## NCERT solutions for class 10 Maths

**Chapter -1**

**Real Numbers**

**Exercise 1**

**Ex 1.1 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:**

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

**Ex 1.1 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 a be any positive integer and b = 6.

Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

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

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer 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 expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5 .

**Ex 1.1 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 (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the HCF.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

**Ex 1.1 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:**

Let a be any positive integer and b = 3.

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

And r = 0, 1, 2 because 0 ≤ r < 3

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

a² = (3q)² or (3q + 1)² or (3q + 2)²

= (3q)² or 9q² + 6q + 1 or 9q² + 12q + 4

= 3 × (3q²) or 3 × (3q² + 2q) + 1 or 3 × (3q² + 4q + 1) + 1

= 3k1 or 3k2 + 1 or 3k3 + 1

Where k1, k2, and k3 are some positive integers

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

**Ex 1.1 Question 5.**

**Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.**

**Solution:**

Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

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

Therefore, every number can be represented as these three forms.

There are three cases.

Case 1: When a = 3q, a³ = (3q)³ = 27q³ = 9(3q³ )= 9m

Where m is an integer such that m = 3q3

Case 2: When a = 3q + 1, a³ = (3q +1)³ a³ = 27q³ + 27q² + 9q + 1 a³ = 9(3q³ + 3q² + q) + 1 a³ = 9m + 1

Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3: When a = 3q + 2, a³ = (3q +2)³ a³ = 27q³ + 54q² + 36q + 8 a³ = 9(3q³ + 6q² + 4q) + 8 a³ = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

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