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.

To find the remainder when \(2^{50}\) is divided by 5, we use the Fermat's Little Theorem approach by finding the pattern of powers of 2 modulo 5.

Step 1: Find the pattern of powers of 2 (mod 5)

Calculate successive powers of 2 and their remainders when divided by 5:

The pattern repeats every 4 powers.

Step 2: Apply the cycle to find \(2^{50} \pmod{5}\)

Since the remainder repeats with period 4, divide the exponent by 4:

This means \(2^{50}\) has the same remainder as \(2^2\).

Step 3: Calculate the remainder

Answer: The remainder is \(4\) (Option D)

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

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

Therefore, other number = 2700 ÷ 45 = 60.

144 = 2⁴ × 3². 196 = 2² × 7².

Common prime factors: 2².

Therefore, HCF = 4.

Let numbers be 4k and 5k. HCF = k = 6.

Therefore, numbers are 24 and 30.

The larger number is 30.

56 = 2³ × 7, 84 = 2² × 3 × 7, 140 = 2² × 5 × 7.

Common prime factors: 2 × 7 = 14.

16 = 2⁴, 24 = 2³ × 3, 36 = 2² × 3². LCM = 2⁴ × 3² = 16 × 9 = 144.

HCF × LCM = Product of numbers. 18 × 324 = 36 × other.

Therefore, other = 5832 ÷ 36 = 162.

110 = 2 × 5 × 11, 165 = 3 × 5 × 11, 220 = 2² × 5 × 11.

Common factors: 5 × 11 = 55.