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Sequences problem
BBNoDollar   1
N 16 minutes ago by ICE_CNME_4
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
4 hours ago
ICE_CNME_4
16 minutes ago
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Concurrency in Parallelogram
amuthup   89
N an hour ago by happypi31415
Source: 2021 ISL G1
Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.
89 replies
amuthup
Jul 12, 2022
happypi31415
an hour ago
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deleting multiple or divisor in pairs from 2-50 on a blackboard
parmenides51   1
N 2 hours ago by TheBaiano
Source: 2023 May Olympiad L2 p3
The $49$ numbers $2,3,4,...,49,50$ are written on the blackboard . An allowed operation consists of choosing two different numbers $a$ and $b$ of the blackboard such that $a$ is a multiple of $b$ and delete exactly one of the two. María performs a sequence of permitted operations until she observes that it is no longer possible to perform any more. Determine the minimum number of numbers that can remain on the board at that moment.
1 reply
parmenides51
Mar 24, 2024
TheBaiano
2 hours ago
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at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   9
N 2 hours ago by atdaotlohbh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
9 replies
NJAX
May 31, 2024
atdaotlohbh
2 hours ago
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Easy complete system of residues problem in Taiwan TST
Fysty   6
N 2 hours ago by Primeniyazidayi
Source: 2025 Taiwan TST Round 1 Independent Study 1-N
Find all positive integers $n$ such that there exist two permutations $a_0,a_1,\ldots,a_{n-1}$ and $b_0,b_1,\ldots,b_{n-1}$ of the set $\lbrace0,1,\ldots,n-1\rbrace$, satisfying the condition
$$ia_i\equiv b_i\pmod{n}$$for all $0\le i\le n-1$.

Proposed by Fysty
6 replies
Fysty
Mar 5, 2025
Primeniyazidayi
2 hours ago
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JBMO Shortlist 2022 A2
Lukaluce   13
N 3 hours ago by Rayvhs
Source: JBMO Shortlist 2022
Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that
$$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} + 3 \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$
Proposed by Petar Filipovski, Macedonia
13 replies
Lukaluce
Jun 26, 2023
Rayvhs
3 hours ago
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A very beautiful geo problem
TheMathBob   4
N 3 hours ago by ravengsd
Source: Polish MO Finals P2 2023
Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$Prove that $YI$ is the angle bisector of $XYA$.
4 replies
TheMathBob
Mar 29, 2023
ravengsd
3 hours ago
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Inspired by old results
sqing   6
N 3 hours ago by Jamalll
Source: Own
Let $ a,b>0 , a^2+b^2+ab+a+b=5 . $ Prove that
$$ \frac{ 1 }{a+b+ab+1}+\frac{6}{a^2+b^2+ab+1}\geq \frac{7}{4}$$$$ \frac{ 1 }{a+b+ab+1}+\frac{1}{a^2+b^2+ab+1}\geq \frac{1}{2}$$$$ \frac{41}{a+b+2}+\frac{ab}{a^3+b^3+2} \geq \frac{21}{2}$$
6 replies
sqing
Apr 29, 2025
Jamalll
3 hours ago
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A Duality Operation on Decreasing Integer Sequences
Ritangshu   0
3 hours ago
Let \( S \) be the set of all sequences \( (a_1, a_2, \ldots) \) of non-negative integers such that
(i) \( a_1 \geq a_2 \geq \cdots \); and
(ii) there exists a positive integer \( N \) such that \( a_n = 0 \) for all \( n \geq N \).

Define the dual of the sequence \( (a_1, a_2, \ldots) \in S \) to be the sequence \( (b_1, b_2, \ldots) \), where, for \( m \geq 1 \),
\( b_m \) is the number of \( a_n \)'s which are greater than or equal to \( m \).

(i) Show that the dual of a sequence in \( S \) belongs to \( S \).

(ii) Show that the dual of the dual of a sequence in \( S \) is the original sequence itself.

(iii) Show that the duals of distinct sequences in \( S \) are distinct.
0 replies
Ritangshu
3 hours ago
0 replies
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Property of a function
Ritangshu   0
3 hours ago
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[ f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\} \]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

0 replies
Ritangshu
3 hours ago
0 replies
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Subset of digits to express as a sum
anantmudgal09   46
N 4 hours ago by anudeep
Source: INMO 2020 P3
Let $S$ be a subset of $\{0,1,2,\dots ,9\}$. Suppose there is a positive integer $N$ such that for any integer $n>N$, one can find positive integers $a,b$ so that $n=a+b$ and all the digits in the decimal representations of $a,b$ (expressed without leading zeros) are in $S$. Find the smallest possible value of $|S|$.

Proposed by Sutanay Bhattacharya

Original Wording
46 replies
anantmudgal09
Jan 19, 2020
anudeep
4 hours ago
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Interesting Function
Kei0923   4
N Apr 30, 2025 by CrazyInMath
Source: 2024 JMO preliminary p8
Function $f:\mathbb{Z}_{\geq 0}\rightarrow\mathbb{Z}$ satisfies
$$f(m+n)^2=f(m|f(n)|)+f(n^2)$$for any non-negative integers $m$ and $n$. Determine the number of possible sets of integers $\{f(0), f(1), \dots, f(2024)\}$.
4 replies
Kei0923
Jan 9, 2024
CrazyInMath
Apr 30, 2025
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Interesting Function
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Source: 2024 JMO preliminary p8
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Kei0923
95 posts
Kei0923
#1 h  Jan 9, 2024, 7:36 AM
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Function $f:\mathbb{Z}_{\geq 0}\rightarrow\mathbb{Z}$ satisfies
$$f(m+n)^2=f(m|f(n)|)+f(n^2)$$for any non-negative integers $m$ and $n$. Determine the number of possible sets of integers $\{f(0), f(1), \dots, f(2024)\}$.
This post has been edited 1 time. Last edited by Kei0923, Jan 9, 2024, 8:41 AM
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nguyenloc1712
62 posts
nguyenloc1712
#3 h  Jan 10, 2024, 5:37 AM
Y by
answer :maybe:
sol
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sami1618
902 posts
sami1618
#5 h  May 2, 2024, 12:46 AM
Y by
The answer is $\boxed{992}$. Let $P(m,n)$ be the given assertion. Notice $P(0,0)$ gives $f(0)^2=2f(0)$ so either $f(0)=0$ or $f(0)=2$.

Case 1: $f(0)=0$
The assertion $P(m,0)$ gives $f(m)^2=0\Rightarrow f(m)=0$. Which can be checked to work.

Case 2: $f(0)=2$
The assertion $P(0,n)$ gives $f(n)^2=2+f(n^2)$. The assertion $P(m,0)$ gives $f(m)^2=f(2m)+2$. Combined we get $f(n^2)=f(2n)$. The assertion $P(0,1)$ gives $f(1)^2=2+f(1)$ so either $f(1)=2$ or $f(1)=-1$. If $f(1)=-1$ then $f(2)=-1$ and $P(1,1)$ gives a contradiction. Now assume $f(1)=2\Rightarrow f(2)=2$. We get from $P(m,1)$ that $f(m+1)^2=f(2m)+2\Rightarrow f(2m+2)=f(2m)\Rightarrow$ $f(2m)=2\Rightarrow f(m^2)=2$. Thus we get $f(n)^2=4\Rightarrow f(n)=\pm 2$. Thus $f(x)=2$ whenever $x$ is even or a square and $f(x)$ can be either $2$ or $-2$ when $x$ is not even or a square. It is easy to check that all such functions work.
This post has been edited 1 time. Last edited by sami1618, May 2, 2024, 11:47 AM
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JerryXi
2 posts
JerryXi
#8 h  Oct 13, 2024, 8:01 AM
Y by
I think the correct answer should be 991, because the number of possible sets in Case 2 should be 990. It seems like you may have made a small counting mistake. The number of integers from 3 to 2024 that are neither even nor perfect squares is 990.
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CrazyInMath
457 posts
CrazyInMath
#9 h  Apr 30, 2025, 3:54 AM
Y by
$P(0, 0)\Longrightarrow f(0)=0, 2$
if $f(0)=0$, then $P(m, 0)\Longrightarrow f(m)^2=0\Longrightarrow f(m)=0$

Let $f(0)=2$ from now on
$P(0, 1)\Longrightarrow f(1)^2=2+f(1)\Longrightarrow f(1)=-1, 2$
if $f(1)=-1$ then $P(1, 1)\Longrightarrow f(2)^2=-2$, contradiction, so $f(1)=2$
$P(m, 1)\Longrightarrow f(m+1)^2=f(2m)+2$, $P(m, 0)\Longrightarrow f(m)^2=f(2m)+2$, so $f(m+1)=\pm f(m)$, and so $f(m)=\pm2$
and so the original FE is now $4=f(2m)+f(n^2)$, which means even numbers and perfect squares must have $f(a)=2$, the others can have $f(a)=2, -2$

The number of sets would be just $2$ as sets don't count duplicates. If it is multiset then the answer is $991$.
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