Let the numbers be 4x and 5x. Sum = 4x + 5x = 9x = 180. So x = 20. The larger number = 5x = 5(20) = 100.

Perfect cubes: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000. Between 1 and 1000 (inclusive) there are 10 perfect cubes.

If a number is divisible by 9, then: (A) It must be divisible by 3 (since 9 = 3²). (B) Sum of its digits must be divisible by 9 (divisibility rule). Both A and B are true.

144 = 2⁴ × 3², 180 = 2² × 3² × 5, 216 = 2³ × 3³. GCD = 2² × 3² = 4 × 9 = 36.

Even numbers from 1 to 100: 2, 4, 6, ..., 100. This is an AP with first term 2, last term 100, common difference 2. Number of terms = (100-2)/2 + 1 = 50.

We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.

Let the integers be x and x+1. Sum = x + (x+1) = 51. 2x + 1 = 51. 2x = 50. x = 25. The larger integer = 26.

Using formula for sum of first n natural numbers: S = n(n+1)/2 where n=99. S = 99(100)/2 = 9900/2 = 4950.

Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.

For divisibility by both 6 and 8, the number must be divisible by LCM(6,8) = 24. Check: 48 ÷ 24 = 2 ✓. Option A is correct.