Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Step-by-Step Solution
Key Concept: Use the <b>division method</b> of prime factorisation – repeatedly divide the given number by the smallest possible prime (2, 3, 5, 7, …) until the quotient becomes 1. The primes used in the divisions, together with their multiplicities, give the required product of prime factors.
(i) 140
- Step 1: 140 is even, divide by 2 → $140 \div 2 = 70$.
- Step 2: 70 is even, divide by 2 → $70 \div 2 = 35$.
- Step 3: 35 is not divisible by 2, try 3 (no), try 5 → $35 \div 5 = 7$.
- Step 4: 7 is a prime number.
- Hence $140 = 2 \times 2 \times 5 \times 7 = 2^{2}\,5\,7$.
(ii) 156
- Step 1: 156 is even, divide by 2 → $156 \div 2 = 78$.
- Step 2: 78 is even, divide by 2 → $78 \div 2 = 39$.
- Step 3: 39 is not divisible by 2, try 3 → $39 \div 3 = 13$.
- Step 4: 13 is a prime number.
- Hence $156 = 2 \times 2 \times 3 \times 13 = 2^{2}\,3\,13$.
(iii) 3825
- Step 1: The last digit is 5, so 3825 is divisible by 5 → $3825 \div 5 = 765$.
- Step 2: 765 also ends with 5, divide by 5 again → $765 \div 5 = 153$.
- Step 3: Sum of digits of 153 is $1+5+3=9$, divisible by 3 → $153 \div 3 = 51$.
- Step 4: 51 is divisible by 3 → $51 \div 3 = 17$.
- Step 5: 17 is a prime number.
- Hence $3825 = 5 \times 5 \times 3 \times 3 \times 17 = 5^{2}\,3^{2}\,17$.
(iv) 5005
- Step 1: The number is odd, not divisible by 2. Sum of digits $5+0+0+5=10$ → not divisible by 3. Ends with 5 → not divisible by 5. Try 7: $5005 \div 7 = 715$ (exact).
- Step 2: 715 is odd; sum of digits $7+1+5=13$ → not divisible by 3. Ends with 5 → divisible by 5 → $715 \div 5 = 143$.
- Step 3: 143 is not divisible by 2,3,5. Try 7 → $143 \div 7 = 20.428$ (not integer). Try 11 → $143 \div 11 = 13$.
- Step 4: 13 is prime.
- Hence $5005 = 7 \times 5 \times 11 \times 13 = 5\,7\,11\,13$.
(v) 7429
- Step 1: The number is odd, not divisible by 2. Sum of digits $7+4+2+9 = 22$ → not divisible by 3. Does not end with 5 → not divisible by 5.
- Step 2: Test divisibility by 7: $7429 \div 7 = 1061.285$ (not integer). Test 11: $7429 \div 11 = 675.363$ (no). Test 13: $7429 \div 13 = 571.461$ (no).
- Step 3: Test 17: $7429 \div 17 = 437$ (exact). So 17 is a factor.
- Step 4: Now factor 437. It is odd, sum of digits $4+3+7 = 14$ → not divisible by 3. Ends with 7 → not 5. Test 7: $437 \div 7 = 62.428$ (no). Test 11: $437 \div 11 = 39.727$ (no). Test 13: $437 \div 13 = 33.615$ (no). Test 17: $437 \div 17 = 25.705$ (no). Test 19: $437 \div 19 = 23$ (exact).
- Step 5: 23 is a prime number.
- Hence $7429 = 17 \times 19 \times 23$.
Thus the prime‑factorisations are:
- $140 = 2^{2}\,5\,7$
- $156 = 2^{2}\,3\,13$
- $3825 = 5^{2}\,3^{2}\,17$
- $5005 = 5\,7\,11\,13$
- $7429 = 17\,19\,23$
Correct Answer: 140 = 2^2·5·7; 156 = 2^2·3·13; 3825 = 5^2·3^2·17; 5005 = 5·7·11·13; 7429 = 17·19·23