[b]p1.[/b] James has $8$ Instagram accounts, $3$ Facebook accounts, $4$ QQ accounts, and $3$ YouTube accounts. If each Instagram account has $19$ pictures, each Facebook account has $5$ pictures and $9$ videos, each QQ account has a total of $17$ pictures, and each YouTube account has $13$ videos and no pictures, how many pictures in total does James have in all these accounts?


[b]p2.[/b] If Poonam can trade $7$ shanks for $4$ shinks, and she can trade $10$ shinks for $17$ shenks. How many shenks can Poonam get if she traded all of her $105$ shanks?


[b]p3.[/b] Jerry has a bag with $3$ red marbles, $5$ blue marbles and $2$ white marbles. If Jerry randomly picks two marbles from the bag without replacement, the probability that he gets two different colors can be expressed as a fraction $\frac{m}{n}$ in lowest terms. What is $m + n$?


[b]p4.[/b] Bob's favorite number is between $1200$ and $4000$, divisible by $5$, has the same units and hundreds digits, and the same tens and thousands digits. If his favorite number is even and not divisible by $3$, what is his favorite number?


[b]p5.[/b] Consider a unit cube $ABCDEFGH$. Let $O$ be the center of the face $EFGH$. The length of $BO$ can be expressed in the form $\frac{\sqrt{a}}{b}$, where $a$ and $b$ are simplified to lowest terms. What is $a + b$?


[b]p6.[/b] Mr. Eddie Wang is a crazy rich boss who owns a giant company in Singapore. Even though Mr. Wang appears friendly, he finds great joy in firing his employees. His immediately fires them when they say "hello" and/or "goodbye" to him. It is well known that $1/2$ of the total people say "hello" and/or "goodbye" to him everyday. If Mr. Wang had $2050$ employees at the end of yesterday, and he hires $2$ new employees at the beginning of each day, in how many days will Mr. Wang first only have $6$ employees left?


[b]p7.[/b] In $\vartriangle ABC$, $AB = 5$, $AC = 6$. Let $D,E,F$ be the midpoints of $\overline{BC}$, $\overline{AC}$, $\overline{AB}$, respectively. Let $X$ be the foot of the altitude from $D$ to $\overline{EF}$. Let $\overline{AX}$ intersect $\overline{BC}$ at $Y$ . Given $DY = 1$, the length of $BC$ is $\frac{p}{q}$ for relatively prime positive integers $p, q$: Find $p + q$.


[b]p8.[/b] Given $\frac{1}{2006} = \frac{1}{a} + \frac{1}{b}$ where $a$ is a $4$ digit positive integer and $b$ is a $6$ digit positive integer, find the smallest possible value of $b$.


[b]p9.[/b] Pocky the postman has unlimited stamps worth $5$, $6$ and $7$ cents. However, his post office has two very odd requirements: On each envelope, an odd number of $7$ cent stamps must be used, and the total number of stamps used must also be odd. What is the largest amount of postage money Pocky cannot make with his stamps, in cents?


[b]p10.[/b] Let $ABCDEF$ be a regular hexagon with side length $2$. Let $G$ be the midpoint of side $DE$. Now let $O$ be the intersection of $BG$ and $CF$. The radius of the circle inscribed in triangle $BOC$ can be expressed in the form $\frac{a\sqrt{b}-\sqrt{c}}{d} $ where $a$, $b$, $c$, $d$ are simplified to lowest terms. What is $a + b + c + d$?


[b]p11.[/b] Estimation (Tiebreaker):
What is the total number of characters in all of the participants' email addresses in the Accuracy Round?


PS. You should use hide for answers.  Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the equation \(\frac{1}{2006} = \frac{1}{a} + \frac{1}{b}\) where \(a\) is a 4-digit positive integer and \(b\) is a 6-digit positive integer, we start by clearing the denominators:

1. Multiply both sides by \(2006ab\):
   \[
   ab = 2006a + 2006b
   \]

2. Rearrange the equation:
   \[
   ab - 2006a - 2006b = 0
   \]

3. Add \(2006^2\) to both sides to facilitate factoring:
   \[
   ab - 2006a - 2006b + 2006^2 = 2006^2
   \]

4. Factor the left-hand side using Simon's Favorite Factoring Trick (SFFT):
   \[
   (a - 2006)(b - 2006) = 2006^2
   \]

5. Calculate \(2006^2\):
   \[
   2006^2 = 4024036
   \]

6. Identify the divisors of \(4024036\) to find suitable values for \(a\) and \(b\):
   - The divisors of \(4024036\) include \(1, 2, 4, 17, 34, 59, 118, 236, 1003, 2006, 4012, 17051, 34102, 59177, 118354, 236708, 1006009, 2012018, 4024036\).

7. We need \(a\) to be a 4-digit number and \(b\) to be a 6-digit number. We test the pairs \((a - 2006, b - 2006)\) to find the smallest \(b\):
   - For \(a - 2006 = 1\), \(b - 2006 = 4024036\) (not valid as \(b\) is too large).
   - For \(a - 2006 = 2\), \(b - 2006 = 2012018\) (not valid as \(b\) is too large).
   - For \(a - 2006 = 4\), \(b - 2006 = 1006009\) (not valid as \(b\) is too large).
   - For \(a - 2006 = 17\), \(b - 2006 = 236708\) (not valid as \(b\) is too large).
   - For \(a - 2006 = 34\), \(b - 2006 = 118354\) (valid as \(b\) is a 6-digit number).

8. Calculate the corresponding values of \(a\) and \(b\):
   \[
   a = 34 + 2006 = 2040
   \]
   \[
   b = 118354 + 2006 = 120360
   \]

9. Verify the solution:
   \[
   \frac{1}{2040} + \frac{1}{120360} = \frac{120360 + 2040}{2040 \times 120360} = \frac{122400}{245544000} = \frac{1}{2006}
   \]

Thus, the smallest possible value of \(b\) is \(120360\).

The final answer is \(\boxed{120360}\)