[hide=D stands for Dedekind, Z stands for Zermelo]they had two problem sets under those two names[/hide] 


[u]Set 4[/u]


[b]D16.[/b]  The cooking club at Blair creates $14$ croissants and $21$ danishes. Daniel chooses pastries randomly, stopping when he gets at least one croissant and at least two danishes. How many pastries must he choose to guarantee that he has one croissant and two danishes?


[b]D17.[/b] Each digit in a $3$ digit integer is either $1, 2$, or $4$ with equal probability. What is the probability that the hundreds digit is greater than the sum of the tens digit and the ones digit?


[b]D18 / Z11.[/b] How many two digit numbers are there such that the product of their digits is prime?


[b]D19 / Z9.[/b] In the coordinate plane, a point is selected in the rectangle defined by $-6 \le x \le 4$ and $-2 \le y \le 8$. What is the largest possible distance between the point and the origin, $(0, 0)$?


[b]D20 / Z10.[/b] The sum of two numbers is $6$ and the sum of their squares is $32$. Find the product of the two numbers.


[u]Set 5[/u]


[b]D21 / Z12.[/b] Triangle $ABC$ has area $4$ and $\overline{AB} = 4$. What is the maximum possible value of $\angle ACB$?


[b]D22 / Z13.[/b] Let $ABCD$ be an iscoceles trapezoid with $AB = CD$ and M be the midpoint of $AD$. If $\vartriangle ABM$ and $\vartriangle MCD$ are equilateral, and $BC = 4$, find the area of trapezoid $ABCD$.


[b]D23 / Z14.[/b] Let $x$ and $y$ be positive real numbers that satisfy $(x^2 + y^2)^2 = y^2$. Find the maximum possible value of $x$.


[b]D24 / Z17.[/b] In parallelogram $ABCD$, $\angle A \cdot \angle C - \angle B  \cdot \angle D = 720^o$ where all angles are in degrees. Find the value of $\angle C$.


[b]D25.[/b] The number $12ab9876543$ is divisible by $101$, where $a, b$ represent digits between $0$ and $9$. What is $10a + b$?


[u]Set 6[/u]


[b]D26  / Z26.[/b] For every person who wrote a problem that appeared on the final MBMT tests, take the number of problems they wrote, and then take that number’s factorial, and finally multiply all these together to get $n$. Estimate the greatest integer $a$ such that $2^a$ evenly divides $n$.


[b]D27 / Z27.[/b] Circles of radius $5$ are centered at each corner of a square with side length $6$. If a random point $P$ is chosen randomly inside the square, what is the probability that $P$ lies within all four circles?


[b]D28 / Z28.[/b] Mr. Rose’s evil cousin, Mr. Caulem, has teaches a class of three hundred bees. Every week, he tries to disrupt Mr. Rose’s $4$th period by sending three of his bee students to fly around and make human students panic. Unfortunately, no pair of bees can fly together twice, as then Mr. Rose will become suspicious and trace them back to Mr. Caulem. What’s the largest number of weeks Mr. Caulem can disrupt Mr. Rose’s class?


[b]D29 / Z29. [/b]Two blind brothers Beard and Bored are driving their tractors in the middle of a field facing north, and both are $10$ meters west from a roast turkey. Beard, can turn exactly $0.7^o$ and Bored can turn exactly $0.2^o$ degrees. Driving at a consistent $2$ meters per second, they drive straight until they notice the smell of the turkey getting farther away, and then turn right and repeat until they get to the turkey. 
Suppose Beard gets to the Turkey in about $818.5$ seconds. Estimate the amount of time it will take Bored.


[b]D30 / Z30.[/b] Let a be the probability that $4$ randomly chosen positive integers have no common divisor except for $1$. Estimate $300a$. Note that the integers $1, 2, 3, 4$ have no common divisor except for $1$. 

Remark. This problem is asking you to find $300 \lim_{n\to \infty} a_n$, if $a_n$ is defined to be the probability that $4$ randomly chosen integers from $\{1, 2, ..., n\}$ have greatest common divisor $1$.


PS. You should use hide for answers. D.1-15 / Z.1-8 problems have been collected [url=https://artofproblemsolving.com/community/c3h2916240p26045561]here [/url]and Z.15-25  [url=https://artofproblemsolving.com/community/c3h2916258p26045774]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the problem, we need to determine the digits \(a\) and \(b\) such that the number \(12ab9876543\) is divisible by 101. Let's break down the steps:

1. **Express the number in terms of \(a\) and \(b\):**
   \[
   N = \overline{12ab9876543} = 12009876543 + (10a + b) \times 10^7
   \]
   Since \(N\) is divisible by 101, we can write:
   \[
   N \equiv 0 \pmod{101}
   \]

2. **Simplify the expression modulo 101:**
   \[
   12009876543 \equiv 75 \pmod{101}
   \]
   This is because:
   \[
   12009876543 \div 101 \approx 118910653 \quad \text{with a remainder of } 75
   \]

3. **Consider the term \((10a + b) \times 10^7\):**
   \[
   10^7 \equiv 91 \pmod{101}
   \]
   This is because:
   \[
   10^7 = 10000000 \quad \text{and} \quad 10000000 \div 101 \approx 99009 \quad \text{with a remainder of } 91
   \]

4. **Combine the results:**
   \[
   N \equiv 75 + (10a + b) \times 91 \pmod{101}
   \]
   For \(N\) to be divisible by 101:
   \[
   75 + (10a + b) \times 91 \equiv 0 \pmod{101}
   \]

5. **Simplify the congruence:**
   \[
   75 + 91(10a + b) \equiv 0 \pmod{101}
   \]
   \[
   75 + 91 \times 10a + 91b \equiv 0 \pmod{101}
   \]
   \[
   75 + 910a + 91b \equiv 0 \pmod{101}
   \]
   Since \(910 \equiv 1 \pmod{101}\):
   \[
   75 + a + 91b \equiv 0 \pmod{101}
   \]

6. **Solve for \(a\) and \(b\):**
   \[
   a + 91b \equiv -75 \pmod{101}
   \]
   We need to find \(a\) and \(b\) such that \(0 \le a \le 9\) and \(0 \le b \le 9\). Let's test possible values for \(b\):

   - For \(b = 8\):
     \[
     a + 91 \times 8 \equiv -75 \pmod{101}
     \]
     \[
     a + 728 \equiv -75 \pmod{101}
     \]
     \[
     a + 21 \equiv -75 \pmod{101}
     \]
     \[
     a \equiv -75 - 21 \pmod{101}
     \]
     \[
     a \equiv -96 \pmod{101}
     \]
     \[
     a \equiv 5 \pmod{101}
     \]

   Thus, \(a = 5\) and \(b = 8\) satisfy the conditions.

The final answer is \(10a + b = 10 \times 5 + 8 = \boxed{58}\).