[u]Set 4[/u]


[b]G10.[/b] Let $ABCD$ be a square with side length $1$. It is folded along a line $\ell$ that divides the square into two pieces with equal area. The minimum possible area of the resulting shape is $A$. Find the integer closest to $100A$.


[b]G11.[/b] The $10$-digit number $\underline{1A2B3C5D6E}$ is a multiple of $99$. Find $A + B + C + D + E$. 


[b]G12.[/b] Let $A, B, C, D$ be four points satisfying $AB = 10$ and $AC = BC = AD = BD = CD = 6$. If $V$ is the volume of tetrahedron $ABCD$, then find $V^2$.


[u]Set 5[/u]


[b]G13.[/b]  Nate the giant is running a $5000$ meter long race. His first step is $4$ meters, his next step is $6$ meters, and in general, each step is $2$ meters longer than the previous one. Given that his $n$th step will get him across the finish line, find $n$.


[b]G14.[/b] In square $ABCD$ with side length $2$, there exists a point $E$ such that $DA = DE$. Let line $BE$ intersect side $AD$ at $F$ such that $BE = EF$. The area of $ABE$ can be expressed in the form $a -\sqrt{b}$ where $a$ is a positive integer and $b$ is a square-free integer. Find $a + b$.


[b]G15.[/b] Patrick the Beetle is located at $1$ on the number line. He then makes an infinite sequence of moves where each move is either moving $1$, $2$, or $3$ units to the right. The probability that he does reach $6$ at some point in his sequence of moves is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.


[u]Set 6[/u]


[b]G16.[/b] Find the smallest positive integer $c$ greater than $1$ for which there do not exist integers $0 \le x, y \le9$ that satisfy $2x + 3y = c$.


[b]G17.[/b] Jaeyong is on the point $(0, 0)$ on the coordinate plane. If Jaeyong is on point $(x, y)$, he can either walk to $(x + 2, y)$, $(x + 1, y + 1)$, or $(x, y + 2)$. Call a walk to $(x + 1, y + 1)$ an Brilliant walk. If Jaeyong cannot have two Brilliant walks in a row, how many ways can he walk to the point $(10, 10)$?


[b]G18.[/b]  Deja vu?
Let $ABCD$ be a square with side length $1$. It is folded along a line $\ell$ that divides the square into two pieces with equal area. The maximum possible area of the resulting shape is $B$. Find the integer closest to $100B$.


PS. You should use hide for answers. Sets 1-3 have been posted [url=https://artofproblemsolving.com/community/c3h3131303p28367061]here [/url] and 7-9 [url=https://artofproblemsolving.com/community/c3h3131308p28367095]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. To solve the problem, we need to use the divisibility rules for 9 and 11. A number is divisible by 99 if and only if it is divisible by both 9 and 11.
2. For a number to be divisible by 9, the sum of its digits must be divisible by 9. Let the sum of the digits of the number be \( S \). Then, we have:
   \[
   1 + A + 2 + B + 3 + C + 5 + D + 6 + E = 17 + A + B + C + D + E
   \]
   Therefore, \( 17 + A + B + C + D + E \equiv 0 \pmod{9} \), which simplifies to:
   \[
   A + B + C + D + E \equiv -17 \equiv 1 \pmod{9}
   \]
3. For a number to be divisible by 11, the alternating sum of its digits must be divisible by 11. The alternating sum of the digits of the number is:
   \[
   1 - A + 2 - B + 3 - C + 5 - D + 6 - E = 17 - (A + B + C + D + E)
   \]
   Therefore, \( 17 - (A + B + C + D + E) \equiv 0 \pmod{11} \), which simplifies to:
   \[
   17 - (A + B + C + D + E) \equiv 0 \pmod{11} \implies A + B + C + D + E \equiv 17 \pmod{11}
   \]
   Simplifying further, we get:
   \[
   A + B + C + D + E \equiv 6 \pmod{11}
   \]
4. We now have two congruences:
   \[
   A + B + C + D + E \equiv 1 \pmod{9}
   \]
   \[
   A + B + C + D + E \equiv 6 \pmod{11}
   \]
5. To find the integer that satisfies both congruences, we look for a common solution in the range [0, 45]. We start by listing the possible values for \( A + B + C + D + E \) that satisfy the first congruence:
   \[
   A + B + C + D + E = 1, 10, 19, 28, 37, 46, \ldots
   \]
6. We then check which of these values also satisfy the second congruence:
   \[
   1 \not\equiv 6 \pmod{11}
   \]
   \[
   10 \not\equiv 6 \pmod{11}
   \]
   \[
   19 \not\equiv 6 \pmod{11}
   \]
   \[
   28 \equiv 6 \pmod{11}
   \]
   \[
   37 \not\equiv 6 \pmod{11}
   \]
   \[
   46 \not\equiv 6 \pmod{11}
   \]
7. The only value that satisfies both congruences is \( 28 \).

The final answer is \( \boxed{28} \).