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Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Sol:

(1)         135 and 225

  a = 225, b = 135               {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

  a = bq + r (then)

  225 = 135 ×1 + 90

 135 = 90 ×1 + 45

 90 = 45 × 2 + 0                  {when we get r=0, our computing get stopped}

 b = 45 {b is HCF}

Hence:  HCF = 45

 

Sol:

 (ii)        196 and 38220

 a = 38220, b = 196          {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

 a = bq + r (then)

 38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

 b = 196 {b is HCF}

Hence:  HCF = 196

Sol:

(iii)        867 and 255

a = 867, b = 255               {Greatest number is ‘a’ and smallest number is ‘b’}

Using Euclid’s division algorithm

a = bq + r (then)

38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

b = 196 {b is HCF}

Hence:  HCF = 196

 

2.    Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

 

Sol:

Let a is the positive odd integer

Where b = 6,

When we divide a by 6 we get reminder 0, 1, 2, 3, 4 and 5,          {r < b}

Here a is odd number then reminder will be also odd one.

We get reminders 1, 3, 5 

Using Euclid’s division algorithm

So we get

a = 6q + 1, 6q+3 and 6q+5

 

3.   An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Sol:

      Maximum number of columns = HCF (616, 32)

      a = 616, b = 32  {Greatest number is ‘a’ and smallest number is ‘b’}

      Using Euclid’s division algorithm

      a = bq + r (then)

      616 = 32 ×19 + 8 {when we get r=0, our computing get stopped}

      32 = 8 × 4 + 0

       b = 8 {b is HCF}

       HCF = 8

       Hence: Maximum number of columns = 8

 

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

 

Solution :

To Show :

a² = 3m or 3m + 1

a = bq + r

Let a be any positive integer, where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3

Then a = 3q + r for some integer q ≥ 0

Therefore, a = 3q + 0 or 3q + 1 or 3q + 2

Now we have;

⇒ a² = (3q + 0)² or (3q + 1)² or (3q +2)²

⇒ a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 4

⇒ a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 3 + 1

⇒ a² = 3(3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1

Let m = (3q²) or (3q² + 2q)  or (3q² + 4q + 1)

Then we get;

a² = 3m or 3m + 1 or 3m + 1

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let , a is any positive integer

By using Euclid’s division lemma;

a = bq + r             where; 0 ≤ r < b

Putting b = 9

a = 9q + r             where; 0 ≤ r < 9

when r = 0

a = 9q + 0 = 9q

a³  = (9q)³ = 9(81q³) or 9m where m = 81q³

when r = 1

a = 9q + 1 

a³  = (9q + 1)³ = 9(81q³ + 27q² + 3q) + 1

      = 9m + 1  where m = 81q³ + 27q² + 3q

when r = 2

a = 9q + 2 

a³  = (9q + 2)³ = 9(81q³ + 54q² + 12q) + 8

      = 9m + 2  where m = 81q³ + 54q² + 12q

⇒ The End

EXERCISE 1.2

Q1. Express each number as a product of its prime factors:
(i) 140             

Solution:

         = 2² × 5 × 7 

(ii) 156

Solution:

        = 2² × 3 × 13

(iii) 3825

Solution:

          = 3² × 5² × 17 

(iv) 5005

Solution:

          = 5 × 7 × 11 × 13 

(v) 7429

Solution:

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

Solution:

26 = 2 × 13

91 = 7 × 13 

Common factors = 13 

∴ HCF = 13 

LCM = 2 × 7 × 13 = 182 

Now varification,

product of the two numbers = LCM × HCF 

N₁ × N₂ = LCM × HCF 

26 × 91 = 13 × 182 

    2366 =  2366

Hence Varified,

(ii) 510 and 92

Solution:

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

Common factors = 2

∴ HCF = 2 

LCM = 2 × 2 × 3 × 5 × 17 × 23 =  23460

Now varification,

product of the two numbers = LCM × HCF 

N₁ × N₂ = LCM × HCF 

510 × 92 = 2 × 23460 

    46920 =  46920

Hence Varified,

(iii) 336 and 54

Solution:

336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

Common factors = 2 × 3

∴ HCF = 6 

LCM = 2 × 2 × 2× 2 × 3 × 3 × 3 × 7 =  3024

Now varification,

product of the two numbers = LCM × HCF 

N₁ × N₂ = LCM × HCF 

336 × 54 = 6 × 3024 

    18144 =  18144

Hence Varified,

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Solution:

12 = 2 × 2 × 3

15 = 5 × 3 

21 = 7 × 3

Common Factors = 3 

HCF = 3 

​LCM = 3 × 2 × 2 × 5 × 7 = 420 

(ii) 17, 23 and 29

Solution:

17 = 1 × 17 

23 = 1 × 23 

29 = 1 × 29 

HCF = 1 

LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Solution:

8 = 2 × 2 × 2 

9 = 3 × 3 

25 = 5 × 5 

There is no common factor except 1. 

∴ HCF = 1 

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 

        = 8 × 9 × 25 

        = 1800

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

HCF (306, 657) = 9

 LCM × HCF = ​N₁ × N₂ 

LCM = 22338

Q5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution:

Prime factorisation of 6ⁿ = (2 × 3 )ⁿ

While, Any natural number which end with digit 0 has

the prime factorisation as form of (2 × 5 )ⁿ

Therefore, 6ⁿ will not end with digit 0. 

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:

Let A = 7 × 11 × 13 + 13

        = 13 (7 × 11 + 1)

        = 13 (77 + 1)

        = 13 × 78 

Hence this is composite number because It has at least one positive divisor other than one. 

Similarily,

Let B = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

        = 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)​  

        = 5 × (1008 + 1)  

        = 5  ×  1009    

Hence this is also a composite number because It has at least one positive divisor other than one.    

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: 

Sonia takes 18 minutes in one round.

Ravi takes 12 minutes in one round 

they will meet again at the starting point after LCM(18, 12) minutes

18 = 2 × 3 × 3 

12 = 2 × 2 × 3 

HCF = 2 × 3 = 6 

           = 36 minutes 

 

Exercise 1.3

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