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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.

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[*]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
The return of an inequality
giangtruong13   4
N an hour ago by sqing
Let $a,b,c$ be real positive number satisfy that: $a+b+c=1$. Prove that: $$\sum_{cyc} \frac{a}{b^2+c^2} \geq \frac{3}{2}$$
4 replies
giangtruong13
Mar 18, 2025
sqing
an hour ago
Looking for users and developers
derekli   10
N an hour ago by Jackson0423
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
10 replies
derekli
Yesterday at 12:57 AM
Jackson0423
an hour ago
Polynomial
kellyelliee   1
N an hour ago by Jackson0423
Let the polynomial $f(x)=x^2+ax+b$, where $a,b$ integers and $k$ is a positive integer. Suppose that the integers
$m,n,p$ satisfy: $f(m), f(n), f(p)$ are divisible by k. Prove that:
$(m-n)(n-p)(p-m)$ is divisible by k
1 reply
kellyelliee
Today at 3:57 AM
Jackson0423
an hour ago
Sum of digits is 18
Ecrin_eren   14
N an hour ago by jestrada
How many 5 digit numbers are there such that sum of its digits is 18
14 replies
Ecrin_eren
May 3, 2025
jestrada
an hour ago
Cube Colouring Problems
Saucepan_man02   1
N 4 hours ago by removablesingularity
Could anyone kindly post some problems (and hopefully along the solution thread/final answer) related to combinatorial colouring of cube?
1 reply
Saucepan_man02
May 3, 2025
removablesingularity
4 hours ago
D1026 : An equivalent
Dattier   4
N Today at 4:56 AM by 3ch03s
Source: les dattes à Dattier
Let $u_0=1$ and $\forall n \in \mathbb N, u_{2n+1}=\ln(1+u_{2n}), u_{2n+2}=\sin(u_{2n+1})$.

Find an equivalent of $u_n$.
4 replies
Dattier
May 3, 2025
3ch03s
Today at 4:56 AM
Putnam 1947 B1
sqrtX   3
N Yesterday at 5:38 PM by Levieee
Source: Putnam 1947
Let $f(x)$ be a function such that $f(1)=1$ and for $x \geq 1$
$$f'(x)= \frac{1}{x^2 +f(x)^{2}}.$$Prove that
$$\lim_{x\to \infty} f(x)$$exists and is less than $1+ \frac{\pi}{4}.$
3 replies
sqrtX
Apr 3, 2022
Levieee
Yesterday at 5:38 PM
Double Sum
P162008   2
N Yesterday at 3:01 PM by Etkan
Evaluate $\sum_{a=1}^{\infty} \sum_{b=1}^{a + 2} \frac{1}{ab(a + 2)}.$
2 replies
P162008
Yesterday at 12:19 PM
Etkan
Yesterday at 3:01 PM
Summation
Saucepan_man02   4
N Yesterday at 2:47 PM by Etkan
If $P = \sum_{r=1}^{50} \sum_{k=1}^{r} (-1)^{r-1} \frac{\binom{50}{r}}{k}$, then find the value of $P$.

Ans
4 replies
Saucepan_man02
May 3, 2025
Etkan
Yesterday at 2:47 PM
Trigo + Series
P162008   0
Yesterday at 11:59 AM
$F(r,x) = \tan\left(\frac{\pi}{3} + 3^r x\right)$ and $G(r,x) = \cot\left(\frac{\pi}{6} + 3^r x\right)$

Let $A = \sum_{r=0}^{24} \frac{d(F(r,x))}{dx} \left[\frac{1}{F(r,x) + \frac{1}{F(r,x)}}\right]$

$B = \sum_{r=0}^{24} \frac{-d(G(r,x))}{dx} \left[\frac{1}{G(r,x) + \frac{1}{G(r,x)}}\right]$

$P = \prod_{r=0}^{24} F(r,x), Q = \prod_{r=0}^{24} G(r,x)$

and, $R = \frac{A - B}{\tan x\left[PQ - \frac{1}{3^{25}}\right]}$ where $x \in R$

Find the remainder when $R^{2025}$ is divided by $11.$
0 replies
P162008
Yesterday at 11:59 AM
0 replies
Integration Bee in Czechia
Assassino9931   2
N Yesterday at 9:35 AM by pi_quadrat_sechstel
Source: Vojtech Jarnik IMC 2025, Category II, P3
Evaluate the integral $\int_0^{\infty} \frac{\log(x+2)}{x^2+3x+2}\mathrm{d}x$.
2 replies
Assassino9931
May 2, 2025
pi_quadrat_sechstel
Yesterday at 9:35 AM
Alternating series and integral
jestrada   4
N Yesterday at 4:06 AM by bakkune
Source: own
Prove that for all $\alpha\in\mathbb{R}, \alpha>-1$, we have
$$ \frac{1}{\alpha+1}-\frac{1}{\alpha+2}+\frac{1}{\alpha+3}-\frac{1}{\alpha+4}+\cdots=\int_0^1 \frac{x^{\alpha}}{x+1}  \,dx. $$
4 replies
jestrada
Saturday at 10:56 PM
bakkune
Yesterday at 4:06 AM
Find all continuous functions
bakkune   3
N Yesterday at 3:58 AM by bakkune
Source: Own
Find all continuous function $f, g\colon\mathbb{R}\to\mathbb{R}$ satisfied
$$
(x - k)f(x) = \int_k^x g(y)\mathrm{d}y 
$$for all $x\in\mathbb{R}$ and all $k\in\mathbb{Z}$.
3 replies
bakkune
May 3, 2025
bakkune
Yesterday at 3:58 AM
Equivalent condition of the uniformly continuous fo a function
Alphaamss   2
N Yesterday at 2:05 AM by Alphaamss
Source: Personal
Let $f_{a,b}(x)=x^a\cos(x^b),x\in(0,\infty)$. Get all the $(a,b)\in\mathbb R^2$ such that $f_{a,b}$ is uniformly continuous on $(0,\infty)$.
2 replies
Alphaamss
May 3, 2025
Alphaamss
Yesterday at 2:05 AM
Absolute value
Silverfalcon   8
N Apr 22, 2025 by zhoujef000
This problem seemed to be too obvious.. And I think I"m wrong.. :D

Problem:

Consider the sequence $x_0, x_1, x_2,...x_{2000}$ of integers satisfying

\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]

for $n = 1,2,...2000$.

Find the minimum value of the expression $|x_1 + x_2 + ... x_{2000}|$.

My idea

Pretty sure I'm wrong but where did I go wrong?
8 replies
Silverfalcon
Jun 27, 2005
zhoujef000
Apr 22, 2025
Absolute value
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Silverfalcon
5006 posts
#1 • 2 Y
Y by Adventure10, Mango247
This problem seemed to be too obvious.. And I think I"m wrong.. :D

Problem:

Consider the sequence $x_0, x_1, x_2,...x_{2000}$ of integers satisfying

\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]

for $n = 1,2,...2000$.

Find the minimum value of the expression $|x_1 + x_2 + ... x_{2000}|$.

My idea

Pretty sure I'm wrong but where did I go wrong?
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Palytoxin
490 posts
#2 • 2 Y
Y by Adventure10, Mango247
yes wrong because check this out

$|x_1|=|x_0+1|=|1|$ and this means that $x_1$ can be $x_1=1$ or $x_1=-1$
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Silverfalcon
5006 posts
#3 • 2 Y
Y by Adventure10, Mango247
Ohh..

Then

I'm not sure if that makes the minimum though. No wonder it's the last question now..
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peeta
364 posts
#4 • 2 Y
Y by Adventure10, Mango247
well when u say every even will be 0 u can say for the odd $|x_n|=1$ therefore u can use 1 and -1 alternating coming to the minimum of 0, because there is sequence is periodic with the period 4, and the first 4 $x_n$ besides $x_0$ are 0 and they repeat themselves 250 times. also 0 is the smallest number possible, since $|x|\ge 0$
done :)
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JesusFreak197
1939 posts
#5 • 2 Y
Y by Adventure10, Mango247
Sweet, an Olympiad problem that I can actually solve! :P

Click to reveal hidden text
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K81o7
2417 posts
#6 • 2 Y
Y by Adventure10, Mango247
Peeta...there's a slight problem...
You can't go from 1 to 0...the thing about this problem is that $|x_n|\le|x_{n+1}|$ when $x_n$ is positive. In other words, you can't go from 1 to -1.
This one isn't quite as easy as it looks at first...
I'm thinking of one where the minimum is 12...
Basically you can start with 0, then go
-1, 0, 1, -2, -1, 0, 1, 2, -3, ...
going through sets of sizes which are an odd number each, and by the time you get to 22, you'll be at $x_1976$. 24 more numbers. Then, you go
-23, 22, -23, 22, -23...
But, if there's a way that 2000 - square number=multiple of 23 or 21, then the minimum is 0.
I'm sure there's a solution for that...
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JesusFreak197
1939 posts
#7 • 2 Y
Y by Adventure10, Mango247
Oh, you're right... by a more complicated set, you might be able to get it lower.
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Assassino9931
1321 posts
#8
Y by
Bump this
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zhoujef000
314 posts
#9
Y by
see 2006 AIME I P15 for a similar problem. (post #9 has a good solution)
This post has been edited 1 time. Last edited by zhoujef000, Apr 22, 2025, 7:47 PM
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