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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
Consecutive sum of integers sum up to 2020
NicoN9   2
N a few seconds ago by NicoN9
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
2 replies
NicoN9
6 hours ago
NicoN9
a few seconds ago
Range of a^3+b^3-3c
Kunihiko_Chikaya   1
N 2 minutes ago by Mathzeus1024
Let $a,\ b,\ c$ be real numbers such that $b<\frac{1}{c}<a$ and

$$\begin{cases}a+b+c=1 \ \\ a^2+b^2+c^2=23	

\end{cases}$$
Find the range of $a^3+b^3-3c.$


Proposed by Kunihiko Chikaya/September 23, 2020
1 reply
Kunihiko_Chikaya
Sep 23, 2020
Mathzeus1024
2 minutes ago
equations
kjhgyuio   1
N 9 minutes ago by mashumaro
........
1 reply
kjhgyuio
13 minutes ago
mashumaro
9 minutes ago
Function equation
LeDuonggg   4
N 19 minutes ago by mashumaro
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
4 replies
LeDuonggg
Yesterday at 2:59 PM
mashumaro
19 minutes ago
No more topics!
Outcome related combinatorics problem
egxa   1
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.7
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
1 reply
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
Outcome related combinatorics problem
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Source: All Russian 2025 10.7
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egxa
209 posts
#1
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A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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iliya8788
8 posts
#2
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We claim that the maximum $k$ is equal to $24$.
We correspond each athlete to a number from $1$ to $25$.
Define Podium as the $n$-tuple $(a_{1},a_{2},...,a_{25})$ where $a_{i}$ is the amount of medals won by athlete $i$.
First notice that none of the experts can predict the $n$-tuple $(1,1,...,1)$ because then if the podium is equal to $(25,0,0,...,0)$ then the most possible amount of judges predicting correctly is equal to $24$. From the previous statement it follows that one of the judges predicts a $n$-tuple which has a $0$. So we proceed with the following algorithm. Pick one of the prediction $n$-tuples and convert it with the following algorithm to a new $n$-tuple: turn anything that isn't equal to $0$ to $0$ and then spread the 25 medals among the zero's.
Now if the podium $n$-tuple is the same as this $n$-tuple then at least one of the experts isn't competent and so maximum $k$ is equal to $24$.
Now we will prove that no matter what the podium looks like we can achieve $k=24$.
So we propose the following predictions: $(24,1,0,0,...,0),(24,0,1,0,0,...,0),(24,0,0,1,0,0,...0),...,(24,0,0,...,1),(1,1,...,1,1,1)$.
Define the count of zeros of the podium as $C$.
Now 3 cases might happen:
$C>2$: then it's easy to notice that every expert except the last one is competent.
$C=2$: an expert isn't competent if and only if both zeros are in the same place as the $1$ and $24$. notice that no $2$ of the predictions have these 2 numbers in the same spot so worst case possible $24$ of the experts are competent.
$C=1$: notice that the podium $n$-tuple must have $23$, $1$'s and a single $2$ and a single $0$. an expert isn't competent if and only if one of the following 2 cases occur:
First case: the $1$ in the prediction is in the same place as the $0$ of the podium.
Second case: the $24$ in the prediction is in the same place as the $0$ of the podium and the $1$ in the prediction is in the same place as the $2$ of the podium.
notice that the position of the $1$ differs among the predictions so for a fixed podium only one expert can be not competent because of the first case. The same applies for the second case as well. Since the union of the positions of $1$ and $24$ differs among the predictions there can't be 2 experts that are incompetent because of the second case. Also both cases can't make experts incompetent simultaneously since in case of both cases happening the $1$ and $24$ must basically switch places but in each of the first $24$ predictions $24$ is not in the first spot and all of the $1$'s are in the first spot. So only one of these $24$ experts can be incompetent and the last expert is competent so at least $24$ of the experts are competent. $\blacksquare$
This post has been edited 5 times. Last edited by iliya8788, Apr 29, 2025, 12:11 PM
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