Question 1
(i) \(140\) (ii) \(156\) (iii) \(3825\) (iv) \(5005\) (v) \(7429\)
📋 Given Information
- Five composite numbers: \(140, 156, 3825, 5005, 7429\)
- We need to express each as a product of prime factors
🎯 To Find
Prime factorisation of each given number using the division method
📐 Concept Used
Any composite number can be expressed as a product of prime numbers in a unique way
Method: Divide the number successively by prime numbers (2, 3, 5, 7, 11, 13, …) starting from the smallest until we get 1. This is based on the Fundamental Theorem of Arithmetic.
📝 Step-by-Step Solution
Divide by smallest prime 2:
\[ \begin{aligned} 140 &= 2 \times 70\\ &= 2 \times 2 \times 35\\ &= 2 \times 2 \times 5 \times 7\\ &= 2^2 \times 5 \times 7 \end{aligned} \]
Answer: \(140 = 2^2 \times 5 \times 7\)
Divide by 2 repeatedly, then by 3:
\[ \begin{aligned} 156 &= 2 \times 78\\ &= 2 \times 2 \times 39\\ &= 2 \times 2 \times 3 \times 13\\ &= 2^2 \times 3 \times 13 \end{aligned} \]Answer: \(156 = 2^2 \times 3 \times 13\)
Divide by 3, then by 5:
\[ \begin{aligned} 3825 &= 3 \times 1275\\ &= 3 \times 3 \times 425\\ &= 3 \times 3 \times 5 \times 85\\ &= 3 \times 3 \times 5 \times 5 \times 17\\ &= 3^2 \times 5^2 \times 17 \end{aligned} \]Answer: \(3825 = 3^2 \times 5^2 \times 17\)
Divide by 5, then by consecutive primes:
\[ \begin{aligned} 5005 &= 5 \times 1001\\ &= 5 \times 7 \times 143\\ &= 5 \times 7 \times 11 \times 13 \end{aligned} \]Answer: \(5005 = 5 \times 7 \times 11 \times 13\)
Testing divisibility by primes 17 and 19:
\[ \begin{aligned} 7429 &= 17 \times 437\\ &= 17 \times 19 \times 23 \end{aligned} \]Answer: \(7429 = 17 \times 19 \times 23\)
💡 Alternative Method: Factor Tree (Click to expand)
Method 2: Using Factor Tree
Instead of systematic division, we can use a factor tree by breaking numbers into any two factors and continuing until all factors are prime.
Example for 140:
Start with any factorisation: \(140 = 10 \times 14\)
\[ \begin{aligned} 140 &= 10 \times 14\\ &= (2 \times 5) \times (2 \times 7)\\ &= 2^2 \times 5 \times 7 \end{aligned} \]This gives the same result, confirming the uniqueness of prime factorisation as stated in the Fundamental Theorem of Arithmetic.
(i) \(140 = 2^2 \times 5 \times 7\)
(ii) \(156 = 2^2 \times 3 \times 13\)
(iii) \(3825 = 3^2 \times 5^2 \times 17\)
(iv) \(5005 = 5 \times 7 \times 11 \times 13\)
(v) \(7429 = 17 \times 19 \times 23\)
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Stopping the division when a composite number is reached (e.g., stopping at 35 for 140 instead of continuing to \(5 \times 7\))
✅ Correct Approach: Continue dividing until all factors are prime numbers only
❌ Mistake 2: Not checking if larger numbers are prime (e.g., assuming 143 is prime)
✅ Correct Approach: Always verify by testing divisibility with primes up to the square root of the number
❌ Mistake 3: Writing factors in different orders and thinking they’re different factorisations
✅ Correct Approach: Remember that \(2 \times 5 \times 7 = 5 \times 2 \times 7 = 7 \times 2 \times 5\) are all the same due to commutativity
💡 Key Points to Remember
- Every composite number has a unique prime factorisation (Fundamental Theorem of Arithmetic)
- Always start division with the smallest prime (2) and proceed systematically
- Prime factors can be written in exponential form: \(2 \times 2 = 2^2\)
- The order of prime factors doesn’t matter, but convention is to write in ascending order
- Prime factorisation is the basis for finding HCF and LCM
📝 Practice Similar Problems
Master prime factorisation by solving:
- Express 1260, 2520, and 9240 as products of prime factors
- Find the smallest number divisible by all numbers from 1 to 10
- Determine if 1024 is a perfect power of 2
- Find the prime factorisation of 32760 (Example from textbook)
