source: ncert_exercises
topic: numbers
class_level: class_10
chapter: ch1
exercise: ex1.1
difficulty: intermediate
Class 10 | Ch1: Real Numbers | Exercise 1.1
Key Concepts: Fundamental Theorem of Arithmetic: Every integer > 1 is either prime or can be expressed uniquely as a product of primes. HCF using prime factorisation: product of smallest powers of common prime factors. LCM using prime factorisation: product of greatest powers of all prime factors. HCF × LCM = Product of two numbers (for two numbers only).
Q1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Answer: (i) 140 = 2²×5×7. ✅ (ii) 156 = 2²×3×13. ✅ (iii) 3825 = 3²×5²×17. ✅ (iv) 5005 = 5×7×11×13. ✅ (v) 7429 = 17×19×23. ✅
Q2. Find LCM and HCF using prime factorisation method: (i) 12,15,21 (ii) 17,23,29 (iii) 8,9,25 (iv) 72,108 (v) 306,657 Answer: (i) 12=2²×3, 15=3×5, 21=3×7. HCF=3. LCM=2²×3×5×7=420. ✅ (ii) 17,23,29 all prime. HCF=1. LCM=17×23×29=11339. ✅ (iii) 8=2³, 9=3², 25=5². HCF=1. LCM=2³×3²×5²=1800. ✅ (iv) 72=2³×3², 108=2²×3³. HCF=2²×3²=36. LCM=2³×3³=216. ✅ (v) 306=2×3²×17, 657=3²×73. HCF=3²=9. LCM=306×657/9=22338. ✅
Q3. Check whether 6ⁿ can end with digit 0 for any natural number n. Answer: For 6ⁿ to end with 0, it must be divisible by 10=2×5. 6ⁿ=2ⁿ×3ⁿ. Prime factorisation of 6ⁿ has factors 2 and 3 only — NO factor of 5. By Fundamental Theorem of Arithmetic, 6ⁿ CANNOT end with 0 for any n. ✅
Q4. Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers. Answer: 7×11×13+13 = 13×(7×11+1) = 13×78 = 13×2×3×13. Has factors other than 1 and itself → composite. ✅ 7×6×5×4×3×2×1+5 = 5×(7×6×4×3×2×1+1) = 5×1009. Has factor 5 → composite. ✅
Q5. How will you show that 17×11×2+17×11×5 is composite? Explain. Answer: 17×11×2+17×11×5 = 17×11×(2+5) = 17×11×7. Has factors 7,11,17 → composite. ✅