Number and Polynomial Division

Using the sum of cubes formula or modular arithmetic: 16^3 + 17^3 + 18^3 + 19^3. 16+19=35 and 17+18=35. Since a^3+b^3 is divisible by a+b, the sum is divisible by 35. Also, the sum of four consecutive cubes is divisible by 4. Thus it is divisible by 70, leaving remainder 0.

Let the number be N. When divided by 36, remainder is 19, so N = 36k + 19 for some integer k. N = 36k + 19 = 12(3k) + 12 + 7 = 12(3k+1) + 7. When N is divided by 12, the remainder is 7. (192 would mean N > 36, which is incorrect for a remainder).

If n divided by 7 leaves remainder 3, then n = 7k + 3. Then (n-3) = 7k. Squaring both sides gives (n-3)^2 = (7k)^2 = 49k^2. Since 49k^2 is a multiple of 49, (n-3)^2 is exactly divisible by 49.

A number's remainder when divided by 9 is the same as the remainder of the sum of its digits divided by 9. The sum of 7+6+5+4+3+2+9+8+7+6+5+4+3+2+1+8 is 80. 80 divided by 9 is 8 with a remainder of 8.

Using Fermat's Little Theorem: for prime p=101, a^100 ≡ 1 (mod 101) when a is not divisible by 101. Therefore, 205^205 ≡ 205^(2×100 + 5) ≡ (205^100)^2 × 205^5 ≡ 1^2 × 205^5. Simplify: 205 ≡ 3 (mod 101), so 3^5 = 243 ≡ 41 (mod 101). The remainder is 41.

Since a > b, a! = a * (a-1) * ... * (b+1) * b!. Therefore, a! is a multiple of b!, and the remainder of any multiple divided by its factor is always 0.

The number 11 satisfies all conditions: 11/2 leaves remainder 1, 11/3 leaves remainder 2, and 11/4 leaves remainder 3. It is the smallest such positive integer.

Let the number be $10x + y$. From the first condition, $10x + y = 4(x + y) + 3$, which simplifies to $2x - y = 1$. From the second, $10x + y = 3xy + 5$. Substituting $y = 2x - 1$ into the second equation gives $10x + 2x - 1 = 3x(2x - 1) + 5$, leading to $6x^2 - 15x + 6 = 0$. Solving gives $x=2$, so $y=3$. The number is 23.

If a number leaves remainder 3 when divided by 17 and 23, then that number minus 3 is divisible by both 17 and 23. Since 3 itself satisfies this condition (3-3=0, which is divisible by any number), 3 is the least positive integer with this property. The LCM of 17 and 23 is 391, so other solutions are 394, 785, etc.

We need 7^2001 mod 14. Note that 7 mod 14 = 7, and 7² = 49, which gives 49 mod 14 = 7 (since 49 = 3×14 + 7). Similarly, 7³ mod 14 = 7² × 7 mod 14 = 7 × 7 mod 14 = 7. By induction, all positive powers of 7 give remainder 7 when divided by 14. Therefore, 7^2001 mod 14 = 7. Option D is correct. Option A would be true if dividing by 6, and Option B incorrectly assumes divisibility.

(3264) in base 7 is 3(7^3) + 2(7^2) + 6(7^1) + 4(7^0) = 1029 + 98 + 42 + 4 = 1173 in base 10. 1173 divided by 8 is 146 with a remainder of 5. The other options are incorrect remainders.

Using the division algorithm: Dividend = Divisor × Quotient + Remainde r. Here: 4150 = Divisor × 55 + 25. So 4150 - 25 = 4125 = Divisor × 55. Therefore: Divisor = 4125 ÷ 55 = 75.

When a number is divided by 36 and leaves remainder 19, we can write: Number = 36k + 19 for some integer k. Since 36 = 12 × 3, when we divide by 12, we get: Number = 12(3k) + 12 + 7 = 12(3k+1) + 7. Thus the remainder is 7. Option C (192) is incorrect as it's greater than the divisor 12.