Numbers are 10, 15, 20. 10 = 2 × 5, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60 × 10 = 600.
12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 72.
Smallest 4-digit multiple: ⌈1000 ÷ 72⌉ × 72 = 14 × 72 = 1008.
Let numbers be 2k, 3k, 4k. HCF = k = 5.
Numbers are 10, 15, 20.
Sum = 10 + 15 + 20 = 45.
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.
For 28 and 35: HCF = 7, LCM = 140.
Verify: 28 = 2² × 7, 35 = 5 × 7. HCF = 7, LCM = 2² × 5 × 7 = 140. ✓
100 = 2² × 5².
Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.
Sum excluding 100 = 217 - 100 = 117.
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.
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.
Last digits of powers of 3: 3¹=3, 3²=9, 3³=27(7), 3⁴=81(1), 3⁵=243(3)...
Pattern: 3,9,7,1 repeats every 4. 2023 = 4×505 + 3, so 3^2023 has same last digit as 3³, which is 7.
Units place: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 (10 times).
Tens place: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 (10 times).
Total = 20 (note: 77 contains two 7s).