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.
Auto comment: topic has been updated by Misa-Misa (previous revision, new revision, compare).
When posting problems you did not create, you should credit the original source. (The first four problems come from the 2023 AIME I and the last comes from the 2023 USAJMO.)
Ok i will do that, I starting practising some maths after reading your blog. Thought it would be good idea to post problems here, for others and my own reference too.
Makes sense, good luck with your training!
Hi