**KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.1** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.1

Question 1.

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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Solution:

(i) 135 and 225

∴ 90 = 45 × 2 + 0

∴ HCF of 135 and 225 is 45.

(ii) 196 and 38220

∴ 38220 = 196 × 195 + 0

∴ HCF of 196 and 38220 is 196.

(iii) 867 and 2552

2 = 1 × 2 + 0

∴ HCF of 867 and 2552 is 1.

Question 2.

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

Solution:

Let us consider a positive odd integer as ‘a’.

On dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.

∴ Using Euclid’s lemma, we get a = 6q + r

where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,

a = 6q + 0 = 6q or a = 6q + 1

or a = 6q + 2 or a = 6q + 3

or a = 6q + 4 or a = 6q + 5

But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ‘a’.

[∵ 6q = 2(3q) = 2m_{1} 6q + 2 = 2(3q + 1) = 2m_{2},

6q + 4 = 2(3 q + 2) = 2m_{3}]

But ‘a’ being an odd integer, we have :

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

Question 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 ?

Solution:

HCF of 32 and 616

∴ 32 = 8 × 4 + 0

∴ HCF of 32 and 616 is 8.

∴ Both groups can march in 8 columns.

Question 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 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]

Solution:

Let us consider an arbitrary positive integer as ‘x’ such that it is of the form

3q, (3q + 1) or (3q + 2)

For x = 3q, we have x^{2} = (3q)^{2}

⇒ x^{2} = 9q^{2} = 3(3q^{2}) = 3m ………. (1)

Putting 3q^{2} = m, where m is an integer.

For x = 3q + 1,

x^{2} = (3q + 1)^{2} = 9q^{2} + 6q + 1

= 3(3q^{2} + 2q) + 1 = 3m + 1 ………… (2)

Putting 3q^{2} + 2q = m, where m is an integer.

For x = 3q + 2,

x^{2} = (3q + 2)^{2}

= 9q^{2} + 12q + 4 = (9q^{2} + 12q + 3) + 1

= 3(3q^{2} + 4q + 1) + 1 = 3m + 1 ……….. (3)

Putting 3q^{2} + 4q +1 = m, where m is an integer.

From (1), (2) and (3),

x^{2} = 3m or 3m + 1

Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question 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 and b be two positive integers and a > b

a = (b × q) + r where q and r are positive integers and 0 ≤ r < b.

Let b = 3 (if 9 is multiplied by 3 a perfect cube number is obtained)

a = 3q + r where 0 ≤ r < 3

(i) if r = 0, a = 3q

(ii) if r = 1, a = 3q + 1

(iii) if r = 2 a = 3q + 2

Consider cubes of these,

case i) a = 3q

a^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3})

a^{3} = 9m

where m= 3q^{3} & m is an integer

case ii) a = 3q + 1

a^{3} = (3q + 1)^{3}

(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

a^{3} = 27q^{3} + 1 + 27q^{3} + 9q

a^{3} = 27 q^{3} + 27 q^{2} + 9q + 1

a^{3} = 9 (3q^{3} + 3q^{2} + q) + 1

a^{3} = 9m + 1

where m = 3q^{3} + 3q^{2} + q & m is an integer

case iii) a = 3q + 2

a^{3} = (3q + 2)^{3} = 27q^{3} + 8 + 54q^{2} + 36q

a^{3} = 27q^{3} + 54q^{2} + 36q + 8

= 9 (3q^{3} + 6q^{2} + 4q) + 8

a^{3} = 9m + 8

where m = 3q^{3} + 6q^{2} + 4q and m is an integer.

∴ Cube of any positive integer is either of the form 9m, 9m + 1 (or) 9m + 8 for some integer m.

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