Perfect squares from 1 to 1000 are 1², 2², 3², ..., n² where n² ≤ 1000. So n ≤ √1000 ≈ 31.62. Therefore n can be 1, 2, 3, ..., 31. Total = 31 perfect squares.
Let the two consecutive even numbers be n and n+2. Then n(n+2) = 528. So n² + 2n - 528 = 0. Using quadratic formula or factoring: (n+24)(n-22) = 0. So n = 22 (taking positive value). The two numbers are 22 and 24. Larger = 24.
Let number = 10a + b. Reversed = 10b + a. Given: (10b + a) - (10a + b) = 45, so 9b - 9a = 45, thus b - a = 5. Also |a - b| = 5 or a - b = 5. From b - a = 5 and a - b could be -5 or 5. Testing: if b - a = 5 and digits sum conditions... Let a = 2, b = 7: number = 27. Reversed = 72. 72 - 27 = 45. ✓
Using Euclidean algorithm: 180 = 144×1 + 36; 144 = 36×4 + 0. Therefore HCF = 36.
Since 9 and 11 are coprime (HCF=1), a number divisible by both must be divisible by their product: 9×11 = 99.
Digit sum = 9 + 8 + 7 + 6 + 5 + 4 = 39.
360 = 2³×3²×5¹. Number of divisors = (3+1)(2+1)(1+1) = 4×3×2 = 24.
Let numbers be x and y. x + y = 50 and x - y = 10. Adding: 2x = 60, so x = 30.
A number is divisible by 8 if its last 3 digits form a number divisible by 8. 456 ÷ 8 = 57. So 2456 is divisible by 8.
1071 = 462×2 + 147; 462 = 147×3 + 21; 147 = 21×7 + 0. Therefore HCF = 21.