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Function on positive integers with two inputs
Assassino9931   2
N 2 minutes ago by Assassino9931
Source: Bulgaria Winter Competition 2025 Problem 10.4
The function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ is such that $f(a,b) + f(b,c) = f(ac, b^2) + 1$ for any positive integers $a,b,c$. Assume there exists a positive integer $n$ such that $f(n, m) \leq f(n, m + 1)$ for all positive integers $m$. Determine all possible values of $f(2025, 2025)$.
2 replies
Assassino9931
Jan 27, 2025
Assassino9931
2 minutes ago
Tiling rectangle with smaller rectangles.
MarkBcc168   61
N 30 minutes ago by YaoAOPS
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
61 replies
MarkBcc168
Jul 10, 2018
YaoAOPS
30 minutes ago
A magician has one hundred cards numbered 1 to 100
Valentin Vornicu   49
N 35 minutes ago by YaoAOPS
Source: IMO 2000, Problem 4, IMO Shortlist 2000, C1
A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn.

How many ways are there to put the cards in the three boxes so that the trick works?
49 replies
+1 w
Valentin Vornicu
Oct 24, 2005
YaoAOPS
35 minutes ago
Nice inequality
sqing   2
N 39 minutes ago by Seungjun_Lee
Source: WYX
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be real numbers . Prove that : There exist positive integer $k\in \{1,2,\cdots,n\}$ such that $$\sum_{i=1}^{n}\{kx_i\}(1-\{kx_i\})<\frac{n-1}{6}.$$Where $\{x\}=x-\left \lfloor x \right \rfloor.$
2 replies
sqing
Apr 24, 2019
Seungjun_Lee
39 minutes ago
Concurrency
Dadgarnia   27
N 41 minutes ago by zuat.e
Source: Iranian TST 2020, second exam day 2, problem 4
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
27 replies
Dadgarnia
Mar 12, 2020
zuat.e
41 minutes ago
nice geo
Melid   1
N an hour ago by Melid
Source: 2025 Japan Junior MO preliminary P9
Let ABCD be a cyclic quadrilateral, which is AB=7 and BC=6. Let E be a point on segment CD so that BE=9. Line BE and AD intersect at F. Suppose that A, D, and F lie in order. If AF=11 and DF:DE=7:6, find the length of segment CD.
1 reply
Melid
an hour ago
Melid
an hour ago
product of all integers of form i^3+1 is a perfect square
AlastorMoody   3
N an hour ago by Assassino9931
Source: Balkan MO ShortList 2009 N3
Determine all integers $1 \le m, 1 \le n \le 2009$, for which
\begin{align*} \prod_{i=1}^n \left( i^3 +1 \right) = m^2 \end{align*}
3 replies
AlastorMoody
Apr 6, 2020
Assassino9931
an hour ago
IMO ShortList 1998, combinatorics theory problem 1
orl   44
N an hour ago by YaoAOPS
Source: IMO ShortList 1998, combinatorics theory problem 1
A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)
44 replies
orl
Oct 22, 2004
YaoAOPS
an hour ago
A game optimization on a graph
Assassino9931   2
N an hour ago by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[
  \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
  \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bobby has a winning strategy.
2 replies
Assassino9931
Apr 8, 2025
dgrozev
an hour ago
Composite sum
rohitsingh0812   39
N an hour ago by YaoAOPS
Source: INDIA IMOTC-2006 TST1 PROBLEM-2; IMO Shortlist 2005 problem N3
Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers and let $ S = a+b+c+d+e+f$.
Suppose that the number $ S$ divides $ abc+def$ and $ ab+bc+ca-de-ef-df$. Prove that $ S$ is composite.
39 replies
rohitsingh0812
Jun 3, 2006
YaoAOPS
an hour ago
Problem 1
SpectralS   145
N an hour ago by IndexLibrorumProhibitorum
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
145 replies
SpectralS
Jul 10, 2012
IndexLibrorumProhibitorum
an hour ago
beautiful functional equation problem
Medjl   6
N Apr 6, 2025 by Sadigly
Source: Netherlands TST for BxMO 2017 problem 2
Let define a function $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that :
$i)$$f(p)=1$ for all prime numbers $p$.
$ii)$$f(xy)=xf(y)+yf(x)$ for all positive integers $x,y$
find the smallest $n \geq 2016$ such that $f(n)=n$
6 replies
Medjl
Feb 1, 2018
Sadigly
Apr 6, 2025
beautiful functional equation problem
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G H BBookmark kLocked kLocked NReply
Source: Netherlands TST for BxMO 2017 problem 2
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Medjl
757 posts
#1 • 4 Y
Y by Muradjl, Adventure10, Mango247, PikaPika999
Let define a function $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that :
$i)$$f(p)=1$ for all prime numbers $p$.
$ii)$$f(xy)=xf(y)+yf(x)$ for all positive integers $x,y$
find the smallest $n \geq 2016$ such that $f(n)=n$
This post has been edited 2 times. Last edited by Medjl, Feb 1, 2018, 3:16 PM
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Math-Ninja
3749 posts
#2 • 2 Y
Y by Adventure10, PikaPika999
Why is this named beautiful problem. Titles are meant for description please.
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ThE-dArK-lOrD
4071 posts
#3 • 7 Y
Y by rkm0959, Muradjl, Math-Ninja, Com10atorics, Adventure10, Mango247, PikaPika999
Not hard to prove by induction on $s\in \mathbb{Z}^+$ that $$f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$$where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ with $\sum_{i=1}^{k}{\alpha_i} =s$.
Hence, $f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$ where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ is true for all positive integer $n$.
We want to find $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ that $f(n)=n\iff \sum_{i=1}^{k}{\frac{\alpha_i }{p_i}} =1$.

If $k\geq 2$, we get that $\alpha_i \leq p_i$ for all $i\in \{ 1,2,...,k\}$.
But we've $\sum_{i=1}^{k}{\alpha_i \times \frac{P}{p_i} }=P$ where $P=\prod_{i=1}^{k}{p_i}$. Modulo $p_1$ gives us $p_1\mid \alpha_1 \times \frac{P}{p_1}$, impossible since all $p_i$s are distinct.

So, $k=1$. This gives $n=p^p$ for some prime number $p$.
Easy to see that the smallest such $n$ that not smaller than $2016$ is $3125=5^5$.
This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Feb 1, 2018, 4:58 PM
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soryn
5331 posts
#4 • 2 Y
Y by Adventure10, PikaPika999
Very nice! I like this problem, I like this solution! Congratulations!:))))))
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User335559
472 posts
#5 • 3 Y
Y by Adventure10, Mango247, PikaPika999
ThE-dArK-lOrD wrote:
Not hard to prove by induction on $s\in \mathbb{Z}^+$ that $$f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$$where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ with $\sum_{i=1}^{k}{\alpha_i} =s$.
Hence, $f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$ where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ is true for all positive integer $n$.
We want to find $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ that $f(n)=n\iff \sum_{i=1}^{k}{\frac{\alpha_i }{p_i}} =1$.

If $k\geq 2$, we get that $\alpha_i \leq p_i$ for all $i\in \{ 1,2,...,k\}$.
But we've $\sum_{i=1}^{k}{\alpha_i \times \frac{P}{p_i} }=P$ where $P=\prod_{i=1}^{k}{p_i}$. Modulo $p_1$ gives us $p_1\mid \alpha_1 \times \frac{P}{p_1}$, impossible since all $p_i$s are distinct.

So, $k=1$. This gives $n=p^p$ for some prime number $p$.
Easy to see that the smallest such $n$ that not smaller than $2016$ is $3125=5^5$.

How did you find the function? What's the motivation behind that?
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liekkas
370 posts
#6 • 3 Y
Y by Adventure10, Mango247, PikaPika999
It's 2015 Austria MO (maybe not posted yet)
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Sadigly
147 posts
#7 • 1 Y
Y by PikaPika999
Easy to see that $f(0)=0$ and $f(1)=0$, and them being the only zeroes of this function. One can also say $f$ doesn't take any negative numbers.

Induct to get $f(p^x)=xp^{x-1}$

Let $n=p^x\cdot A$, where $p\nmid A$, and assume this number satisfies the given condition

$p^x\cdot A=n=f(n)=f(p^x\cdot A)=p^xf(A)+Af(p^x)=p^xf(A)+xAp^{x-1}$

$LHS\equiv 0~(mod~p^x)\Rightarrow RHS\equiv0~(mod~p^x).$ We get that $p\mid x$. Rewriting $x=p\cdot k$ gives $$(1-k)A=f(A)$$It should be obvious that $k=1$ and $f(A)=0$. Since $A\neq0$ (Or else $n=0$), we have $n=p^p$. All that is left to find the smallest prime that satisfies $p^p\geq 2016$, which is $5$
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