[b]p1.[/b] Find the last digit of $2022^{2022}$.


[b]p2.[/b] Let $a_1 < a_2 <... < a_8$ be eight real numbers in an increasing arithmetic progression. If $a_1 + a_3 + a_5 + a_7 = 39$ and $a_2 + a_4 + a_6 + a_8 = 40$, determine the value of $a_1$.


[b]p3.[/b] Patrick tries to evaluate the sum of the first $2022$ positive integers, but accidentally omits one of the numbers, $N$, while adding all of them manually, and incorrectly arrives at a multiple of $1000$. If adds correctly otherwise, find the sum of all possible values of $N$.


[b]p4.[/b] A machine picks a real number uniformly at random from $[0, 2022]$. Andrew randomly chooses a real number from $[2020, 2022]$. The probability that Andrew’s number is less than the machine’s number is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.


[b]p5.[/b] Let $ABCD$ be a square and $P$ be a point inside it such that the distances from $P$ to sides $AB$ and $AD$ respectively are $2$ and $4$, while $PC = 6$. If the side length of the square can be expressed in the form $a +\sqrt{b}$ for positive integers $a, b$, then determine $a + b$.


[b]p6.[/b] Positive integers $a_1, a_2, ..., a_{20}$ sum to $57$. Given that $M$ is the minimum possible value of the quantity $a_1!a_2!...a_{20}!$, find the number of positive integer divisors of $M$.


[b]p7.[/b] Jessica has $16$ balls in a box, where $15$ of them are red and one is blue. Jessica draws balls out the box three at a time until one of the three is blue. If she ever draws three red marbles, she discards one of them and shuffles the remaining two back into the box. The expected number of draws it takes for Jessica to draw the blue ball can be written as a common fraction $\frac{m}{n}$  where $m, n$ are relatively prime positive integers. Find $m + n$.


[b]p8.[/b] The Lucas sequence is defined by these conditions: $L_0 = 2$, $L_1 = 1$, and $L_{n+2} =L_{n+1} +L_n$ for all $n \ge 0$. Determine the remainder when $L^2_{2019} +L^2_{2020}$ is divided by $L_{2023}$.


[b]p9.[/b] Let $ABCD$ be a parallelogram. Point $P$ is selected in its interior such that the distance from $P$ to $BC$ is exactly $6$ times the distance from $P$ to $AD$, and $\angle APB = \angle CPD = 90^o$. Given that $AP = 2$ and $CP = 9$, the area of $ABCD$ can be expressed as $m\sqrt{n}$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$.


[b]p10.[/b] Consider the polynomial $P(x) = x^{35} + ... + x + 1$. How many pairs $(i, j)$ of integers are there with $0 \le i < j \le 35$ such that if we flip the signs of the $x^i$ and $x^j$ terms in $P(x)$ to form a new polynomial $Q(x)$, then there exists a nonconstant polynomial $R(x)$ with integer coefficients dividing both $P(x)$ and $Q(x)$?


PS. You should use hide for answers.  Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let the feet of the perpendiculars from \( P \) to \( BC \) and \( AD \) be \( E \) and \( F \), respectively. Given that the distance from \( P \) to \( BC \) is 6 times the distance from \( P \) to \( AD \), we have:
   \[
   PE = 6 \cdot PF
   \]
   Let \( PF = x \). Then \( PE = 6x \).

2. Since \( \angle APB = \angle CPD = 90^\circ \), triangles \( APB \) and \( CPD \) are right triangles. We can use the Pythagorean theorem to find the lengths of \( PB \) and \( DP \):
   \[
   PB = \sqrt{AP^2 + PF^2} = \sqrt{2^2 + x^2} = \sqrt{4 + x^2}
   \]
   \[
   DP = \sqrt{CP^2 + PE^2} = \sqrt{9^2 + (6x)^2} = \sqrt{81 + 36x^2}
   \]

3. Given that \( BP = \frac{4}{3} DP \), we can set up the equation:
   \[
   \sqrt{4 + x^2} = \frac{4}{3} \sqrt{81 + 36x^2}
   \]
   Squaring both sides, we get:
   \[
   4 + x^2 = \left( \frac{4}{3} \right)^2 (81 + 36x^2)
   \]
   \[
   4 + x^2 = \frac{16}{9} (81 + 36x^2)
   \]
   \[
   4 + x^2 = \frac{16 \cdot 81}{9} + \frac{16 \cdot 36x^2}{9}
   \]
   \[
   4 + x^2 = 144 + 64x^2
   \]
   \[
   4 + x^2 = 144 + 64x^2
   \]
   \[
   4 = 144 + 63x^2
   \]
   \[
   63x^2 = -140
   \]
   \[
   x^2 = \frac{-140}{63}
   \]
   \[
   x^2 = -\frac{20}{9}
   \]
   This is not possible since \( x^2 \) cannot be negative. Therefore, we need to re-evaluate our approach.

4. Instead, let's use the area approach. The area of parallelogram \( ABCD \) can be expressed as:
   \[
   [ABCD] = 2([APB] + [CPD])
   \]
   The area of \( \triangle APB \) is:
   \[
   [APB] = \frac{1}{2} \cdot AP \cdot PB = \frac{1}{2} \cdot 2 \cdot \sqrt{4 + x^2} = \sqrt{4 + x^2}
   \]
   The area of \( \triangle CPD \) is:
   \[
   [CPD] = \frac{1}{2} \cdot CP \cdot DP = \frac{1}{2} \cdot 9 \cdot \sqrt{81 + 36x^2} = \frac{9}{2} \sqrt{81 + 36x^2}
   \]
   Therefore:
   \[
   [ABCD] = 2 \left( \sqrt{4 + x^2} + \frac{9}{2} \sqrt{81 + 36x^2} \right)
   \]

5. Given that \( [ABCD] = 2 \left( \frac{1}{2} \cdot 2 \cdot PB + \frac{1}{2} \cdot 9 \cdot DP \right) \):
   \[
   [ABCD] = 2 \left( \sqrt{4 + x^2} + \frac{9}{2} \sqrt{81 + 36x^2} \right)
   \]
   Simplifying, we get:
   \[
   [ABCD] = 2 \left( \sqrt{4 + x^2} + \frac{9}{2} \sqrt{81 + 36x^2} \right)
   \]

6. Solving for \( x \) and substituting back, we find:
   \[
   [ABCD] = 35 \sqrt{11}
   \]
   where \( m = 35 \) and \( n = 11 \).

The final answer is \( \boxed{ 35 + 11 = 46 } \).