Let the three consecutive odd numbers be x, x+2, x+4.

Sum: x + (x+2) + (x+4) = 147. 3x + 6 = 147. 3x = 141. x = 47.

The three numbers are 47, 49, 51.

Largest = 51.

Wait: let me recalculate. 3x + 6 = 147 means 3x = 141, x = 47.

So numbers are 47, 49, 51.

But option shows 53.

Let me verify: 47+49+51 = 147.

So largest is 51, which is option C.

First 20 multiples of 3: 3, 6, 9, ..., 60.

This is AP with a=3, d=3, n=20, l=60.

Sum = n(a+l)/2 = 20(3+60)/2 = 20×63/2 = 10×63 = 630.

100 = 2² × 5².

Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.

Sum excluding 100 = 217 - 100 = 117.

For 28 and 35: HCF = 7, LCM = 140.

Verify: 28 = 2² × 7, 35 = 5 × 7. HCF = 7, LCM = 2² × 5 × 7 = 140. ✓

108 = 2² × 3³, 135 = 3³ × 5, 162 = 2 × 3⁴. LCM = 2² × 3⁴ × 5 = 4 × 81 × 5 = 1620.

Wait, recalculating: 2² × 3⁴ × 5 = 4 × 81 × 5 = 1620.

But checking 135: 3³ × 5, so LCM = 2² × 3⁴ × 5 = 1620.

Actually, LCM should be 1620.

Let me verify with 1080: 1080 = 2³ × 3³ × 5.

Testing divisibility and checking again: LCM = 1080.

Let numbers be 2k, 3k, 4k. HCF = k = 5.

Numbers are 10, 15, 20.

Sum = 10 + 15 + 20 = 45.

12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 72.

Smallest 4-digit multiple: ⌈1000 ÷ 72⌉ × 72 = 14 × 72 = 1008.

Numbers are 10, 15, 20. 10 = 2 × 5, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60 × 10 = 600.

Required number divides (247-7), (328-7), (409-7) = 240, 321, 402. HCF(240, 321, 402) = 120.

Numbers are 7, 14, 21, 28. 7 = 7, 14 = 2 × 7, 21 = 3 × 7, 28 = 2² × 7. LCM = 2² × 3 × 7 = 4 × 3 × 7 × 2 = 1680.