[b]p4.[/b] For how many three-digit multiples of $11$ in the form $\underline{abc}$ does the quadratic $ax^2 + bx + c$ have real roots?


[b]p5.[/b] William draws a triangle $\vartriangle ABC$ with $AB =\sqrt3$, $BC = 1$, and $AC = 2$ on a piece of paper and cuts out $\vartriangle ABC$. Let the angle bisector of $\angle ABC$ meet $AC$ at point $D$. He folds $\vartriangle ABD$ over $BD$. Denote the new location of point $A$ as $A'$. After William folds $\vartriangle A'CD$ over $CD$, what area of the resulting figure is covered by three layers of paper?


[b]p6.[/b] Compute $(1)(2)(3) + (2)(3)(4) + ... + (18)(19)(20)$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the problem, we need to compute the sum of the products \((1)(2)(3) + (2)(3)(4) + \ldots + (18)(19)(20)\).

1. **Express the general term**:
   Each term in the sum can be written as \((n-1)n(n+1)\). This is a product of three consecutive integers.

2. **Simplify the general term**:
   Notice that \((n-1)n(n+1)\) can be rewritten using the identity for the product of three consecutive integers:
   \[
   (n-1)n(n+1) = n^3 - n
   \]
   This identity can be derived by expanding the product:
   \[
   (n-1)n(n+1) = n(n^2 - 1) = n^3 - n
   \]

3. **Sum the series**:
   We need to sum the series from \(n=2\) to \(n=19\):
   \[
   \sum_{n=2}^{19} (n^3 - n)
   \]
   This can be split into two separate sums:
   \[
   \sum_{n=2}^{19} n^3 - \sum_{n=2}^{19} n
   \]

4. **Evaluate the sum of cubes**:
   The sum of cubes from \(n=1\) to \(n=19\) is given by:
   \[
   \left( \sum_{n=1}^{19} n \right)^2 = \left( \frac{19 \cdot 20}{2} \right)^2 = 190^2 = 36100
   \]
   We need to subtract the cube of 1 since our sum starts from \(n=2\):
   \[
   \sum_{n=2}^{19} n^3 = 36100 - 1^3 = 36100 - 1 = 36099
   \]

5. **Evaluate the sum of integers**:
   The sum of integers from \(n=2\) to \(n=19\) is:
   \[
   \sum_{n=2}^{19} n = \sum_{n=1}^{19} n - 1 = \frac{19 \cdot 20}{2} - 1 = 190 - 1 = 189
   \]

6. **Combine the results**:
   Subtract the sum of integers from the sum of cubes:
   \[
   \sum_{n=2}^{19} (n^3 - n) = 36099 - 189 = 35910
   \]

The final answer is \(\boxed{35910}\).