NCERT Class 10 Maths – Real Numbers Solutions

Class 10 Maths

Real Numbers • Exercise 1.1

💡 Key Concepts

[Image of real number system diagram]

Fundamental Theorem of Arithmetic: Every composite number can be written as a product of primes uniquely.

$$ \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b $$

Question 1

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

(i) 140

$$140 = 2 \times 70 = 2 \times 2 \times 35 = 2^2 \times 5 \times 7$$

(ii) 156

$$156 = 2 \times 78 = 2 \times 2 \times 39 = 2^2 \times 3 \times 13$$

(iii) 3825

$$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 5^2 \times 17$$

(iv) 5005

$$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$$

(v) 7429

$$7429 = 17 \times 437 = 17 \times 19 \times 23$$

Question 2

Find LCM and HCF of the pairs and verify that LCM × HCF = Product of numbers.

(i) 26 and 91

$$26 = 2 \times 13, \quad 91 = 7 \times 13$$

HCF = 13, LCM = 182

Verification: $26 \times 91 = 2366$, $13 \times 182 = 2366$. ✔️

(ii) 510 and 92

$$510 = 2 \cdot 3 \cdot 5 \cdot 17, \quad 92 = 2^2 \cdot 23$$

HCF = 2, LCM = 23460

Verification: $510 \times 92 = 46920$, $2 \times 23460 = 46920$. ✔️

Question 3

Find LCM and HCF by applying prime factorization method.

(i) 12, 15 and 21

$$12=2^2 \cdot 3, \quad 15=3 \cdot 5, \quad 21=3 \cdot 7$$

HCF = 3, LCM = 420

(ii) 17, 23 and 29

All are Prime Numbers

HCF = 1, LCM = 11339

Question 4

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

$$\text{LCM} = \frac{306 \times 657}{9} = 34 \times 657$$

✔️ LCM = 22338

Question 5

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

For a number to end with digit 0, its prime factorization must contain at least one pair of 2 and 5.

$$6^n = (2 \times 3)^n = 2^n \times 3^n$$

The factorization contains 2 but no 5. By the Fundamental Theorem of Arithmetic, this factorization is unique.

Therefore, $6^n$ can never end with 0.

Contents

Class 10 Maths

💡 Key Concepts