[b]p1.[/b] While computing $7 - 2002 \cdot x$, John accidentally evaluates from left to right $((7 - 2002)  \cdot x)$ instead of correctly using order of operations $(7 - (2002  \cdot x))$. If he gets the correct answer anyway, what is $x$?


[b]p2.[/b] Given that
$$x^2 + y^2 + z^2 = 6$$
$$ \left( \frac{x}{y} + \frac{y}{x} \right)^2 + \left( \frac{y}{z} + \frac{z}{y} \right)^2 + \left( \frac{z}{x} + \frac{x}{z} \right)^2 = 16.5,$$
what is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}$ ?


[b]p3.[/b] Evaluate
$$\frac{tan \frac{\pi}{4}}{4}+\frac{tan \frac{3\pi}{4}}{8}+\frac{tan \frac{5\pi}{4}}{16}+\frac{tan \frac{7\pi}{4}}{32}+ ...$$


[b]p4.[/b] Note that $2002 = 22 \cdot 91$, and so $2002$ is a multiple of the number obtained by removing its middle $2$ digits. Generalizing this, how many $4$-digit palindromes, $abba$, are divisible by the $2$-digit palindrome, $aa$?


[b]p5.[/b] Let $ABCDE$ be a pyramid such that $BCDE$ is a square with side length $2$, and $A$ is $2$ units above the center of $BCDE$. If $F$ is the midpoint of $\overline{DE}$ and $G$ is the midpoint of $\overline{AC}$, what is the length of $\overline{DE}$?


[b]p6.[/b] Suppose $a_1, a_2,..., a_{100}$ are real numbers with the property that $$i(a_1 + a_2 +... + a_i) = 1 + (a_{i+1} + a_{i+2} + ... + a_{100})$$ for all $i$. Compute $a_{10}$.


[b]p7.[/b] A bug is sitting on one corner of a $3' \times 4' \times 5'$ block of wood. What is the minimum distance nit needs to travel along the block’s surface to reach the opposite corner?


[b]p8.[/b] In the number game, a pair of positive integers $(n,m)$ is written on a blackboard. Two players then take turns doing the following:
1. If $n \ge m$, the player chooses a positive integer $c$ such that $n - cm \ge 0$, and replaces $(n,m)$ with $(n - cm,m)$.
2. If $m > n$, the player chooses a positive integer $c$ such that $m - cn \ge 0$, and replaces $(n,m)$ with $(n,m - cn)$.
If $m$ or $n$ ever become $0$, the game ends, and the last player to have moved is declared the winner. If $(n,m)$ are originally $(20021000, 2002)$, what choices of $c$ are winning moves for the first player?


PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the given problem, we need to evaluate the infinite series:

\[
\frac{\tan \frac{\pi}{4}}{4} + \frac{\tan \frac{3\pi}{4}}{8} + \frac{\tan \frac{5\pi}{4}}{16} + \frac{\tan \frac{7\pi}{4}}{32} + \cdots
\]

First, we note the values of the tangent function at the specified angles:
\[
\tan \frac{\pi}{4} = 1, \quad \tan \frac{3\pi}{4} = -1, \quad \tan \frac{5\pi}{4} = 1, \quad \tan \frac{7\pi}{4} = -1, \quad \text{and so on.}
\]

Thus, the series alternates between \(1\) and \(-1\). We can rewrite the series as:
\[
\frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \cdots
\]

This is an alternating geometric series with the first term \(a = \frac{1}{4}\) and common ratio \(r = -\frac{1}{2}\).

The sum \(S\) of an infinite geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)) is given by:
\[
S = \frac{a}{1 - r}
\]

Substituting \(a = \frac{1}{4}\) and \(r = -\frac{1}{2}\), we get:
\[
S = \frac{\frac{1}{4}}{1 - (-\frac{1}{2})} = \frac{\frac{1}{4}}{1 + \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{3}{2}} = \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{6}
\]

Therefore, the sum of the series is:
\[
\boxed{\frac{1}{6}}
\]