NCERT Solutions for Class 10 Mathematics Chapter 1: Real Numbers

Chapter 1 revisits the properties of real numbers built up in earlier classes and develops two central results: Euclid's Division Lemma and the Fundamental Theorem of Arithmetic. Together these tools are used across the chapter's two exercises to find the HCF and LCM of integers using Euclid's division algorithm and prime factorisation.

Chapter Overview

Chapter 1 of NCERT Class 10 Mathematics builds two ideas on the number system studied in earlier classes: Euclid's Division Lemma, which underlies a fast algorithm for finding the HCF of two positive integers, and the Fundamental Theorem of Arithmetic, which says every composite number factorises uniquely into primes. Together these two results are used throughout Exercise 1.1 and Exercise 1.2 below to find HCF and LCM, and to reason about integers without factorising them from scratch every time.

Euclid's Division Lemma

Given positive integers a and b, there exist unique whole numbers q and r satisfying a = bq + r, where 0 ≤ r < b.

Exercise 1.1

Q1. Use Euclid's division algorithm to find the HCF of: (i) 900 and 270 (ii) 196 and 38220 (iii) 1651 and 2032

Answer

(i) 900 and 270

Since 900 > 270, apply the division lemma to 900 and 270, then keep applying it to each new divisor and remainder until the remainder is 0.

Steps

900 = 270 × 3 + 90
270 = 90 × 3 + 0
The remainder is now 0, so the algorithm stops here.

\text{HCF}(900, 270) = 90

(ii) 196 and 38220

Since 38220 > 196, apply the division lemma to 38220 and 196.

Steps

38220 = 196 × 195 + 0
The remainder is 0 on the very first step, so the algorithm stops here.

\text{HCF}(196, 38220) = 196

(iii) 1651 and 2032

Since 2032 > 1651, apply the division lemma to 2032 and 1651, then continue with each new divisor and remainder.

Steps

2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
The remainder is now 0, so the algorithm stops here.

\text{HCF}(1651, 2032) = 127

Q2. Show that any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5, where q is some integer.

Number line marking integers of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5 — Positive integers grouped by remainder on division by 6

Answer

Let a be any positive integer and b = 6. By Euclid's division lemma, a can be written as:

a = 6q + r, \quad 0 \le r < 6

Here r can only take the values 0, 1, 2, 3, 4 or 5, so a is always one of the six forms 6q, 6q+1, 6q+2, 6q+3, 6q+4 or 6q+5.

Form	Parity
6q	Even
6q + 1	Odd
6q + 2	Even
6q + 3	Odd
6q + 4	Even
6q + 5	Odd

6q, 6q + 2 and 6q + 4 are all divisible by 2, so they are even. That leaves exactly 6q + 1, 6q + 3 and 6q + 5 as the possible forms of a positive odd integer.

Q3. 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?

Answer

For both groups to march in the same number of columns with no member left out, the number of columns must divide both 616 and 32 exactly. The maximum possible number of columns is therefore HCF(616, 32), found using Euclid's algorithm.

Steps

616 = 32 × 19 + 8
32 = 8 × 4 + 0
The remainder is now 0, so the algorithm stops here.

\text{HCF}(616, 32) = 8

So the maximum number of columns in which the two groups can march is 8.

Q4. 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.

Answer

Let x be any positive integer and b = 3. By Euclid's division lemma, x = 3q + r for some integer q and 0 ≤ r < 3, so x is of the form 3q, 3q + 1 or 3q + 2. Squaring each case:

Proof

If x = 3q, then x² = 9q² = 3(3q²), which is of the form 3m with m = 3q². If x = 3q + 1, then x² = 9q² + 6q + 1 = 3(3q² + 2q) + 1, which is of the form 3m + 1 with m = 3q² + 2q. If x = 3q + 2, then x² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1, which is of the form 3m + 1 with m = 3q² + 4q + 1. In every case x² is of the form 3m or 3m + 1, so the square of any positive integer is never of the form 3m + 2.

Q5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer

Let x be any positive integer and b = 3, so by Euclid's division lemma x is of the form 3q, 3q + 1 or 3q + 2. Cubing each case:

Proof

If x = 3q, then x³ = 27q³ = 9(3q³), which is of the form 9m with m = 3q³. If x = 3q + 1, then x³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1, which is of the form 9m + 1 with m = 3q³ + 3q² + q. If x = 3q + 2, then x³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8, which is of the form 9m + 8 with m = 3q³ + 6q² + 4q. So the cube of any positive integer is always of the form 9m, 9m + 1 or 9m + 8.

Exercise 1.2

Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Q1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Answer

(i) 140

Steps

140 = 2 × 70
70 = 2 × 35
35 = 5 × 7
7 is prime, so the factorisation stops here.

140 = 2^2 \times 5 \times 7

(ii) 156

Steps

156 = 2 × 78
78 = 2 × 39
39 = 3 × 13
13 is prime, so the factorisation stops here.

156 = 2^2 \times 3 \times 13

Factor tree showing 156 broken down into 2, 2, 3 and 13

Alternative factor tree arriving at the same prime factorisation of 156

(iii) 3825

Steps

3825 = 3 × 1275
1275 = 3 × 425
425 = 5 × 85
85 = 5 × 17
17 is prime, so the factorisation stops here.

3825 = 3^2 \times 5^2 \times 17

(iv) 5005

Steps

5005 = 5 × 1001
1001 = 7 × 143
143 = 11 × 13
13 is prime, so the factorisation stops here.

5005 = 5 \times 7 \times 11 \times 13

(v) 7429

Steps

7429 = 17 × 437
437 = 19 × 23
23 is prime, so the factorisation stops here.

7429 = 17 \times 19 \times 23

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 (ii) 510 and 92 (iii) 336 and 54

Answer

(i) 26 and 91

26 = 2 \times 13, \qquad 91 = 7 \times 13

\text{HCF} = 13, \qquad \text{LCM} = 2 \times 7 \times 13 = 182

Verification:

\text{LCM} \times \text{HCF} = 182 \times 13 = 2366 = 26 \times 91

(ii) 510 and 92

510 = 2 \times 3 \times 5 \times 17, \qquad 92 = 2^2 \times 23

\text{HCF} = 2, \qquad \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460

Verification:

\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920 = 510 \times 92

(iii) 336 and 54

336 = 2^4 \times 3 \times 7, \qquad 54 = 2 \times 3^3

\text{HCF} = 2 \times 3 = 6, \qquad \text{LCM} = 2^4 \times 3^3 \times 7 = 3024

Verification:

\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144 = 336 \times 54

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method: (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Answer

(i) 12, 15 and 21

12 = 2^2 \times 3, \qquad 15 = 3 \times 5, \qquad 21 = 3 \times 7

\text{HCF} = 3, \qquad \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420

(ii) 17, 23 and 29

All three numbers are prime and none of them repeats, so they share no common factor other than 1.

\text{HCF} = 1, \qquad \text{LCM} = 17 \times 23 \times 29 = 11339

(iii) 8, 9 and 25

8 = 2^3, \qquad 9 = 3^2, \qquad 25 = 5^2

The three numbers share no common prime factor.

\text{HCF} = 1, \qquad \text{LCM} = 2^3 \times 3^2 \times 5^2 = 1800

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

Answer

For two positive integers, HCF × LCM equals the product of the numbers, so:

\text{LCM}(306, 657) = \dfrac{306 \times 657}{\text{HCF}(306, 657)} = \dfrac{201042}{9}

\text{LCM}(306, 657) = 22338

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

Answer

Proof

If 6ⁿ were to end with the digit 0, it would have to be divisible by 10, and hence by both 2 and 5, so 5 would have to appear in the prime factorisation of 6ⁿ. But 6 = 2 × 3, so 6ⁿ = 2ⁿ × 3ⁿ, and by the Fundamental Theorem of Arithmetic this prime factorisation is unique — it contains only the primes 2 and 3, never 5. Therefore 6ⁿ can never be divisible by 5, and so 6ⁿ can never end with the digit 0, for any natural number n.

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

Answer

7 × 11 × 13 + 13

7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78 = 1014

This is a product of 13 and 78, both greater than 1, so 1014 is composite.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009 = 5045

This is a product of 5 and 1009, both greater than 1, so 5045 is composite as well.

Q7. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field, while Ravish 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?

Answer

Priya and Ravish are both back at the starting point only at multiples of their own round times, 18 and 12 minutes. They meet again at the starting point at the first time that is a common multiple of both — that is, at LCM(18, 12).

18 = 2 \times 3^2, \qquad 12 = 2^2 \times 3

\text{LCM}(18, 12) = 2^2 \times 3^2 = 36

So Priya and Ravish will meet again at the starting point after 36 minutes.

Key Results from This Chapter

Euclid's Division Lemma: for positive integers a and b, a = bq + r with 0 ≤ r < b, for unique q and r.
Euclid's Division Algorithm repeatedly applies this lemma to find the HCF of two positive integers.
Fundamental Theorem of Arithmetic: every composite number factorises into primes uniquely, up to the order of the factors.
HCF(a, b) × LCM(a, b) = a × b holds only for exactly two positive integers, not for three or more.

NCERT Solutions for Class 10 Mathematics Chapter 1: Real Numbers

Chapter overview

Chapter Overview

Exercise 1.1

Exercise 1.2

Key Results from This Chapter

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