Real Numbers

No, because 6ⁿ = (2×3)ⁿ = 2ⁿ×3ⁿ has no factor of 5

Step 1: For a number to end in 0, it must be divisible by 10 = 2 × 5. Step 2: Find prime factorization of 6ⁿ. 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. Step 3: By Fundamental Theorem of Arithmetic, the only prime factors of 6ⁿ are 2 and 3. Step 4: Since 5 is not a prime factor of 6ⁿ, it cannot be divisible by 10. Step 5: Therefore, 6ⁿ can never end with digit 0 for any natural number n.

3² × 5² × 17

Step 1: Divide by smallest prime. 3825 ÷ 3 = 1275. Step 2: Continue dividing by 3. 1275 ÷ 3 = 425. Step 3: 425 is not divisible by 3, try 5. 425 ÷ 5 = 85. Step 4: Continue with 5. 85 ÷ 5 = 17. Step 5: 17 is prime. Therefore, 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17.

HCF = 1, LCM = 1800

Step 1: Find prime factorizations. 8 = 2³, 9 = 3², 25 = 5². Step 2: Find HCF by taking smallest powers of common prime factors. There are no common prime factors. Step 3: Therefore, HCF(8, 9, 25) = 1. Step 4: Find LCM by taking highest powers of all prime factors. LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800. Step 5: Since the numbers are coprime (HCF = 1), LCM equals their product.

Because it can be written as 13(7 × 11 + 1) = 13 × 78

Step 1: Look for common factors in the expression 7 × 11 × 13 + 13. Step 2: Factor out the common factor 13. 7 × 11 × 13 + 13 = 13(7 × 11) + 13. Step 3: Apply distributive property. 13(7 × 11) + 13 = 13(7 × 11 + 1) = 13(77 + 1) = 13 × 78. Step 4: Since this number can be written as a product of two integers greater than 1, it is composite. Step 5: The number has 13 and 78 as factors, proving it's not prime.