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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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
May 1, 2025
0 replies
f(a + b) = f(a) + f(b) + f(c) + f(d) in N-{O}, with 2ab = c^2 + d^2
parmenides51   8
N 32 minutes ago by TiagoCavalcante
Source: RMM Shortlist 2016 A1
Determine all functions $f$ from the set of non-negative integers to itself such that $f(a + b) = f(a) + f(b) + f(c) + f(d)$, whenever $a, b, c, d$, are non-negative integers satisfying $2ab = c^2 + d^2$.
8 replies
parmenides51
Jul 4, 2019
TiagoCavalcante
32 minutes ago
Functional Inequality Implies Uniform Sign
peace09   33
N an hour ago by ezpotd
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
33 replies
peace09
Jul 17, 2024
ezpotd
an hour ago
Labelling edges of Kn
oVlad   1
N 2 hours ago by TopGbulliedU
Source: Romania Junior TST 2025 Day 2 P3
Let $n\geqslant 3$ be an integer. Ion draws a regular $n$-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the $n$-gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
1 reply
oVlad
May 6, 2025
TopGbulliedU
2 hours ago
c^a + a = 2^b
Havu   8
N 2 hours ago by MathematicalArceus
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
8 replies
Havu
May 10, 2025
MathematicalArceus
2 hours ago
No more topics!
floor of (an+b)/(cn+d) is surjective
Miquel-point   3
N Apr 11, 2025 by Rohit-2006
Source: Romanian NMO 2021 grade 10 P2
Let $a,b,c,d\in\mathbb{Z}_{\ge 0}$, $d\ne 0$ and the function $f:\mathbb{Z}_{\ge 0}\to\mathbb Z_{\ge 0}$ defined by
\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]Prove that the following are equivalent:
[list=1]
[*] $f$ is surjective;
[*] $c=0$, $b<d$ and $0<a\le d$.
[/list]

Tiberiu Trif
3 replies
Miquel-point
Apr 15, 2023
Rohit-2006
Apr 11, 2025
floor of (an+b)/(cn+d) is surjective
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G H BBookmark kLocked kLocked NReply
Source: Romanian NMO 2021 grade 10 P2
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Miquel-point
499 posts
#1
Y by
Let $a,b,c,d\in\mathbb{Z}_{\ge 0}$, $d\ne 0$ and the function $f:\mathbb{Z}_{\ge 0}\to\mathbb Z_{\ge 0}$ defined by
\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]Prove that the following are equivalent:
  1. $f$ is surjective;
  2. $c=0$, $b<d$ and $0<a\le d$.

Tiberiu Trif
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Rohit-2006
245 posts
#2
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One side proved....doing the other side
Attachments:
This post has been edited 1 time. Last edited by Rohit-2006, Apr 11, 2025, 4:14 PM
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Rohit-2006
245 posts
#3
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I guess there is a little flaw in the third contradiction....
Suppose $a>d$, say {$\frac{a}{d}$}=$0.d_1d_2\cdots$
Hence there exists a $N$ such that $N\times0.d_1d_2\cdots>1$. We can make $\frac{b}{d}$ sufficiently small such that $\frac{b}{d}+0.d_1d_2\cdots<<1$ so that no integer is skipped. So in case of that $N$, $f(N)-f(N-1)\geq2$ and done!
Z K Y
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Rohit-2006
245 posts
#4
Y by
For the remaining part, we have $c=0$,$b<d$ and $0<a\leq d$. Putting into $f(n)$ we get,
$$f(n)=\lfloor\frac{an+b}{d}\rfloor$$I will do this by induction,
So base case is $n=0$ we have $\lfloor\frac{b}{d}\rfloor=0$ since $b<d$.
Suppose we can get all non negative integers uptil $k$ which attains at $n=m$, where $m$ is the maximum value for which $f(m)=k$.
For $n=m+1$,
$$f(m+1)=\lfloor\frac{am+b}{d}+\frac{1}{d}\rfloor$$Now $\frac{1}{d}\leq1$. For equality case it's too trivial. For inequality, $\lfloor\frac{am+b}{d}+\frac{1}{d}\rfloor=k+1$ since $m$ was maximum and $\frac{1}{d}\leq1$. And the induction is complete.
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