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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Geo challenge on finding simple ways to solve it
Assassino9931   4
N 20 minutes ago by iv999xyz
Source: Bulgaria Spring Mathematical Competition 2025 9.2
Let $ABC$ be an acute scalene triangle inscribed in a circle \( \Gamma \). The angle bisector of \( \angle BAC \) intersects \( BC \) at \( L \) and \( \Gamma \) at \( S \). The point \( M \) is the midpoint of \( AL \). Let \( AD \) be the altitude in \( \triangle ABC \), and the circumcircle of \( \triangle DSL \) intersects \( \Gamma \) again at \( P \). Let \( N \) be the midpoint of \( BC \), and let \( K \) be the reflection of \( D \) with respect to \( N \). Prove that the triangles \( \triangle MPS \) and \( \triangle ADK \) are similar.
4 replies
Assassino9931
Mar 30, 2025
iv999xyz
20 minutes ago
Nested function expression for positive integers
Equinox8   3
N an hour ago by AshAuktober
Source: IrMO 2024 #10
Let $\mathbb{Z}_+=\{1,2,3,4...\}$ be the set of all positive integers. Find, with proof, all functions $f : \mathbb{Z}_+ \mapsto \mathbb{Z}_+$ with the property that $$f(x+f(y)+f(f(z)))=z+f(y)+f(f(x))$$for all positive integers $x,y,z$.
3 replies
Equinox8
Feb 18, 2025
AshAuktober
an hour ago
Totally normal inequality
giangtruong13   10
N an hour ago by Mathzeus1024
Let $a,b,c>0$ and $a^2+b^2+c^2+2ab=3(a+b+c)$. Find the minimum value:$$P=a+b+c+\frac{20}{\sqrt{a+c}}+\frac{20}{\sqrt{b+2}}$$
10 replies
giangtruong13
Feb 13, 2025
Mathzeus1024
an hour ago
An almost identity polynomial
nAalniaOMliO   5
N an hour ago by AshAuktober
Source: Belarusian National Olympiad 2025
Let $n$ be a positive integer and $P(x)$ be a polynomial with integer coefficients such that $P(1)=1,P(2)=2,\ldots,P(n)=n$.
Prove that $P(0)$ is divisible by $2 \cdot 3 \cdot \ldots \cdot n$.
5 replies
nAalniaOMliO
Mar 28, 2025
AshAuktober
an hour ago
Inequalities
sqing   0
Today at 3:53 AM
Let $ a,b,c,d\geq 0 ,a-b+d=21 $ and $ a+3b+4c=101 $. Prove that
$$ - \frac{1681}{3}\leq   ab - cd \leq 820$$$$ - \frac{16564}{9}\leq   ac -bd \leq 420$$$$ - \frac{10201}{48}\leq ad- bc \leq\frac{1681}{3}$$
0 replies
sqing
Today at 3:53 AM
0 replies
law of log
Miranda2829   18
N Today at 1:53 AM by RandomMathGuy500
5log (5²) + 8 ˡºᵍ₈4 =

is this answer 6?
18 replies
Miranda2829
Yesterday at 2:12 AM
RandomMathGuy500
Today at 1:53 AM
Hard number theory
td12345   7
N Yesterday at 9:29 PM by td12345
Let $q$ be a prime number. Define the set
\[
M_q = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2q^{2025} x} \in \mathbb{Q} \right\}. 
\]
Find the number of elements of \(M_2  \cup M_{2027}\).
7 replies
td12345
Wednesday at 11:32 PM
td12345
Yesterday at 9:29 PM
Pythagorean triples vs sine ratio?
Miranda2829   6
N Yesterday at 8:45 PM by anticodon
I'm a bit confused about the

right angle 3 4 5 have a sine ratio of 0.6 and cosine of 0.8,

Do different lengths of right-angle triangles have different ratios?

how to get an actual angle of sine ?

thanks

6 replies
Miranda2829
Feb 27, 2025
anticodon
Yesterday at 8:45 PM
Plane geometry problem with inequalities
ReticulatedPython   1
N Yesterday at 7:50 PM by soryn
Let $A$ and $B$ be points on a plane such that $AB=1.$ Let $P$ be a point on that plane such that $$\frac{AP^2+BP^2}{(AP)(BP)}=3.$$Prove that $$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$
Source: Own
1 reply
ReticulatedPython
Yesterday at 3:59 PM
soryn
Yesterday at 7:50 PM
Sequences and Series
SomeonecoolLovesMaths   4
N Yesterday at 7:49 PM by Alex-131
Prove that $x_n = \frac{1}{\sqrt{3} + 1} + \frac{1}{ \sqrt{7} + \sqrt{5}} + \cdots ( \text{ up to n terms })$ is bounded.

My Progress
4 replies
SomeonecoolLovesMaths
Yesterday at 3:36 PM
Alex-131
Yesterday at 7:49 PM
lcm(1,2,3,...,n)
lgx57   4
N Yesterday at 7:14 PM by td12345
Let $M=\operatorname{lcm}(1,2,3,\cdots,n)$.Estimate the range of $M$.
4 replies
lgx57
Apr 9, 2025
td12345
Yesterday at 7:14 PM
Challenging Trigonometric Sums - AoPS Volume 2 Problem 277
Shiyul   5
N Yesterday at 7:06 PM by vanstraelen
Problem #277 (Source: Mu Alpha Theta 1992)

Find $\color[rgb]{0.35,0.35,0.35}\displaystyle\sum_{n=0}^\infty\frac{\sin (nx)}{3^n}$ if $\color[rgb]{0.35,0.35,0.35}\sin x=1/3$ and $\color[rgb]{0.35,0.35,0.35} 0\le x\le \pi/2$.

I know what cosine of x is also positive because of the value of x. I've also tried to see if the value of sin(nx) ever repeats, but it doesn't. Can anyone give me a hint (not the full solution) on how to start on solving this problem? Thank you.
5 replies
Shiyul
Yesterday at 4:44 AM
vanstraelen
Yesterday at 7:06 PM
JEE Related ig?
mikkymini2   1
N Yesterday at 3:49 PM by SomeonecoolLovesMaths
Hey everyone,

Just wanted to see if there are any other JEE aspirants on this forum currently prepping for it[mention year if you can]

I am actually entering 10th this year and have decided to try for it...So this year is just going to go in me strengthening my math (IOQM level (heard its enough till Mains part, so will start from there) for the problem solving part, and learn some topics from 11th and 12th as well)

It would be great to connect with others who are going through the same thing - share study strategies, tips, resources, discuss, and maybe even form study groups(not sure how to tho :maybe: ) and motivate each other ig?. :D
So yea, cya later
1 reply
mikkymini2
Yesterday at 2:54 PM
SomeonecoolLovesMaths
Yesterday at 3:49 PM
Classic Invariant
Mathdreams   1
N Yesterday at 2:24 PM by Lankou
Source: 2025 Nepal Mock TST Day 1 Problem 1

Prajit and Kritesh challenge each other with a marble game. In a bag, there are initially $2024$ red marbles and $2025$ blue marbles. The rules of the game are as follows:

Move: In each turn, a player (either Prajit or Kritesh) removes two marbles from the bag.

If the two marbles are of the same color, they are both discarded and a red marble is added to the bag.
If the two marbles are of different colors, they are both discarded and a blue marble is added to the bag.

The game continues by repeating the above move.

Prove that no matter what sequence of moves is made, the process always terminates with exactly one marble left. In addition, find the possible colors of the marble remaining.
1 reply
Mathdreams
Yesterday at 1:28 PM
Lankou
Yesterday at 2:24 PM
The last nonzero digit of factorials
Tintarn   4
N Apr 6, 2025 by Sadigly
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
4 replies
Tintarn
Mar 17, 2025
Sadigly
Apr 6, 2025
The last nonzero digit of factorials
G H J
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
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Tintarn
9031 posts
#1
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For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
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AshAuktober
981 posts
#2
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This digit has to be even from $\nu_p$ analysis, and thus it suffices to look at the last digit mod t. Then we can verify the sequence repeats, and this gives (I think?) All even numbers.
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VideoCake
9 posts
#3
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Unusually difficult for problem 2, unless I missed something. Here is a quick sketch of my solution:
Basically you first show that \(\nu_2(n!) > \nu_5(n!)\) for every integer \(n \ge 2\) to reduce it down to the digits 2, 4, 6, 8. Now, using Legendre and the fact that \(1 \cdot 2 \cdot 3 \cdot 4 \equiv -1 \pmod 5\) one can show that if \(n = 5^{\ell}\) for some positive integer \(\ell\), then
\[\frac{n!}{10^{\nu_5(n!)}} \equiv 2^{\ell} \pmod 5\]which is enough, as \(2, 4, 6, 8\) all have different and non-zero residues modulo 5.
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MyobDoesMath
51 posts
#4
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My initial solution: Let $a_n$ be the given sequences. Use $\nu_2(n!)>\nu_5(n!)$ as above to show that all members of the sequence must be even.

Now as a motivation for the construction, one can see that in $16!$, $17!$, ... up to $24!$ all even digits appear once (you can find this if you simply start to calculate the first couple terms of the sequence). When $n$ is not divisible by $5$, it is easy to see that $a_n \equiv n \cdot a_{n-1} \mod 10$. However, dealing with $5 \mid n$ is not so easy. In the example above we have $a_{20} \equiv 2 a_{19}$ as the ending $0$ does not change anything.
This motivates us to look at $(100k+16)!$ up to $(100k+24)!$ where we have $a_n \equiv n \cdot a_{n-1} \mod 10$ for $n \not=100k+20$ and $a_{100k+20} \equiv 2 a_{100k+19}$ by the same argument as before. Here, we can compute all $a_n$ if we know $a_{100k+16}$, and we can see that all even numbers appear in the sequence regardless of which even digit $a_{100k+16}$ is, as we have one of the sequences (if I didn't miscalculate):

$2$, $4$, $2$, $8$, $6$, $6$, $2$, $6$, $4$
$4$, $8$, $4$, $6$, $2$, $2$, $4$, $2$, $8$
$6$, $2$, $6$, $4$, $8$, $8$, $6$, $8$, $2$
$8$, $6$, $8$, $2$, $4$, $4$, $8$, $4$, $6$

(And of course, it is no surprise that all even digits appear in the latter sequences if they all appear in the first. So maybe as an addition to the motivation, it is also clear that this works for all $k$ if it works for $k=0$.)

As there are infinitely many $k \in \mathbb{N}$, this shows that $2,4,6,8$ all appear infinitely often.

Remark
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Sadigly
136 posts
#5
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Claim: Odd numbers can't appear

Proof: Let $n!=\overline{Am\underbrace{000..0}_{k~times}}$ where $m$ is an odd number

$$\frac{n!}{10^k}=\overline{Am}$$
Since $\overline{Am}\in\{1;3;5;7;9\}~(mod ~10)$, we have $10\nmid\frac{5\times n!}{10^k}$, which implies $v_2(n!)\leq v_5(n!)$, which is absurd.


Proving the other one is a pain in the butt. Let $\lfloor n!\rfloor$ denote the nonzero part of $n!$. It should be obvious that $(n+1)\lfloor n!\rfloor\equiv\lfloor (n+1)!\rfloor~(mod~10)$ This works perfectly until $5\mid n+1$. In that case,since $5=\frac{10}2$, we can say $\lfloor (n+1)!\rfloor=\lfloor \frac{n!}{2}\rfloor$ (Keep in note that $\lfloor \overline{blahblah8}\rfloor\in\{4;9\}~(mod~10)$, but since $9$ is odd, it can't appear). Using these, with enough patience , one can find a loop
This post has been edited 2 times. Last edited by Sadigly, Apr 7, 2025, 12:45 PM
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