#### 1↵
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ ↵
↵
<spoiler summary="Answer">↵
191↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
Total Ways = $\binom{14}{5}$. Out of $14$ places choose $5$ places for men. Women will follow.<br>↵
There are total $7$ pairs of seats. Choose $5$ out of them. For each pair of seats there are $2$ ways.<br>↵
So required ways = $\binom{7}{5} \cdot 2^5$<br>↵
Required Probability = $\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$.↵
</spoiler>↵
↵
#### 2↵
A plane contains $40$ lines, no $2$ of which are parallel. Suppose there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.↵
↵
<spoiler summary="Answer">↵
607↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
Two lines can only intersect once.<br>↵
Maximum number of intersections = $\binom{n}{2}$ (Choose any two lines and they intersect)<br>↵
Maximum number of intersections occur when for each intersection point there are only two lines intersecting at that point. Lets label these $T2$ intersection points.<br>↵
If there are intersection points where there are $x$ ($x$ > $2$) lines intersecting at some point, then we will lose $T2$ points.<br>↵
Amount of point we lost = Amount of points $x$ lines could have contributed = $\binom{x}{2}$<br>↵
So the final answer becomes <br>↵
$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$↵
</spoiler>↵
↵
#### 3↵
The sum of all positive integers $m$ for which $\tfrac{13!}{m}$ is a perfect square can be written as $2^{a}3^{b}5^{c}7^{d}11^{e}13^{f}$, where $a, b, c, d, e,$ and $f$ are positive integers. Find $a+b+c+d+e+f$.↵
↵
<spoiler summary="Answer">↵
12↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
Power of $x$ in $n!$ = $\lfloor \frac{n}{x} \rfloor + \lfloor \frac{n}{x^{2}} \rfloor+ \lfloor \frac{n}{x^3{}} \rfloor + ...$↵
Using this formula we get..↵
$13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
For a number to be perfect square it prime factorisation should consist of even powers.<br>↵
Let's write it even and odd powers seperately↵
$13! = 2^{10} \cdot 3^{4} \cdot 5^{2} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$ <br>↵
Now we want sum of all numbers↵
$2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
such that $0 \leq x \leq 10 $ and $0 \leq y \leq 4$ and $0 \leq z \leq 2$ and $x, y, z$ are even<br>↵
The resultant number will be $\sum\limits_{x} \sum\limits_{y} \sum\limits_{z}2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
which is <br>↵
$(\sum\limits_{x} (2^{x}))\cdot (\sum\limits_{y} (3^{y})) \cdot (\sum\limits_{z} (5^{z})) \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
This can be computed using GP series.<br>↵
final answer = $ 2^{1} \cdot 3^{2} \cdot 5^{1} \cdot 7^{3} \cdot 11^{1} \cdot 13^{4}$↵
</spoiler>↵
↵
↵
#### 4↵
There exists a unique positive integer $a$ for which the sum $\[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\]$ is an integer strictly between $-1000$ and $1000$. For that unique $a$, find $a+U$.↵
↵
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)<br>↵
↵
<spoiler summary="Answer">↵
944↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
[AOPS solution link](https://artofproblemsolving.com/community/c5h3011232p27049499)↵
</spoiler>↵
↵
#### 5↵
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.↵
↵
↵
<spoiler summary="Answer">↵
$\frac{(n+1)^{2}}{2}$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
[AOPS solution link](https://artofproblemsolving.com/community/c5h3038308p27349768)↵
</spoiler>↵
↵
<br>↵
<br>↵
<br>↵
↵
↵
<spoiler summary="What is Misa-Math">↵
Math problems that aim to aid competitive programming skills.<br>↵
Target audience : Experts and below.↵
</spoiler>↵
↵
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ ↵
↵
<spoiler summary="Answer">↵
191↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
Total Ways = $\binom{14}{5}$. Out of $14$ places choose $5$ places for men. Women will follow.<br>↵
There are total $7$ pairs of seats. Choose $5$ out of them. For each pair of seats there are $2$ ways.<br>↵
So required ways = $\binom{7}{5} \cdot 2^5$<br>↵
Required Probability = $\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$.↵
</spoiler>↵
↵
#### 2↵
A plane contains $40$ lines, no $2$ of which are parallel. Suppose there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.↵
↵
<spoiler summary="Answer">↵
607↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
Two lines can only intersect once.<br>↵
Maximum number of intersections = $\binom{n}{2}$ (Choose any two lines and they intersect)<br>↵
Maximum number of intersections occur when for each intersection point there are only two lines intersecting at that point. Lets label these $T2$ intersection points.<br>↵
If there are intersection points where there are $x$ ($x$ > $2$) lines intersecting at some point, then we will lose $T2$ points.<br>↵
Amount of point we lost = Amount of points $x$ lines could have contributed = $\binom{x}{2}$<br>↵
So the final answer becomes <br>↵
$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$↵
</spoiler>↵
↵
#### 3↵
The sum of all positive integers $m$ for which $\tfrac{13!}{m}$ is a perfect square can be written as $2^{a}3^{b}5^{c}7^{d}11^{e}13^{f}$, where $a, b, c, d, e,$ and $f$ are positive integers. Find $a+b+c+d+e+f$.↵
↵
<spoiler summary="Answer">↵
12↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
Power of $x$ in $n!$ = $\lfloor \frac{n}{x} \rfloor + \lfloor \frac{n}{x^{2}} \rfloor+ \lfloor \frac{n}{x^3{}} \rfloor + ...$↵
Using this formula we get..↵
$13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
For a number to be perfect square it prime factorisation should consist of even powers.<br>↵
Let's write it even and odd powers seperately↵
$13! = 2^{10} \cdot 3^{4} \cdot 5^{2} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$ <br>↵
Now we want sum of all numbers↵
$2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
such that $0 \leq x \leq 10 $ and $0 \leq y \leq 4$ and $0 \leq z \leq 2$ and $x, y, z$ are even<br>↵
The resultant number will be $\sum\limits_{x} \sum\limits_{y} \sum\limits_{z}2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
which is <br>↵
$(\sum\limits_{x} (2^{x}))\cdot (\sum\limits_{y} (3^{y})) \cdot (\sum\limits_{z} (5^{z})) \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$<br>↵
This can be computed using GP series.<br>↵
final answer = $ 2^{1} \cdot 3^{2} \cdot 5^{1} \cdot 7^{3} \cdot 11^{1} \cdot 13^{4}$↵
</spoiler>↵
↵
↵
#### 4↵
There exists a unique positive integer $a$ for which the sum $\[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\]$ is an integer strictly between $-1000$ and $1000$. For that unique $a$, find $a+U$.↵
↵
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)<br>↵
↵
<spoiler summary="Answer">↵
944↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
[AOPS solution link](https://artofproblemsolving.com/community/c5h3011232p27049499)↵
</spoiler>↵
↵
#### 5↵
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.↵
↵
↵
<spoiler summary="Answer">↵
$\frac{(n+1)^{2}}{2}$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
[AOPS solution link](https://artofproblemsolving.com/community/c5h3038308p27349768)↵
</spoiler>↵
↵
<br>↵
<br>↵
<br>↵
↵
↵
<spoiler summary="What is Misa-Math">↵
Math problems that aim to aid competitive programming skills.<br>↵
Target audience : Experts and below.↵
</spoiler>↵
↵
