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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
Exponents of integer question
Dheckob   4
N 16 minutes ago by LeYohan
Find the smallest positive integer $m$ such that $5m$ is an exact 5th power, $6m$ is an exact 6th power, and $7m$ is an exact 7th power.
4 replies
Dheckob
Apr 12, 2017
LeYohan
16 minutes ago
ISI 2025
Zeroin   0
32 minutes ago
Let $\mathbb{N}$ denote the set of natural numbers and let $(a_i,b_i),1 \leq i \leq 9$ denote $9$ ordered pairs in $\mathbb{N} \times \mathbb{N}$. Prove that there exist $3$ distinct elements in the set $2^{a_i}3^{b_i}$ for $1 \leq i \leq 9$ whose product is a perfect cube.
0 replies
Zeroin
32 minutes ago
0 replies
Inequalities
sqing   3
N an hour ago by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
3 replies
sqing
May 13, 2025
sqing
an hour ago
Pertenacious Polynomial Problem
BadAtCompetitionMath21420   5
N 2 hours ago by BadAtCompetitionMath21420
Let the polynomial $P(x) = x^3-x^2+px-q$ have real roots and real coefficients with $q>0$. What is the maximum value of $p+q$?

This is a problem I made for my math competition, and I wanted to see if someone would double-check my work (No Mike allowed):

solution
Is this solution good?
5 replies
1 viewing
BadAtCompetitionMath21420
Yesterday at 3:13 AM
BadAtCompetitionMath21420
2 hours ago
Max and min of ab+bc+ca-abc
Tiira   5
N 2 hours ago by sqing
a, b and c are three non-negative reel numbers such that a+b+c=1.
What are the extremums of
ab+bc+ca-abc
?
5 replies
Tiira
Jan 29, 2021
sqing
2 hours ago
Inequalities
sqing   12
N 2 hours ago by sqing
Let $a,b,c >2 $ and $ ab+bc+ca \leq 75.$ Show that
$$\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2}\geq 1$$Let $a,b,c >2 $ and $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{6}{7}.$ Show that
$$\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2}\geq 2$$
12 replies
sqing
May 13, 2025
sqing
2 hours ago
2017 DMI Individual Round - Downtown Mathematics Invitational
parmenides51   14
N 3 hours ago by SomeonecoolLovesMaths
p1. Compute the smallest positive integer $x$ such that $351x$ is a perfect cube.


p2. A four digit integer is chosen at random. What is the probability all $4$ digits are distinct?


p3. If $$\frac{\sqrt{x + 1}}{\sqrt{x}}+ \frac{\sqrt{x}}{\sqrt{x + 1}} =\frac52.$$Solve for $x$.


p4. In $\vartriangle ABC$, $AB = 13$, $BC = 14$, and $AC = 15$. Let $D$ be the point on $BC$ such that $AD \perp BC$, and let $E$ be the midpoint of $AD$. If $F$ is a point such that $CDEF$ is a rectangle, compute the area of $\vartriangle AEF$.


p5. Square $ABCD$ has a sidelength of $4$. Points $P$, $Q$, $R$, and $S$ are chosen on $AB$, $BC$, $CD$, and $AD$ respectively, such that $AP$, $BQ$, $CR$, and $DS$ are length $1$. Compute the area of quadrilateral $P QRS$.


p6. A sequence $a_n$ satisfies for all integers $n$, $$a_{n+1} = 3a_n - 2a_{n-1}.$$If $a_0 = -30$ and $a_1 = -29$, compute $a_{11}$.


p7. In a class, every child has either red hair, blond hair, or black hair. All but $20$ children have black hair, all but $17$ have red hair, and all but $5$ have blond hair. How many children are there in the class?


p8. An Akash set is a set of integers that does not contain two integers such that one divides the other. Compute the minimum positive integer $n$ such that the set $\{1, 2, 3, ..., 2017\}$ can be partitioned into n Akash subsets.


PS. You should use hide for answers. Collected here.
14 replies
parmenides51
Oct 2, 2023
SomeonecoolLovesMaths
3 hours ago
p+2^p-3=n^2
tom-nowy   0
4 hours ago
Let $n$ be a natural number and $p$ be a prime number. How many different pairs $(n, p)$ satisfy the equation:
$$p + 2^p - 3 = n^2 .$$
Inspired by https://artofproblemsolving.com/community/c4h3560823
0 replies
tom-nowy
4 hours ago
0 replies
Range of a function
Pscgylotti   1
N 6 hours ago by Mathzeus1024
Try to get the range of function $f(x)=cosx+\sqrt{cos^{2}x-4\sqrt{2}cosx+4sinx+9}$ :
1 reply
Pscgylotti
Jul 22, 2019
Mathzeus1024
6 hours ago
Inequalities
sqing   17
N 6 hours ago by sqing
Let $ a,b,c>0 , a+b+c +abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$Let $ a,b,c>0 , ab+bc+ca+abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$
17 replies
sqing
May 15, 2025
sqing
6 hours ago
Calculate the function
Arkham   1
N 6 hours ago by Mathzeus1024
Consider $ y = f (x) = \arcsin (- \sqrt {1 + 10x}) $, $ x \in [-1 / 10,0] $. Calculate the function where $ g $ is the inverse function of $ f $

Note: $ g (y) = f ^ {- 1} (y) $]
1 reply
Arkham
Apr 29, 2021
Mathzeus1024
6 hours ago
Ez comb proposed by ME
IEatProblemsForBreakfast   0
6 hours ago
A and B play a game on two table:
1.At first one table got $n$ different coloured marbles on it and another one is empty
2.At each move player choose set of marbles that hadn't choose either players before and all chosen marbles from same table, and move all the marbles in that set to another table
3.Player who can not move lose
If A starts and they move alternatily who got the winning strategy?
0 replies
IEatProblemsForBreakfast
6 hours ago
0 replies
Incircle concurrency
niwobin   3
N Today at 8:37 AM by sunken rock
Triangle ABC with incenter I, incircle is tangent to BC, AC, and AB at D, E and F respectively.
DT is a diameter for the incircle, and AT meets the incircle again at point H.
Let DH and EF intersect at point J. Prove: AJ//BC.
3 replies
niwobin
May 11, 2025
sunken rock
Today at 8:37 AM
Combination
AnhIsGod   1
N Today at 7:47 AM by alexheinis
A group consists of 21 people. A and B are said to have a "known relationship" if A knows B or A knows C_1, C_1 knows C_2, ..., C_n knows B. It is known that among any 6 people, there are always 2 who have a known relationship. Prove that in this group, there always exists a group of 5 people who all have a "known relationship" with each other.
1 reply
AnhIsGod
Today at 7:29 AM
alexheinis
Today at 7:47 AM
Generating Functions
greenplanet2050   7
N Apr 30, 2025 by rchokler
So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)
7 replies
greenplanet2050
Apr 29, 2025
rchokler
Apr 30, 2025
Generating Functions
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G H BBookmark kLocked kLocked NReply
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greenplanet2050
1324 posts
#1
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So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)
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Yiyj
26 posts
#2
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Note that $1+2x+3x^2+\cdots$ is an arithmetico-geometric sequence. Then, we have the formula \[\sum_{k=1}^{\infty} x_k = \dfrac{dg_2}{(1-r)^2} + \dfrac{x_1}{1-r},\]where $d$ is the common difference of the arithmetic sequence, $r$ is the common ratio of the geometric sequence, $g_2$ is the second term of the geometric sequence, and $x_k$ are the terms of the arithmetico-geometric sequence.

Plugging in $d=1, r=x, g_2=x, x_1=1$, we get \[1+2x+3x^2+\cdots=\dfrac{x}{(1-x)^2}+\dfrac{1}{1-x} = \dfrac{x}{(1-x)^2}+\dfrac{1-x}{(1-x)^2} = \boxed{\dfrac{1}{(1-x)^2}}.\]Hope that helped!
This post has been edited 1 time. Last edited by Yiyj, Apr 29, 2025, 10:55 PM
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Shan3t
381 posts
#3
Y by
greenplanet2050 wrote:
So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)

Let $1+2x+3x^2\cdots = S.\quad(1)$

Now multiply $S,$ by $x,$ to get:

$x+2x^2+3x^3+4x^4+\cdots = S\cdot x\quad(2)$

Just subtract equation $2$ from equation $1,$ to get $1+x+x^2+x^3\cdots = \frac{1}{1-x} = S-S\cdot x.$ Simplify this, gives $S(1-x)=\frac1{1-x}\implies S=\frac{1}{(1-x)^2}.$
This post has been edited 1 time. Last edited by Shan3t, Apr 29, 2025, 11:02 PM
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martianrunner
207 posts
#4
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Notice that $(1+x+x^2+x^3...)^2 = 1+2x+3x^2+4x^3...$ (this can be elementarily proven with induction by counting the pairs for each coefficient's term)

Since the value of a geometric sequence that goes $1+x+x^2+x^3...$ is $\frac{1}{1-x}$, we square that to get our answer of $\frac{1}{(1-x)^2}$
This post has been edited 2 times. Last edited by martianrunner, Apr 30, 2025, 2:36 AM
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greenplanet2050
1324 posts
#5
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Thank you all!!
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Shan3t
381 posts
#6
Y by
greenplanet2050 wrote:
Thank you all!!

np :D

also @2bove sol very clean :D
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ohiorizzler1434
786 posts
#7 • 1 Y
Y by compoly2010
It's a highly technical concept that combines convergence of geometric sequences with calculus, to represent the power series of a function around a point!
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rchokler
2975 posts
#8
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In general, the ring $\mathbb{C}[x]$ of all polynomials with complex coefficients is spanned by the basis $\{p_n\}_{n=0}^\infty$ where $p_n(x)=(x+1)(x+2)\cdots(x+n)$ under finite linear combinations. Note that $p_0(x)=1$ since it is the empty product.

Using this and the power rule for derivatives, we can find a formula for $\sum_{n=0}^\infty p(n)x^n$ for any polynomial $p$ and any $x\in(-1,1)$.

The idea is that $S_k(x)=\sum_{n=0}^\infty p_k(n)x^n=\sum_{n=0}^\infty\frac{d^k}{dx^k}x^{n+k}=\sum_{n=-k}^\infty\frac{d^k}{dx^k}x^{n+k}=\sum_{n=0}^\infty\frac{d^k}{dx^k}x^n=\sum_{n=0}^\infty\frac{d^k}{dx^k}x^n=\frac{d^k}{dx^k}\sum_{n=0}^\infty x^n=\frac{d^k}{dx^k}\frac{1}{1-x}=\frac{k!}{(1-x)^{k+1}}$.

So all you have to do is write $p$ of degree $n$, as $p=\sum_{k=0}^nc_kp_k$.

Example:
Find $\sum_{n=0}^\infty(n^3+5n^2-3n+4)x^n$.

Solution:
Use $p_0(n)=1$, $p_1(n)=n+1$, $p_2(n)=(n+1)(n+2)=n^2+3n+2$, and $p_3=(n+1)(n+2)(n+3)=n^3+6n^2+11n+6$.

$c_0p_0(n)+c_1p_1(n)+c_2p_2(n)+c_3p_3(n)=c_3n^3+(c_2+6c_3)n^2+(c_1+3c_2+11c_3)n+(c_0+c_1+2c_2+6c_3)\equiv n^3+5n^2-3n+4$
$\implies\begin{cases}c_3=1\\c_2+6c_3=5\\c_1+3c_2+11c_3=-3\\c_0+c_1+2c_2+6c_3=4\end{cases}\implies(c_0,c_1,c_2,c_3)=(11,-11,-1,1)$

Therefore $\sum_{n=0}^\infty(n^3+5n^2-3n+4)x^n=11S_0(x)-11S_1(x)-S_2(x)+S_3(x)=\frac{11}{1-x}-\frac{11}{(1-x)^2}-\frac{2}{(1-x)^3}+\frac{6}{(1-x)^4}=\frac{-11x^3+22x^2-9x+4}{(1-x)^4}$.
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