Plan ahead for the next school year. Schedule your class today!

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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
[PMO27 Areas] I.13 are you sure
BinariouslyRandom   4
N a few seconds ago by Kyj9981
The sequence of real numbers $x_1, x_2, \dots$, satisfies the recurrence relation
\[ \frac{x_{n+1}}{x_n} = \frac{(x_{n+1})^2 + 27}{x_n^2 + 27} \]for all positive integers $n$. Suppose that $x_{20} = x_{25} = 3$. Let $M$ be the maximum value of
\[ \sum_{n=1}^{2025} x_n. \]What is $M \pmod{1000}$?
4 replies
BinariouslyRandom
Jan 25, 2025
Kyj9981
a few seconds ago
Challenge: Make as many positive integers from 2 zeros
Biglion   25
N 14 minutes ago by littleduckysteve
How many positive integers can you make from at most 2 zeros, any math operation and cocatination?
New Rule: The successor function can only be used at most 3 times per number
Starting from 0, 0=0
25 replies
Biglion
Jul 2, 2025
littleduckysteve
14 minutes ago
Last Three digit of Fibonacci
Kyj9981   0
15 minutes ago
Source: Hong Kong Preliminary Selection Contest 2020 #20

Consider the fibonacci sequence $1,1,2,3,5,8,13,\dots$. What are the last three digits (from left to right) of the $2020$th term?
0 replies
Kyj9981
15 minutes ago
0 replies
10 Problems
Sedro   9
N 17 minutes ago by fruitmonster97
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An increasing sequence of positive integers $u_1, u_2, \dots, u_8$ has the property that the sum of its first $n$ terms is divisible by $n$ for every positive integer $n\le 8$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2: Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3: Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4: Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5: Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6: There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7: Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9: Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10: Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
9 replies
Sedro
Jul 10, 2025
fruitmonster97
17 minutes ago
NT By Probabilistic Method
EthanWYX2009   0
4 hours ago
Source: 2024 March 谜之竞赛-6
Given a positive integer \( k \) and a positive real number \( \varepsilon \), prove that there exist infinitely many positive integers \( n \) for which we can find pairwise coprime integers \( n_1, n_2, \cdots, n_k \) less than \( n \) satisfying
\[\text{gcd}(\varphi(n_1), \varphi(n_2), \cdots, \varphi(n_k)) \geq n^{1-\varepsilon}.\]Proposed by Cheng Jiang from Tsinghua University
0 replies
EthanWYX2009
4 hours ago
0 replies
a sequence of a polynomial
truongphatt2668   3
N Today at 3:28 AM by truongphatt2668
Let a sequence of polynomial defined by: $P_0(x) = x$ and $P_{n+1}(x) = -2xP_n(x) + P'_n(x), \forall n \in \mathbb{N}$.
Find: $P_{2017}(0)$
3 replies
truongphatt2668
Yesterday at 2:22 PM
truongphatt2668
Today at 3:28 AM
Minimum value
Martin.s   5
N Today at 2:52 AM by aaravdodhia
What is the minimum value of
$$
\frac{|a + b + c + d| \left( |a - b| |b - c| |c - d| + |b - a| |c - a| |d - a| \right)}{|a - b| |b - c| |c - d| |d - a|}
$$over all triples $a, b, c, d$ of distinct real numbers such that
$a^2 + b^2 + c^2 + d^2 = 3(ab + bc + cd + da).$

5 replies
Martin.s
Oct 17, 2024
aaravdodhia
Today at 2:52 AM
Aproximate ln(2) using perfect numbers
YLG_123   7
N Today at 12:04 AM by vincentwant
Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]where the sums are taken over all prime divisors \(p\) of \(n\).
7 replies
YLG_123
Oct 12, 2024
vincentwant
Today at 12:04 AM
Putnam 2003 B3
btilm305   35
N Yesterday at 2:50 PM by SomeonecoolLovesMaths
Show that for each positive integer n, \[n!=\prod_{i=1}^n \; \text{lcm} \; \{1, 2, \ldots, \left\lfloor\frac{n}{i} \right\rfloor\}\](Here lcm denotes the least common multiple, and $\lfloor x\rfloor$ denotes the greatest integer $\le x$.)
35 replies
btilm305
Jun 23, 2011
SomeonecoolLovesMaths
Yesterday at 2:50 PM
Simple integuration
obihs   1
N Yesterday at 2:21 PM by Litvinov
Source: Own
Find the value of
$$\int_1^2\dfrac{\ln x}{(x^2-2x+2)^2}dx$$
1 reply
obihs
Yesterday at 8:44 AM
Litvinov
Yesterday at 2:21 PM
Estimating the Density
zqy648   0
Yesterday at 1:11 PM
Source: 2024 May 谜之竞赛-6
Given non-empty subset \( I \) of the set of positive integers, a positive integer \( n \) is called good if for every prime factor \( p \) of \( n \), \( \nu_p(n) \in I \). For a positive real number \( x \), let \( S(x) \) denote the number of good numbers not exceeding \( x \).

Determine all positive real numbers \( C \) and \( \alpha \) such that \(
\lim\limits_{x \to +\infty} \dfrac{S(x)}{x^\alpha} = C.
\)

Proposed by Zhenqian Peng, High School Affiliated to Renmin University of China
0 replies
zqy648
Yesterday at 1:11 PM
0 replies
Show that if \( d_3 < \frac{d_1}{3} \), then there exist two other positive diag
Martin.s   3
N Yesterday at 12:26 PM by Martin.s
Let
\[
D = \begin{bmatrix}
d_1 & 0 & 0 \\
0 & d_2 & 0 \\
0 & 0 & d_3
\end{bmatrix}, \quad
T = \begin{bmatrix}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2
\end{bmatrix}
\]where \( d_1, d_2, d_3 \) are positive and \( d_1 \ge d_3 \).

Show that if \( d_3 < \frac{d_1}{3} \), then there exist two other positive diagonal matrices \( D_1 \) and \( D_2 \) such that \( D, D_1, D_2 \) are distinct but \( DT, D_1T, D_2T \) have the same eigenvalues.

Show also that if \( d_3 > \frac{d_1}{3} \) and \( D_1 \) is a positive diagonal matrix distinct from \( D \), then \( DT \) and \( D_1T \) must have different eigenvalues.
3 replies
Martin.s
Jun 23, 2025
Martin.s
Yesterday at 12:26 PM
Inequality
Martin.s   4
N Yesterday at 12:25 PM by Martin.s


For \( n = 2, 3, \dots \), the following inequalities hold:

\[
-\frac{1}{3} \leq \frac{\sin(n\theta)}{n \sin \theta} \leq \frac{\sqrt{6}}{9}
\quad \text{for } \frac{\pi}{n} \leq \theta \leq \pi - \frac{\pi}{n},
\]
and

\[
-\frac{1}{3} \leq \frac{\sin(n\theta)}{n \sin \theta} \leq \frac{1}{5}
\quad \text{for } \frac{\pi}{n} \leq \theta \leq \frac{\pi}{2}.
\]
4 replies
Martin.s
Jun 23, 2025
Martin.s
Yesterday at 12:25 PM
Analytic Number Theory
zqy648   1
N Yesterday at 11:59 AM by zqy648
Source: 2024 December 谜之竞赛-6
For positive integer \( n \), define
\[S_n = \{(a, b) \in \mathbb{N}_+^2 \mid a, b < \sqrt{n} \text{ and } n \mid a^2 + b^3 + 1\}. \]Prove that there exists a positive real number \(\varepsilon\) such that for all integers \(n \geq 10\), \(
\left| S_n \right| < n^{\frac{1}{2} - \frac{\varepsilon}{\ln \ln n}}.
\)

Proposed by Yuxing Ye
1 reply
zqy648
Yesterday at 9:36 AM
zqy648
Yesterday at 11:59 AM
Function and Quadratic equations help help help
Ocean_MathGod   1
N May 20, 2025 by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
May 20, 2025
Function and Quadratic equations help help help
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Ocean_MathGod
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Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
Z K Y
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Mathzeus1024
1057 posts
#2
Y by
Part (I):

Completing the square on this parabola yields:

$y = \left(x+\frac{2m+1}{2}\right)^2 + m(m-3) - \frac{(2m+1)^2}{4} \Rightarrow y = \left(x+\frac{2m+1}{2}\right)^2 -\frac{16m+1}{4}$;

with vertex $\textcolor{red}{P\left(-\frac{2m+1}{2}, -\frac{16m+1}{4}\right)}$.

Part (II):

If the parabola intersects the line $y=x-km$ in exactly one point, then:

$x^2 + (2m+1)x + m(m-3) = x-km \Rightarrow x^2 +2mx + (m^2-3m+km) = 0 \Rightarrow x = \frac{-2m \pm \sqrt{4m^2-4(1)[m^2+(k-3)m]}}{2}$ (i);

of which we require the discriminant in (i) to equal zero. This occurs $\Leftrightarrow \textcolor{red}{k=3}$.

Part (III):

For $m \in [-1,4]$ we have the points $A(-m-1, -4m); B\left(\frac{m}{2}, \frac{9m^2-10m}{4}\right); C(-m,-4m)$ on the parabola. Checking $-4m=\frac{9m^2-10m}{4} \Rightarrow 9m^2-6m=m(9m+6)=0 \Rightarrow m = -\frac{2}{3}, 0$ gives us the orderings:

$y_{2}>y_{1}=y_{3}$ for $m \in \left[-1,-\frac{2}{3}\right) \cup (0,4]$;

$y_{2}=y_{1}=y_{3}$ for $m = -\frac{2}{3}, 0$;

$y_{2} < y_{1} = y_{3}$ for $m \in \left(-\frac{2}{3}, 0\right)$.

Rotating the parabola's axis of symmetry (i.e. $x= -\frac{2m+1}{2}$) $90^{\circ}$ counterclockwise about the origin gives us the horizontal line $y = -\frac{2m+1}{2}$ from which $|PH| = \left|-\frac{16m+1}{4} - \left(-\frac{2m+1}{2}\right)\right| = \frac{|1-12m|}{4}$.

If $1 < |PH| \le 6$, then $1 < \frac{|1-12m|}{4}\le 6 \Rightarrow m \in \left[-\frac{23}{12},-\frac{1}{4}\right) \cup \left(\frac{5}{12},\frac{25}{12}\right]$. This results in the following orderings:

$\textcolor{red}{y_{2}>y_{1}=y_{2}}$ for $\textcolor{red}{m \in \left[-1, -\frac{2}{3}\right) \cup \left(\frac{5}{12},\frac{25}{12}\right]}$;

$\textcolor{red}{y_{2}=y_{1}=y_{3}}$ for $\textcolor{red}{m = -\frac{2}{3}}$;

$\textcolor{red}{y_{2}<y_{1}=y_{3}}$ for $\textcolor{red}{m \in \left(-\frac{2}{3}, -\frac{1}{4}\right)}$.
This post has been edited 6 times. Last edited by Mathzeus1024, May 31, 2025, 1:03 PM
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