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. ✓
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.
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.
A number is divisible by 9 if sum of its digits is divisible by 9. Here sum is 12, which is not divisible by 9. However, any number divisible by 9 is also divisible by 3. But the given condition states sum of digits is 12, and divisible by 9, which is contradictory. Re-reading: if divisible by 9, then sum must be divisible by 9. Since sum is 12 and divisible by 3, the number is divisible by 3.
Let number be n. n ≡ 4 (mod 7) and n ≡ 6 (mod 11). From first: n = 7k + 4. Substituting in second: 7k + 4 ≡ 6 (mod 11), so 7k ≡ 2 (mod 11). Testing values: k = 4 gives 7(4) + 4 = 32 ≡ 10 (mod 11). Try k = 5: 7(5) + 4 = 39 ≡ 6 (mod 11). Yes, 39 works.
2^50 = 1,125,899,906,842,624. Sum of digits = 1+1+2+5+8+9+9+9+0+6+8+4+2+6+2+4 = 76. (Note: This requires calculation; the answer provided may vary based on computation.)
By Fermat's Little Theorem, since 13 is prime and gcd(5,13)=1, we have 5^12 ≡ 1 (mod 13). 100 = 12×8 + 4. So 5^100 ≡ 5^4 (mod 13). 5^4 = 625 = 48×13 + 1 ≡ 1 (mod 13).
Using Legendre's formula: floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24.
Divisors of 20: 1, 2, 4, 5, 10, 20. Proper divisors (excluding 20): 1, 2, 4, 5, 10. Sum = 1 + 2 + 4 + 5 + 10 = 22, not 48.
We need LCM(6, 8, 9). 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Smallest multiple of 72 greater than 100: 72 × 2 = 144.