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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
10 Problems
Sedro   47
N 7 minutes ago by littleduckysteve
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 sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms, and the sum of its first $n$ terms is a multiple of $n$. 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 (solved by fruitmonster97): 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 (solved by Math-lover1): 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 (solved by CubeAlgo15): 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 (solved by maromex): 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 (solved by Math-lover1): 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 (solved by aaravdodhia): 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 $995$.
47 replies
Sedro
Jul 10, 2025
littleduckysteve
7 minutes ago
System of Equations
4everwise   16
N 9 minutes ago by neeyakkid23
Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz=1,$ $x+\frac{1}{z}=5,$ and $y+\frac{1}{x}=29.$ Then $z+\frac{1}{y}=\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
16 replies
+1 w
4everwise
Jan 5, 2007
neeyakkid23
9 minutes ago
Inequalities
sqing   5
N 17 minutes ago by sqing
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
5 replies
1 viewing
sqing
Jul 9, 2025
sqing
17 minutes ago
Ez inequality with AM-GM
Tofa7a._.36   4
N 4 hours ago by DAVROS
Prove that for all positive real numbers $x,y,z$ such that $xy+yz+zx = 3$.
We have:
$$\dfrac{1}{xyz} \ge \dfrac{2}{3}(\sqrt{x} + \sqrt{y} + \sqrt{z}) -1$$
4 replies
Tofa7a._.36
Jul 15, 2025
DAVROS
4 hours ago
No more topics!
Combinatorics
AlexCenteno2007   2
N May 11, 2025 by Royal_mhyasd
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
2 replies
AlexCenteno2007
May 9, 2025
Royal_mhyasd
May 11, 2025
Combinatorics
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AlexCenteno2007
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Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
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AlexCenteno2007
166 posts
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Could you help me?
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Royal_mhyasd
103 posts
#3 • 1 Y
Y by AlexCenteno2007
Nice one. There's quite a few solutions as well but I'll just leave the one I thought of
We're going to prove that Adrian wins for n=3 and n=even and Bertrand wins for n=odd, n>3.
n=3 is obvious - A wins by dividing it into a pile of 1 and a pile of 2.
For n>=4, we're gonna use induction. (our induction is that A wins when n=3 and n=even and B wins when n=odd, n>3)
If n=even, A can simply divide his pile into a pile of 1 and a pile of n-1. n-1 is odd, so the player who is about to make a move (player B) is gonna lose, so A wins.
If n=odd, A is gonna divide the pile into two piles of a and b stones, a<b, a+b = n so either a is odd and b is even or a is even and b is odd.
If b-a isn't 3 :
Bertrand is gonna divide the current piles into 3 piles of : b-a, a, a stones. We define "the first zone" as any move that can be traced back to the pile of b-a stones and "the second zone" as any love that can be traced back to the piles of a stones. If A makes a move in the second zone, B will simply mirror him in the other sub-zone (the one that can be traced back to the other pile of a stones). Therefore, it's pretty obvious that B will make the last move in the second zone, seeing as he can always mirror A. b-a is odd because a and b arent equal mod 2, so if A makes a move in the first zone then B can simply apply his winning strategy for b-a (it's odd and it isn't 3 so the second player wins), ensuring that he doesn't lose. Also, as long as the second zone still hasn't been reduced to piles of at most 2 stones, moves can still be made in the first zone even if B already technically won since there will still be piles of 2 stones. But it's impossible for A to make the last move in the first zone since that would mean that an odd number of total moves were made, so there's an even number of piles of 1 and no other piles, so b-a is even which isn't true. Therefore, B makes the last move in the first zone. So B wins.
If b-a = 3:
If the odd number out of a and b isn't 3 :
Then B can simply split the 2 piles into 2 piles of x/2, x/2 and y where x is the even number and y is the odd number. He can apply the same strategy as before since y isn't 3.
If the odd number out of a and b is 3:
Then we have n=9, b=6, a=3.
In this case, B will divide the pile of 6 into two piles of 3. This forces A to divide one of the piles of 3 into 2 piles of 1 and 2 so we will have
2 3 3 (the pile of 1 doesn't matter)
B divides the pile of 2 into 2 piles of 1,so we have
3 3
A divides one of the piles of 3 into 2 piles of 1 and 2,and then B will do the same for the other pile of 3 so B wins, which concludes our proof.
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