1. Real Numbers

Class X

NCERT Text Book Solution

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

Solution:

(i)

From question it is clear that

225 > 135.

Apply Euclid's division lemma, to 135 and 225.

225 = 135x1 + 90

135 = 90x1 + 45

90 = 45x2 + 0

The remainder has now become zero.

The HCF of 225 and 135 is 45 Ans.

(ii)

From question it is clear that 

38220>196

Apply Euclid's division lemma, to 38220 and 196.

38220 = 196x195 + 0

The remainder has now become zero.

The HCF of 38220 and 196 is 195 Ans.

(iii)

From question it is clear that

867>255

Apply Euclid's division lemma, to 867 and 255.

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0 

The remainder has now become zero.

The HCF of 867 and 255 is 51 Ans

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 a be any positive integer and b = 6.

Now, by Euclid's division algorithm,

a = 6q + r for some integer q ≥ 0,

r = 0, 1, 2, 3, 4, 5

Because 0 ≤ r < b

So, a can be

6q,or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5.

Since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 ( since they are all divisible by 2)

Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Proved.

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:

For this, it will be find out the HCF of 616 and 32.

Apply Euclid's algorithm, to 616 and 32.

616 = 32x19 + 8

32 = 8x4 + 0

So, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which they can march = 8 Ans.

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.

Solution:

Let a be any positive integer and b = 3.

Now, Euclid's division lemma,

a = 3q + r for some integer q ≥ 0,

Since, 0 ≤ r < 3

a can be 3q, or 3q + 1, or 3q + 2

Therefore,

a² = 9q², or 9q² + 6q + 1, or 9q² + 12q + 1.

or, a² = 3x3q², or 3( 3q² + 2q) + 1, or 3(3q² + 4q) + 1

So, the square of any positive integer is either of the form 3m, or 3m + 1. Proved

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 be any positive integer and b = 9.

Now, Euclid's division lemma,

a = 9q + r for some integer q ≥ 0,

Since, 0 ≤ r < 9

a can be 9q, or 9q + 1, or 9q + 2, 9q + 3, or ...... 9q + 8.

Therefore,

a³ = 9x81q³, or 9(81q³ + 27q² + 3q)+ 1, or 9(81q³ + 54q² + 6q) + 8 ..................

So, the cube of any positive integer is of the form 9m,or 9m + 1 or 9m + 8. Proved.