Real Numbers

Answer

Step 1: The number must be a multiple of 5, so test 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. Step 2: It must leave remainder 3 when divided by 4, so check 5, 15, 25, 35, 45. Step 3: It must leave remain…

Answer

For division by 2: 35 = 2 × 17 + 1 For division by 3: 35 = 3 × 11 + 2 For division by 4: 35 = 4 × 8 + 3 For division by 5: 35 = 5 × 7 + 0 Check: In each case, the remainder is less than the divisor.

Answer

Step 1: Divide 13 by 3. Step 2: 3 × 4 = 12, and 13 - 12 = 1. Step 3: So q = 4 and r = 1. Check: 13 = 3 × 4 + 1, and 0 ≤ 1 < 3.

Answer

Step 1: 100 = 60 × 1 + 40 Step 2: 60 = 40 × 1 + 20 Step 3: 40 = 20 × 2 + 0 Step 4: The last non-zero remainder is 20. Answer: HCF(60, 100) = 20.

Answer

It works because the HCF does not change when the larger number is replaced by its remainder. Key idea: HCF(c, d) = HCF(d, r) If c = dq + r, then any common factor of c and d also divides r, and any c…

Answer

Step 1: 12 = 2 × 2 × 3 = 2^2 × 3 Step 2: 18 = 2 × 3 × 3 = 2 × 3^2 Step 3: Take the smallest power of each common prime factor. HCF = 2^1 × 3^1 = 6 Answer: 6.

Answer

Step 1: 12 = 2^2 × 3 Step 2: 18 = 2 × 3^2 Step 3: Take the greatest power of each prime factor. LCM = 2^2 × 3^2 = 36 Answer: 36.

Answer

Step 1: Use HCF(a,b) × LCM[a,b] = a × b Step 2: 20 × LCM = 60 × 100 Step 3: 20 × LCM = 6000 Step 4: LCM = 6000 ÷ 20 = 300 Answer: LCM(60, 100) = 300.