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 $$$m+n.$$$
SolutionTotal Ways = $$$\binom{14}{5}$$$. Out of $$$14$$$ places choose $$$5$$$ places for men. Women will follow.
There are total $$$7$$$ pairs of seats. Choose $$$5$$$ out of them. For each pair of seats there are $$$2$$$ ways.
So required ways = $$$\binom{7}{5} \cdot 2^5$$$
Required Probability = $$$\frac{\binom{7}{5} \cdot 2^5}{\binom{14}{5}}= \frac{48}{143}$$$.
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.
SolutionTwo lines can only intersect once.
Maximum number of intersections = $$$\binom{n}{2}$$$ (Choose any two lines and they intersect)
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.
If there are intersection points where there are $$$x$$$ ($$$x$$$ > $$$2$$$) lines intersecting at some point, then we will lose $$$T2$$$ points.
Amount of point we lost = Amount of points $$$x$$$ lines could have contributed = $$$\binom{x}{2}$$$
So the final answer becomes
$$$\binom{40}{2} - 3\binom{3}{2} - 4\binom{4}{2}- 5\binom{5}{2}- 6\binom{6}{2} = \boxed{607}.$$$
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 $$$a+b+c+d+e+f$$$.
SolutionPower 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.. $$$13! = 2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
For a number to be perfect square it prime factorisation should consist of even powers.
Let's write it even and odd powers seperately $$$13! = 2^{10} \cdot 3^{4} \cdot 5^{2} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
Now we want sum of all numbers $$$2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
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
The resultant number will be $$$\sum\limits_{x} \sum\limits_{y} \sum\limits_{z}2^{x} \cdot 3^{y} \cdot 5^{z} \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
which is
$$$(\sum\limits_{x} (2^{x}))\cdot (\sum\limits_{y} (3^{y})) \cdot (\sum\limits_{z} (5^{z})) \cdot 3^{1} \cdot 7^{1} \cdot 11^{1} \cdot 13^{1}$$$
This can be computed using GP series.
final answer = $$$ 2^{1} \cdot 3^{2} \cdot 5^{1} \cdot 7^{3} \cdot 11^{1} \cdot 13^{4}$$$
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 $$$x$$$.)
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 $$$n$$$.
Answer$$$\frac{(n+1)^{2}}{4}$$$ , The Optimal Solution would be a spiral (first along the boundaries then curled inside).
Sources2023 AIME and 2023 USAJMO
What is Misa-MathMath problems that aim to aid competitive programming skills.
Target audience : Experts and below.
