Here is the solution of NCERT Class 10 Maths Chapter 1 Real Number Exercise 1.1. There are 5 Questions in Class 10 NCERT Maths Chapter 1 Real Numbers Exercise 1.1. All 5 Questions under Exercise 1.1 are based on Euclid's Division Lemma. You can go through each one of them one by one.

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

**(i)135 and 225**

** Solution 1 (i):** Since 225 > 135,

225 = 135 × 1 + 90

Also,

135 = 90 × 1 + 45

90 = 2 × 45 + 0

Therefore, the HCF of 135 and 225 is 45.

**(ii)196 and 38220**

** Solution 1 (ii):** Since 38220 > 196,

38220 = 196 × 195 + 0

Therefore, HCF of 196 and 38220 is 196.

**(iii)867 and 255**

** Solution 1 (iii):** Since 867 > 255,

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

Therefore, HCF of 867 and 255 is 51.

**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 2:** By Euclid’s algorithm,

a = 6q + r, and r = 0, 1, 2, 3, 4, 5

Hence, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these numbers are odd numbers.

**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 3:** Euclid’s algorithm

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF of (616, 32) is 8.

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

**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.**

** Solution 4:** a = 3q + r, r = 0, 1, 2

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

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

**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 5:** a = 3q + r, 0 ≤ r < 3

a = 3q, 3q + 1, 3q + 2

There are three cases.

Case 1: When a = 3q,

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

Case 2: When a = 3q + 1,

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

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

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

a^{3} = 9m+ 1

Case 3: When a = 3q + 2,

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

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

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

a^{3}=9m+ 8

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

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