# Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

**Solution:**

Let's prime factorize the given numbers by the prime factorization method and then multiply the obtained prime numbers to get the product of the prime numbers.

(i) 140

Prime factorization of 140 = 2 × 2 × 5 × 7

= 2² × 5 × 7

(ii) 156

Prime factorization of 156 = 2 × 2 × 3 × 13

= 2² × 3 × 13

(iii) 3825

Prime factorization of 3825 = 3 × 3 × 5 × 5 × 17

= 3² × 5² × 17

(iv) 5005

Prime factorization of 5005 = 5 × 7 × 11 × 13

(v) 7429

Prime factorization of 7429 = 17 × 19 × 23

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 1

**Video Solution:**

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

NCERT Solutions Class 10 Maths - Chapter 1 Exercise 1.2 Question 1

**Summary:**

The numbers 140, 156, 3825, 5005 and 7429 can be expressed as the product of their prime factors as 2 × 2 × 5 × 7, 2 × 2 × 3 × 13, 3 × 3 × 5 × 5 × 17, 5 × 7 × 11 × 13 and 17 × 19 × 23 respectively.

**☛ Related Questions:**

- 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
- 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
- Given that HCF (306, 657) = 9, find LCM (306, 657).
- Check whether 6n can end with the digit 0 for any natural number n.

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