[u]Part 1 [/u]

[b]p1.[/b] Calculate $$(4!-5!+2^5 +2^6) \cdot \frac{12!}{7!}+(1-3)(4!-2^4).$$


[b]p2.[/b] The expression $\sqrt{9!+10!+11!}$ can be expressed as $a\sqrt{b}$ for positive integers $a$ and $b$, where $b$ is squarefree. Find $a$.


[b]p3.[/b] For real numbers $a$ and $b$, $f(x) = ax^{10}-bx^4+6x +10$ for all real $x$. Given that $f(42) = 11$, find $f (-42)$.


[u]Part 2[/u]


[b]p4.[/b] How many positive integers less than or equal to $2023$ are divisible by $20$, $23$, or both?


[b]p5.[/b] Larry the ant crawls along the surface of a cylinder with height $48$ and base radius $\frac{14}{\pi}$ . He starts at point $A$ and crawls to point $B$, traveling the shortest distance possible. What is the maximum this distance could be?


[b]p6.[/b] For a given positive integer $n$, Ben knows that $\lfloor 20x \rfloor = n$, where $x$ is real. With that information, Ben determines that there are $3$ distinct possible values for $\lfloor 23x \rfloor$. Find the least possible value of $n$.


[u]Part 3 [/u]


[b]p7.[/b] Let $ABC$ be a triangle with area $1$. Points $D$, $E$, and $F$ lie in the interior of $\vartriangle ABC$ in such a way that $D$ is the midpoint of $AE$, $E$ is the midpoint of $BF$, and $F$ is the midpoint of $CD$. Compute the area of $DEF$.


[b]p8.[/b] Edwin and Amelia decide to settle an argument by running a race against each other. The starting line is at a given vertex of a regular octahedron and the finish line is at the opposite vertex. Edwin has the ability to run straight through the octahedron, while Amelia must stay on the surface of the octahedron. Given that they tie, what is the ratio of Edwin’s speed to Amelia’s speed?


[b]p9.[/b] Jxu is rolling a fair three-sided die with faces labeled $0$, $1$, and $2$. He keeps going until he rolls a $1$, immediately followed by a $2$. What is the expected number of rolls Jxu makes?


[u]Part 4 [/u]


[b]p10.[/b] For real numbers $x$ and $y$, $x +x y = 10$ and $y +x y = 6$. Find the sum of all possible values of $\frac{x}{y}$.


[b]p11.[/b] Derek is thinking of an odd two-digit integer $n$. He tells Aidan that $n$ is a perfect power and the product of the digits of $n$ is also a perfect power. Find the sum of all possible values of $n$.


[b]p12.[/b] Let a three-digit positive integer $N = \overline{abc}$ (in base $10$) be stretchable with respect to $m$ if $N$ is divisible by $m$, and when $N$‘s middle digit is duplicated an arbitrary number of times, it‘s still divisible by $m$. How many three-digit positive integers are stretchable with respect to $11$? (For example, $432$ is stretchable with respect to $6$ because $433...32$ is divisible by $6$ for any positive integer number of $3$s.)


[u]Part 5 [/u]


[b]p13.[/b] How many trailing zeroes are in the base-$2023$ expansion of $2023!$ ?


[b]p14.[/b] The three-digit positive integer $k = \overline{abc}$ (in base $10$, with a nonzero) satisfies $\overline{abc} = c^{2ab-1}$. Find the sum of all possible $k$.


[b]p15.[/b] For any positive integer $k$, let $a_k$ be defined as the greatest nonnegative real number such that in an infinite grid of unit squares, no circle with radius less than or equal to $a_k$ can partially cover at least $k$ distinct unit squares. (A circle partially covers a unit square only if their intersection has positive area.) Find the sumof all positive integers $n \le 12$ such that $a_n \ne a_{n+1}$.


PS. You should use hide for answers. Rounds 6-9 have been posted [url=https://artofproblemsolving.com/community/c3h3267915p30057005]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To find the number of trailing zeroes in the base-$2023$ expansion of $2023!$, we need to determine the number of factors of $2023$ in $2023!$. Since $2023 = 7 \cdot 17^2$, we need to find the number of factors of $7$ and $17$ in $2023!$.

Notice that the number of factors of $17$ in $2023!$ is less than the number of factors of $7$ in $2023!$. Also, $2023$ has two factors of $17$, making the total possible number of factors of $2023$ even less. Because of this, we just need to calculate the number of factors of $17$ in $2023!$ and divide it by $2$.

1. **Calculate the number of factors of $17$ in $2023!$:**
   \[
   \left\lfloor \frac{2023}{17} \right\rfloor + \left\lfloor \frac{2023}{17^2} \right\rfloor + \left\lfloor \frac{2023}{17^3} \right\rfloor + \cdots
   \]
   \[
   \left\lfloor \frac{2023}{17} \right\rfloor = \left\lfloor 119.0 \right\rfloor = 119
   \]
   \[
   \left\lfloor \frac{2023}{17^2} \right\rfloor = \left\lfloor \frac{2023}{289} \right\rfloor = \left\lfloor 7.0 \right\rfloor = 7
   \]
   \[
   \left\lfloor \frac{2023}{17^3} \right\rfloor = \left\lfloor \frac{2023}{4913} \right\rfloor = \left\lfloor 0.411 \right\rfloor = 0
   \]
   Therefore, the total number of factors of $17$ in $2023!$ is:
   \[
   119 + 7 + 0 = 126
   \]

2. **Calculate the number of factors of $2023$ in $2023!$:**
   Since $2023 = 7 \cdot 17^2$, we need to divide the number of factors of $17$ by $2$:
   \[
   \left\lfloor \frac{126}{2} \right\rfloor = 63
   \]

Therefore, the number of trailing zeroes in the base-$2023$ expansion of $2023!$ is $\boxed{63}$.