[b]p1.[/b] What is the smallest positive prime divisor of $101^2 -99^2$?


[b]p2.[/b] The product $81 \cdot 13579$ equals $1099A99$ for some digit $A$. What is $A$?


[b]p3.[/b] On a MMATHS team of $6$ students, all students forget which of the $5$ MMATHS sites (Yale, UVA, UF, UM, and Columbia) they are supposed to attend. On competition day, each student chooses a random site uniformly and independently of each other to attend, and they each take a flight to the one they picked. What is the probability that all of the sites end up with at least one person from that team?


[b]p4.[/b] Square $ABCD$ has area $12$. Let $E$ be the midpoint of $AD$, and let $F$ be the point of intersection of $BE$ and $AC$. Find the area of quadrilateral $EFCD$.


[b]p5.[/b] Alexander is eating sliced olives for dinner. Olives are sliced into thirds so that each olive is composed of two indistinguishable “end“ pieces and one “middle“ piece. For each olive slice that Alexander puts on his plate, there is a $\frac23$ chance that it is an end piece and a $\frac13$ chance that it is a middle piece. After placing a randomly selected olive slice on his plate, he checks to see if he can rearrange the slices on his plate so that he has at least one whole olive, which consists of any middle piece and any two end pieces. He can successfully rearrange the olive slices on his plate into at least one whole olive after randomly selecting $n$ olive slices with probability of at least $0.8$. What is the minimum n for which this is true?


[b]p6.[/b] What is the smallest positive integer $n$ whose digits multiply to $8000$?


[b]p7.[/b] Mitchell has a bunch of $2\times 1$ rectangular tiles. He needs to arrange them in such a way that they exactly cover a $2\times 10$ grid of $1\times 1$ squares. How many ways are there for him to do this? Two such “tilings“ are considered distinct from each other if not all of the $2$-by-$1$ rectangular tiles in the two tilings are placed in exactly the same position and orientation.


[b]p8.[/b] Trapezoid $ABCD$ is inscribed in a circle of radius $100$. If $AB = 200$ and $BC = 50$, what is the perimeter of $ABCD$?


[b]p9.[/b] How many ordered pairs of nonzero integers $(a, b)$ are there such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{24}$?


[b]p10.[/b] Suppose $a$ and $b$ are complex numbers such that $|a|= |b| = 1$ and $a + b = \frac13 + \frac14 i$. Find the product $ab$.


[b]p11.[/b] Find the number of ordered triples of positive integers $(a_1, a_2, a_3)$ such that $a_1 + a_2 + a_3 = 2017$ and $2$ does not divide $a_1$, $3$ does not divide $a_2$, and $4$ does not divide $a_3$.


[b]p12.[/b] Circles $\Gamma$ and $\Omega$ have radii of $20$ and $17$, respectively, and their centers are separated by a distance of $5$. Let $X$ be a fixed intersection point of these two circles, and let $Z$ be some point on $\Gamma$ for which segment $XZ$ intersects $\Omega$ at a second point $Y$ . (If $XZ$ is tangent to $\Omega$, then define $Y$ to be $X$.) Find the maximum possible length of segment $Y Z$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. We start with the given equation:
   \[
   \frac{1}{a} + \frac{1}{b} = \frac{1}{24}
   \]
2. Multiply both sides by \(24ab\) to clear the denominators:
   \[
   24b + 24a = ab
   \]
3. Rearrange the equation to facilitate factoring:
   \[
   ab - 24a - 24b = 0
   \]
4. Add \(576\) (which is \(24^2\)) to both sides to complete the square:
   \[
   ab - 24a - 24b + 576 = 576
   \]
5. Factor the left-hand side:
   \[
   (a - 24)(b - 24) = 576
   \]
6. Determine the number of integer solutions for \((a-24)(b-24) = 576\). The prime factorization of \(576\) is:
   \[
   576 = 2^6 \cdot 3^2
   \]
7. The number of divisors of \(576\) is calculated by adding 1 to each of the exponents in the prime factorization and then multiplying:
   \[
   (6+1)(2+1) = 7 \cdot 3 = 21
   \]
8. Each divisor \(d\) of \(576\) gives a pair \((a-24, b-24)\) and \((b-24, a-24)\), so we have \(2 \times 21 = 42\) pairs.
9. Since \(a\) and \(b\) are nonzero integers, we must exclude the pair \((0,0)\) which is not valid in this context. Therefore, we subtract 1 from the total:
   \[
   42 - 1 = 41
   \]

The final answer is \(\boxed{41}\)