two numbers having only 1 as a common factor are called co-prime numbers.

  1. a)18 and 35

Solution

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

  1. b)15 and 37

Solution

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.

  1. c)30 and 415

Solution

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

  1. 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.

  1. 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.

  1. 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.

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

Solution

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.

  1. A number is divisible by 12. By what other numbers will that number be divisible?

Solution

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

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

False, 12 is divisible by 3 but not by 9.

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

True

  1. 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 not by 18.

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

True

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

False, 14 and 27 are co-primes and are composite numbers.

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

False, 12 is divisible by 4 but not by 8.

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

True

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

True

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

False, 3 divides 18 but not 16 and 2.

  1. Here are two different factor trees for 60. Write the missing numbers.
  1. a)

 

Solution

6 × 10 = 60

Now, 2 × 3 = 6 and 5 × 2 = 10

So, the missing number are 3 and 2.

  1. b)  

Solution

60 = 30 × 2

30 = 10 × 3

10 = 2 × 5

The missing numbers are 2, 3 and 5

  1. Which factors are not included in the prime factorisation of a composite number?

Solution

1 and the number itself.

  1. Write the greatest 4-digit number and express it in terms of its prime factors.

Solution

The greatest 4-digit number is 9999.






So, the prime factors are 3 × 3 × 11 × 101

  1. Write the smallest 5-digit number and express it in the form of its prime factors.

Solution

The smallest 5-digit number is 10,000.










The prime factors of 10,000 is 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

  1. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Solution

1729 = 7 × 13 × 19

The difference between the prime factors is 6.

  1. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Solution

12 × 13 × 14 = 2184; is divisible by 6.

25 × 26 × 27 = 17550; is divisible by 6

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

Solution

3 + 5 = 8, is divisible by 4.

13 + 15 = 28, is divisible by 4.

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

(a)    24 = 2 × 3 × 4

4 is not a prime factor.

(b)   56 = 7 × 2 × 2 × 2

The numbers are prime factorised.

(c)    70 = 2 × 5 × 7

The numbers are prime factorised.

(d)   54 = 2 × 3 × 9

9 is not a prime factor.

  1. 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].

Solution

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.

  1. 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.

Solution

No, not necessary.

For example, 36 is divisible by 4 and 6 but not by 24.

  1. I am the smallest number, having four different prime factors. Can you find me?

Solution

As the number is the smallest number, it will have the smallest prime factors, 2, 3, 5 and 7.

So, 2 × 3 × 5 × 7 = 210 is the required smallest number.

Exercise 3.6

  1. Find the HCF of the following numbers:

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

  1. a)18, 48

Solution

                                    

18 = 2 × 3 × 3

48 = 3 × 2 × 2 × 2 × 2

The common factors are 2 and 3. So, HCF = 2 × 3 = 6

  1. b)30, 42

Solution

                            

 

30 = 2 × 3 × 5

42 = 2 × 3 × 7

Common factors are 2 and 3. HCF = 2 × 3 = 6

  1. c)18, 60

Solution

 

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

Common factors are 2 and 3. HCF = 2 × 3 = 6

  1. d)27, 63

 

27 = 3 × 3 × 3

63 = 3 × 3 × 7

Common factors are 3(twice). HCF = 3 × 3 = 9

  1. e)36, 84

 

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

Common factors are 2(twice) and 3. HCF is 2 × 2 × 3 = 12

  1. f)34, 102

 

34 = 2 × 17

102 = 2 × 17 × 3

Common factors are 2 and 17. HCF = 2 × 17 = 34

  1. g)70, 105, 175

 

70 = 2 × 5 × 7

105 = 5 × 3 × 7

175 = 5 × 5 × 7

Common factors are 5 and 7. HCF = 5 × 7 = 35

  1. h)91, 112, 49

 

 

 

91 = 13 × 7

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

Common factor is 7. HCF = 7

  1. i)18, 54, 81

 

 

 

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

Common factors are 3 (twice). HCF = 9

  1. j)12, 45, 75

 

12 = 2 × 2 × 3

45 = 5 × 3 × 3

75 = 5 × 3 × 5

Common factor is 3. HCF = 3

  1. What is the HCF of two consecutive
  2.       numbers?

Solution

The HCF of two consecutive numbers is 1. For example, HCF of 4 and 5 is 1.

  1.       even numbers?

The HCF of two consecutive even numbers is 2. For example, HCF of 8 and 10 is 2.

iii.      odd numbers?

The HCF of two consecutive odd numbers is 1. For example, HCF of 13 and 15 is 1.

  1. 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?

Solution

The answer is incorrect.

The HCF of 4 and 15 is 1.

Exercise 3.7

  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.

Solution

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.

  1. 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?

Solution

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.

 

The LCM of the three numbers is 3 × 3 × 7 × 2 × 5 × 11 = 6930 cm

So, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm

  1. 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.

Solution

We require the longest tape that can measure the dimensions exactly. Thus, the maximum value can be obtained by finding the HCF of the numbers.

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.

  1. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Solution

Find the LCM of the numbers 6, 8, 12.

So, LCM is 24.

But, we need to find the smallest three-digit number.

24 × 5 = 120 is the smallest three-digit multiple of 24. So, 120 is the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

  1. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Solution

LCM = 120

Now, 120 × 8 = 960 is the largest three-digit multiple of 120.

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

  1. 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?

Solution

LCM = 432 seconds.

So, the traffic lights at three different road crossings change every 432 seconds, that is, 420 minutes + 12 seconds; 7 min 12 seconds.

Hence they will change simultaneously at 7:07:12 a.m.

  1. 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.

Solution

Maximum capacity is the HCF of the numbers.

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

So, the maximum capacity of the container is 31 litres.

  1. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Solution

Least number is 90.

Now, 90 is the least number which leaves a remainder 0 when divided by the numbers 6, 15 and 18.

But, we need the least number that leaves remainder 5 in each case. Therefore, the required number is 5 more than 90. The required least number = 95.

  1. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Solution

LCM is 288

We require a 4-digit number. So, 288 × 4 = 1152 is the smallest 4-digit multiple of 288.

So, 1152 is the required number.

  1. Find the LCM of the following numbers :
  2. a)9 and 4
  3. b)12 and 5
  4. c)6 and 5
  5. d)15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Solution

LCM = 36

LCM = 60

LCM = 30

LCM = 60

We see that the LCM of the numbers is the products of the given numbers. This is because the given numbers are co-primes.

Thus, when the numbers are co-primes, the LCM of the numbers is their product.

  1. Find the LCM of the following numbers in which one number is the factor of the other.
  2. a)5, 20
  3. b)6, 18
  4. c)12, 48
  5. d)9, 45

What do you observe in the results obtained?

Solution

LCM = 20

LCM = 18

LCM = 48

LCM = 45

We see that the LCM of the numbers is the larger number.


So, when number is a factor of the other number, their LCM will be the larger number. 

error: Content is protected !!