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.

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.

Let the number be n.

Given: n = 8k + 5 for some integer k.

When divided by 4: n = 8k + 5 = 4(2k) + 4 + 1 = 4(2k + 1) + 1.

Therefore, remainder = 1.

Prime factorization of 360: 360 = 2³ × 3² × 5.

The unique prime factors are 2, 3, and 5.

Product = 2 × 3 × 5 = 30.