20 = 2² × 5.

Sum of divisors = (1 + 2 + 4)(1 + 5) = 7 × 6 = 42.

Divisors are: 1, 2, 4, 5, 10, 20.

32 ÷ 7 = 4 remainder 4 (No). 33 ÷ 7 = 4 remainder 5 (No). 34 ÷ 7 = 4 remainder 6 (No). 32 = 7×4 + 5 = 28 + 5 (Yes). 32 ÷ 7 gives remainder 5.

Digital root = 9 + 8 + 7 + 6 = 30 → 3 + 0 = 3.

Alternatively, since the sum is divisible by 9, the digital root is 9.

Wait: 9+8+7+6 = 30, 3+0 = 3.

So digital root is 3.

Using HCF × LCM = Product of two numbers: 15 × 180 = 45 × x.

Therefore, 2700 = 45x, so x = 60.

Perfect squares between 100 and 200: 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225 (exceeds 200).

So there are 4 perfect squares.

Wait: 10² = 100 (not between), so squares are 121, 144, 169, 196.

Count = 4.

Actually checking: we need squares from 11² to 14², which is 4 numbers.

Powers of 2 mod 5: 2¹≡2, 2²≡4, 2³≡3, 2⁴≡1, 2⁵≡2...

Pattern repeats every 4. 50 = 4×12 + 2, so 2⁵⁰ ≡ 2² ≡ 4 (mod 5).

We need LCM(12, 15, 20). 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.

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).

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.

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.