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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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Peru IMO TST 2022
diegoca1   1
N 3 minutes ago by hectorleo123
Source: Peru IMO TST 2022 D1 P4
Let $\Omega$ be the circumcircle of triangle $ABC$, with $\angle BAC > 90^\circ $ and $ AB > AC $. The tangents to $\Omega$ at points $B$ and $C$ intersect at $D$. The tangent to $\Omega$ at point $A$ intersects line $BC$ at $E$. The line through $D$ parallel to $AE$ intersects line $BC$ at $F$. The circumference $\Gamma$ with diameter $EF$ intersects line $AB$ at points $P$ and $Q$, and line $AC$ at points $X$ and $Y$.
Prove that one of the angles $\angle AEB$, $\angle PEQ$, $\angle XEY$ is equal to the sum of the other two.
1 reply
diegoca1
Today at 12:04 AM
hectorleo123
3 minutes ago
A is the smallest angle
Valentin Vornicu   29
N 4 minutes ago by lpieleanu
Source: IMO 1997, Problem 2, IMO Shortlist 1997, Q8
It is known that $ \angle BAC$ is the smallest angle in the triangle $ ABC$. The points $ B$ and $ C$ divide the circumcircle of the triangle into two arcs. Let $ U$ be an interior point of the arc between $ B$ and $ C$ which does not contain $ A$. The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AU$ at $ V$ and $ W$, respectively. The lines $ BV$ and $ CW$ meet at $ T$.

Show that $ AU = TB + TC$.


Alternative formulation:

Four different points $ A,B,C,D$ are chosen on a circle $ \Gamma$ such that the triangle $ BCD$ is not right-angled. Prove that:

(a) The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AD$ at certain points $ W$ and $ V,$ respectively, and that the lines $ CV$ and $ BW$ meet at a certain point $ T.$

(b) The length of one of the line segments $ AD, BT,$ and $ CT$ is the sum of the lengths of the other two.
29 replies
Valentin Vornicu
Oct 27, 2005
lpieleanu
4 minutes ago
diophantine with exponents and factorials
skellyrah   13
N 14 minutes ago by SimplisticFormulas
Find all natural numbers $m$ such that $$ m^k - 3^k= m! - 6 $$for some natural number $k$
13 replies
skellyrah
Jul 22, 2025
SimplisticFormulas
14 minutes ago
2025 IMO Results
ilikemath247365   20
N 22 minutes ago by A7456321
Source: https://www.imo-official.org/year_info.aspx?year=2025
Congrats to China for getting 1st place! Congrats to USA for getting 2nd and congrats to South Korea for getting 3rd!
20 replies
ilikemath247365
Jul 23, 2025
A7456321
22 minutes ago
<QED =? <ABCDE regular pentagon APQ = 90^o, < PAQ = 36^o
parmenides51   1
N 35 minutes ago by SuperBarsh
Source: Mexican Geometry Olympiad 2021 p6 - II Olimpiada de Geometria https://artofproblemsolving.com/community/c2746625_
Let $ABCDE$ be a regular pentagon and let $P$ and $Q$ be points on the sides $BC$ and $CD$ respectively such that $\angle APQ = 90^o$ and $\angle PAQ = 36^o$. What is the measure of the angle $\angle QED$ ?
1 reply
parmenides51
Dec 31, 2021
SuperBarsh
35 minutes ago
New year's question
ylmath123   11
N 39 minutes ago by MathLuis
Source: Own
Let $ABC$ be a triangle inscribed circle $(O)$$D$ is the midpoint of $BC$$P$ on $OD$$PE\perp AC$ at
$E$$PF\perp AB$ at $F$circle $(DEF)$ cut $BC$ at $N$$BE$ cut $CF$ at $K$$NK$ cut circle $(O)$ at $L$
Prove$AL \parallel BC$
11 replies
ylmath123
Feb 14, 2018
MathLuis
39 minutes ago
Symmetric inequality
nexu   7
N an hour ago by nexu
Source: own
Let $x,y,z \ge 0$. Prove that:
$$  \sum_{\mathrm{cyc}}{\left( y-z \right) ^2\left( 7x^2-y^2-z^2 \right) ^2}\ge 112\left( x-y \right) ^2\left( y-z \right) ^2\left( z-x \right) ^2. $$
7 replies
+1 w
nexu
Feb 12, 2023
nexu
an hour ago
A feasible refinement of GMA 567
Rhapsodies_pro   5
N an hour ago by JARP091
Source: Own?
Let \(a_1,a_2,\dotsc,a_n\) (\(n>3\)) be non-negative real numbers fulfilling \[\sum_{k=1}^na_k^2+{\left(n^2-3n+1\right)}\prod_{k=1}^na_k\geqslant{\left(n-1\right)}^2\text.\]Prove or disprove: \[\frac1{n-1}\sum_{1\leqslant i<j\leqslant n}{\left(a_i-a_j\right)}^2\geqslant{\left({\left(n^2-2n-1\right)}\sum_{k=1}^na_k-n{\left(n-1\right)}{\left(n-3\right)}\right)}{\left(n-\sum_{k=1}^na_k\right)}\textnormal.\]
5 replies
Rhapsodies_pro
Jul 21, 2025
JARP091
an hour ago
Find all functions with ...
Math2030   2
N 2 hours ago by MathGuy1729
Problem 3.43 (Czech–Austrian–Polish–Slovak Match 2024).} Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
\[
f(x^2 + y^2) = f(x - y)f(x + y) + \alpha y f(y), \quad \forall x, y \in \mathbb{R}
\](where $\alpha \ne 0$ is a fixed real constant).
2 replies
Math2030
2 hours ago
MathGuy1729
2 hours ago
IMO 2017 Problem 1
cjquines0   159
N 2 hours ago by TigerOnion
Source: IMO 2017 Problem 1
For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as
$$a_{n+1} = 
\begin{cases}
\sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\
a_n + 3 & \text{otherwise.}
\end{cases}
$$Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.

Proposed by Stephan Wagner, South Africa
159 replies
cjquines0
Jul 18, 2017
TigerOnion
2 hours ago
Interesting functional equation
TheUltimate123   15
N 2 hours ago by jasperE3
Source: ELMO Shortlist 2023 A2
Let \(\mathbb R_{>0}\) denote the set of positive real numbers. Find all functions \(f:\mathbb R_{>0}\to\mathbb R_{>0}\) such that for all positive real numbers \(x\) and \(y\), \[f(xy+1)=f(x)f\left(\frac1x+f\left(\frac1y\right)\right).\]
Proposed by Luke Robitaille
15 replies
TheUltimate123
Jun 29, 2023
jasperE3
2 hours ago
IMO ShortList 1998, number theory problem 1
orl   59
N 2 hours ago by SomeonecoolLovesMaths
Source: IMO ShortList 1998, number theory problem 1
Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
59 replies
orl
Oct 22, 2004
SomeonecoolLovesMaths
2 hours ago
ABC is similar to XYZ
Amir Hossein   58
N 2 hours ago by Kempu33334
Source: China TST 2011 - Quiz 2 - D2 - P1
Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.
58 replies
Amir Hossein
May 20, 2011
Kempu33334
2 hours ago
A nice property of triangle with incircle (I)
TUAN2k8   1
N 2 hours ago by Royal_mhyasd
Source: own
Let \( ABC \) be a non-isosceles triangle with incircle (\( I \)). Denote by \( D, E, F \) the points where (\( I \)) touches \( BC, CA, AB \), respectively. The A-excircle of \( ABC \) is tangent to \( BC \) at \( G \). The lines \( IB \) and \( IC \) meet \( AG \) at \( M \) and \( N \), respectively.

a) Prove that the circumcircles of triangles \( MBF \), \( NCE \), and \( BIC \) are concurrent at a point.

b) Let \( L \) and \( K \) be the midpoints of \( AG \) and \( BC \), respectively, and let \( J \) be the orthocenter of triangle \( IMN \). Show that the points \( L, K, J \) are collinear.
1 reply
TUAN2k8
Today at 1:18 AM
Royal_mhyasd
2 hours ago
Integer polynomial commutes with sum of digits
cjquines0   46
N May 21, 2025 by cursed_tangent1434
Source: 2016 IMO Shortlist N1
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
46 replies
cjquines0
Jul 19, 2017
cursed_tangent1434
May 21, 2025
Integer polynomial commutes with sum of digits
G H J
Source: 2016 IMO Shortlist N1
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Pyramix
419 posts
#35
Y by
Idea is to fix $S(n)$.

Take $n=10^k$. Then, $S(n)=1$ and RHS is constant, which is $P(1)$.
Let $P(x) = c_0 + c_1x + \cdots + c_dx^d$. Then, for arbitrarily large $k$, we have
\[S(P(10^k))=S(c_0)+S(c_1)+\cdots+S(c_d)=c_0+c_1+\cdots+c_d\]Similarly, for any $10>c>0$, we have for sufficiently large $k$,
\[S(P(c\cdot 10^k))=S(c_0)+S(cc_1)+\cdots+S(c_dc^d)=c_0+c_1c+\cdots+c_dc^d\]Hence, $c_i=0$ for $i>1$ and $c_1=0,1$.

Only constant polynomials are one digit constants.

The only linear polynomial is $x+t$, and plugging in original equation gives $t=0$. So, $P(x)=x$.
Hence,
\[P(x)\in\left\{x,1,2,\ldots,9\right\}\]
This post has been edited 2 times. Last edited by Pyramix, Apr 3, 2024, 2:27 PM
Z K Y
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thdnder
200 posts
#36
Y by
Answer: $P(x) = x$ or $P(x) = c$, where $c \in \{1, 2, \dots, 9\}$.

Firstly, we'll show that $\deg(P) \le 1$. Assume the contrary, let $P(x) = a_kx^k + a_{k-1}x^{k-1} + \dots + a_1x + a_0$ for some $k \ge 2$. Replacing $P(x) \to -P(x)$, we may assume $a_k > 0$. Let $N$ be a large enough number and take $n = 10^N - 1$. Then $S(n) = 9N$, so $P(S(N)) = P(9N)$ and $P(n) < (a_k+1)n^k < (a_k + 1)10^{Nk}$. Since $PS(P(n)) \le 9(\log_{10}(P(n)) + 1) < 9(\log_{10}(a_k + 1) + Nk + 1)$, we see that $P(9N) < 9Nk + c$ for some constant $c$, contradicting the fact that $N$ is large enough and $\deg(P) \ge 2$. Thus, we may assume $\deg(P) \le 1$.

Case 1: $\deg(P) = 1$.

Let $P(x) = ax + b$ for some $a, b \in \mathbb{Z}$. Take $n = 10^N + k$ for some large $N$ and some $k \ge 2016$. Then, $a + S(ak + b) = a + aS(k) + b = aS(n) + b = S(an + b) = S(a) + S(ak + b)$, so $a = S(a)$, which means that $a \in \{1, 2, \dots, 9\}$. Thus taking $n = 10^N + k$ for some large $N$, we get that $a + aS(k) + b = aS(n) + b = S(an + b) = S(a) + S(ak + b) = a + S(ak + b)$, so $aS(k) + b = S(ak + b)$ for all $k \ge 0$. Plugging $k = 0$ into this, we see that $b = S(b)$, which means that $b \in \{1, \dots, 9\}$. Now for $0 \le k \le 9$, $ak + b = S(ak + b)$, so $9a + b \in \{0, 1, \dots, 9\}$ and $9a + b \ge 9$, so $a = 1$ and $b = 0$, which means that $P(x) = x$.

Case 2: $\deg(P) = 0$.

In this case, we get that $c = P(x) = S(c)$, so $c \in \{0, 1, \dots, 9\}$.

So $P(x) = x$ or $P(x) = c$ for some $c \in \{1, \dots, 9\}$. $\blacksquare$
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ezpotd
1360 posts
#37
Y by
I claim the solutions are $P(x) = c$ for $1 \le c \le 9$ and $P(x) = x$, both of which clearly work.

Let $n = 9999999 \dots$ where the $9$ is repeated $k$ times. Then $P(S(n)) = P(9k)$. Now $S(P(n)) \le (9 \log P(n) + 1) \sim 9k + a$ for some constant $a$. Since the $S(P(n))$ is linear or smaller in $k$, we see that $P(S(n)) = P(9k)$ is linear or smaller in $k$, so $P$ is linear. We then have $cS(n) + d = S(cn + d)$. Substitute $n = 9$, then have $9c + d = S(9c + d)$, this only has single digit solutions (manually observe that $10$ thru $20$ has no solutions, after that all $2$ digits have sum at most $18$ but the value is greater than $20$, after that we generally have $9k < 10^{k- 1}$), so we have $9c + d \ge 9 $, so either we have $c = 0$ and $d$ a single digit, giving the working constant solutions, otherwise we have $c = 1, d = 0$, giving $P(x) = x$, which works.
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Maximilian113
592 posts
#38
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Let $P(x) = a_mx^n+a_{m-1}x^{m-1}+\cdots + a_1x^1+a_0.$ Suppose for the sake of a contradiction that there is some $a_i < 0.$ Letting $n=10^k$ for some arbitrarily large positive integer $k$ gives $P(1)=S(P(10^k)).$ But as $k$ increases $P(10^k)$ gets more and more $9$s and hence its sum of digits goes to infinity, a contradiction. Therefore all $a_i \geq 0.$

Now, notice that letting $n=9 \cdot 10^k$ for some arbitrarily large positive integer $k$ yields $$\sum_{i=0}^n S(a_i9^i) = P(9) = \sum_{i=0}^n a_i9^i.$$However, note that if a positive integer $x$ has $d \geq 3$ digits then $$S(x) \leq 9d < 10^{d-1} \leq x.$$Meanwhile, if $d=2$ and $x > 18$ then $S(x) \leq 18 < x.$ If $10 \leq x \leq 18,$ we have $S(x) \leq 1+8 = 9 < x.$ Therefore, for all $x \geq 10$ we have that $S(x) < x.$ Meanwhile for $1 \leq x \leq 9$ clearly $S(x) = x.$

With this, if $n \geq 2$ we clearly have $$\sum_{i=0}^n S(a_i9^i) < \sum_{i=0}^n a_i9^i,$$which is a contradiction. Thus $n \leq 1,$ in this case equality must hold so we have $S(a_0)=a_0$ and $S(9a_1)=9a_1.$ From the first equation, we have $0 \leq a_0 \leq 9.$ From the second equation, we have $a_1 = 0,1.$

If $a_1=0,$ $P(x)=a_0$ and this clearly works.

If $a_1=1,$ $P(x)=x+a_0.$ If $1 \leq a_0 \leq 9,$ then letting $n=10-a_0$ yields $$a_0+S(10-a_0)=S(10)=1$$which is clearly impossible. Thus $a_0 = 0$ and $P(x)=x.$ This solution clearly works.

Hence, the only solutions are $P(x)=a_0$ where $0 \leq a_0 \leq 9$ or $P(x)=x.$
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onyqz
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#39
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cute one
solution
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HamstPan38825
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#40
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This is not very interesting: the answers are just $P \equiv c \in \{1, 2, \dots, 9\}$ and $P \equiv x$, which work vacuously. Now let $n = 10^N - 1$ for some very large positive integer $N$. We may assume that the leading coefficient $a_d$ of $P$ is positive. For $d = \deg P$, we have \[S(P(n)) \leq 9 \log_{10} P(n) \leq 9 \log_{10} \left(a_d n^d P(1)\right) = 9d \log_{10} n +C < 9dN + C\]where $C$ is a constant. On the other hand, $P(S(n)) = P(9N) > \tfrac 12 a_d(9N)^d$ is strictly greater than the previous if $d \geq 2$.

So we must have $d = 1$; then we have $P(S(n)) = 9a_d N - C$, so $a_d = 1$ is forced too, and we get the desired solution set.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 2, 2025, 3:24 AM
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lelouchvigeo
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#41 • 1 Y
Y by alexanderhamilton124
solution
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scannose
1026 posts
#42 • 1 Y
Y by alexanderhamilton124
cjquines0 wrote:
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
first note that $P(n)$ cannot be constant oops this isn't true, assume $P(n)$ isn't constant for the rest of this proof. let the leading term of $P(n)$ be $ax^b$ for a pair of positive integers $(a, b)$, and let $n = 10^m - 1$ for some arbitrarily large positive integer $m$ satisfying $a(m')^b > 2|f(m') - a(m')^b|$ for all $m' \geq m$ (the proof that such a $m$ exists is omitted).

then, we have $$P(n) < 1.5a \cdot 10^{bm} \implies S(P(n)) < 9bm \log (1.5a)$$; however, on the other hand, we have $$P(S(n)) = P(9m) > 0.5a(9m)^b.$$if $b \geq 2$, there must exist an arbitrarily large value of $m$ such that $0.5a(9m)^b > 9bm \log(1.5a)$, contradiction; therefore, we conclude $b = 1$.

rewrite $P(x)$ as $ax + c$, and take another sufficiently large integer $y$ such that $10^y > a, |c|$. if $c < 0$, comparing $x = 10^y$ and $x = 10^{y+1}$ gives $P(S(10^y)) = P(1) = P(S(10^{y+1}))$, but $S(P(10^{y+1})) = S(P(10^y)) + 9$, reaching a contradiction; therefore, we may assume $c \geq 0$. now, plugging in $x = 10^y$ gives $S(P(10^y)) = S(a) + S(c) = P(1) = a + c$; this implies that $a, c < 10$. then, plugging in $x = 9 \cdot 10^y$ gives $S(9a) + S(b) = 9a + b$, implying $a = 1$. finally, plugging in $x = 2019$ gives $S(2019 + b) = 12 + b$, which only holds when $b = 0$.

therefore, the only non-constant polynomial satisfying the given condition is $P(x) = x$, which obviously works. if $P(x)$ is constant, it is easy to see that the only possible values of $P(x)$ are $1, 2, 3, 4, 5, 6, 7, 8, 9$.

this solution is more messy than i'd like it to be, please lmk if i fakesolved this problem
This post has been edited 1 time. Last edited by scannose, Mar 2, 2025, 5:32 PM
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ihategeo_1969
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#43
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Yah I definitely overcooked this...

We first prove that if $\deg(P) \ge 2$, then there are no solutions.

Claim: There exists a polynomial $Q(x)$ which has exactly $1$ negative coefficient and a leading positive coefficient but $P(Q(x))$ has all non-negative coefficients.
Proof: Key idea is to choose \[Q(x)=Ax^4+Ax^3-x^2+Ax+A \text{ where $A \gg 0$}\]If $Q(x)=Ax^4+Ax^3+AX+A$ then this would have almost worked because if $A$ is too massive then the coefficients of $P$ won't really matter as long as its leading coefficient is positive.

But obviously $Q$ needs to have a negative coefficient, so just adding $-x^2$ will not do anything as we can keep taking $A$ bigger and bigger and the ``resistance" won't matter. $\square$

For large $n$, we have $s(P(Q(10^n)))=s(P(Q(10^{n-1})))$ but $s(Q(10^n))=s(Q(10^{n-1}))+9$. And so we have \[O(1)=s(P(Q(10^n)))=P(s(Q(10^n)))=P(9n+O(1))\]which is a contradiction.

Now we take two cases.
  • If $\deg(P)=0$. Let $P=c$ so $s(c)=c \iff c \in \{1,\dots ,9\}$.
  • If $\deg(P)=1$. Let $P(x)=ax+b$ so $s(an+b)=as(n)+b$. Again we take $n \to 10^n$ (for large $n$), by the same thing above we automatically dismiss the case where $b$ is negative and if $b \ge 0$ then we have $s(a)+s(b)=a+b$ so $a \le 9$. Now put $n \to a10^n$ for large $n$ and again we so again $s(a^2)+s(b)=a^2+b$ so $a^2 \le 9$. Continue doing this so $a^k \le 9$ for all $k$ which forces $a=1$. This also means $b=0$.
Hence only solutions (which obviously work) are \[\boxed{P(x) \equiv x} \text{ AND } \boxed{P(x) \equiv c} \text{ where $c \in \{1,\dots,9 \}$}\]
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joshualiu315
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#44
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The answer is $P(x) = \boxed{c, x}$ where $c \in \{0,1,2\dots,9\}$, which obviously work.

Suppose that $P(x) = a_0+a_1x+\dots+a_nx^n$. Plug in $n = 10^k$, to get

\[P(1) = a_0+a_1+\dots+a_n = S(P(10^k)) \ge 0.\]
Thus, all of the coefficients are nonnegative. Then, plug in $n = 9 \cdot 10^k$ for an extremely large $k$. This gives us

\begin{align*}
P(9) = a_0+9a_1+9^2a_2 + \dots + 9^na_n &= S(a_0 + 9 \cdot 10^k a_1 + 9^2 \cdot 10^{2k}a_2+\dots 9^n10^{nk}a_n) \\
&= S(a_0)+S(9 a_1)+S(9^2a_2)+\dots + S(9^na_n).
\end{align*}
Since $S(n) \le n$, so we need $S(9^ia_i) = 9^ia_i$ for all $i$. Hence, $a_i = 0$ for all $i \ge 2$ and $0 \le a_1 \le 1$.

If $a_1=0$, obviously all values $0 \le a_0 \le 9$ work. Else, we plug in $n=2019$ to get

\[12+a_0 = S(2019+a_0),\]
which can obviously only be satisfied if $a_0 = 0$.
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dolphinday
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#45
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Let $n = 10^k$. Then we have $S(P(10^k)) = P(1)$. If $P$ has any negative coefficients, then $S(P(10^k))$ will increase as $k$ gets sufficiently large(there will be more $9$s in the digits).
Now, let $n = 10^k - 1$. Then $S(P(10^k - 1)) = P(9(k-1))$. Then say $P$ has degree $m$. Then as $k$ gets sufficiently large, $P(10^k-1)$ will have $(k-1)m$ digits, so $S(P(10^k-1)) \leq 9(k-1) = O(k)$. And $P(9(k-1)) = O(k^m)$. $O(k^m) > O(k)$. So eventually, $P(S(n))$ will be much larger than $S(P(n))$ if all of the coefficients are positive, unless $m \leq 1$. If $m = 1$ then $P(x) = x$(skipping this part). Then if $m = 0$ then $P(x) = \{1, \dots, 9\}$.
This post has been edited 3 times. Last edited by dolphinday, May 19, 2025, 5:03 AM
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Ilikeminecraft
734 posts
#46
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Let $P(n) = a_{k} n^k + a_{k - 1}n^{k - 1} + \dots + a_0.$ Suppose $e$ is an integer such that $\log_{10}a_i < e.$ Take $n = 10^{\ell}$ where $\ell\geq e.$ Now, suppose any of them were negative, if we send $\ell$ to some really really big number, we get a massive number of 9's in $P(n)$, but we clearly have $S(P(n)) = S(P(1))$ is constant. Hence, we have that all of $a_i \in \{0, 1, \dots, 9\}.$ Now, take $n = 9\cdot 10^\ell,$ and we get that $S(a_0) + S(9a_1) + \dots = a_0 + 9a_1 + \dots.$ However, by the inequality $S(x)\leq x,$ with equality iff $x < 10,$ we get that $P(x) = x + c$ or $P(x) = c.$ It is trivial to check that $c = 0$ doesn't work for first case.
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Shreyasharma
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#47
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For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Solution. Let $P(x) = \sum_{t = 0}^{m} a_tx^t$. Consider $n = 10^k - 1$ for $k$ large, so that we require,
\begin{align*}
S(P(10^k - 1)) = P(S(10^k - 1)).
\end{align*}The right hand side is near $P(9k) \approx a_m(9k)^m$. However we may bound the left hand side to be at most,
\begin{align*}
S(P(10^k - 1)) \approx S(a_m \cdot 10^{km}) \leq 9 \cdot \left[ \lfloor \log(a_m \cdot 10^{km} + 1) \rfloor + 1 \right] \leq 9 \cdot \left(\log(a_m) + km + 1 \right) \approx 9\log(a_m) + 9km.
\end{align*}However clearly $a_m(9k^m) >> 9 \log(a_m) + 9km$ for $m \geq 2$ by taking $k$ large so $m \in \{0, 1\}$. At this point direct casework yields the solution set.
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anudeep
203 posts
#48
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We claim that is either $P(x)=x$ or $P(x)=c$ with $c=1,\ldots, 9$, which are easy to verify.
Now, we shall show that these are the only solutions. The key insight to the problem is the fact that $S(n)$ is very small compared to $n$ for sufficiently large $n$ (more precisely, it is bounded above by a logarithmic function which grows very slowly) and the entire solution is more or less based on this observation. Firstly let's see what can we tell about $P$, can $\deg P$ be arbitrary? For convenience let Let $P(x)=\sum_{0\le i\le d}a_ix^i$ where $d=\deg P$.
Claim 1. $d\le 1$.
Proof. Using the fact that $S(n)\le 9\lceil \log_{10}n\rceil$, we have the following,
$$9\lceil \log_{10}P(\underbrace{99\cdots9}_{\text{$\ell$ $9$s}})\rceil\ge S(P(99\cdots 9))=P(S(99\cdots 9))=P(9\ell).$$For a sufficiently large $\ell$. Let's find a very convenient upper bound for $P$ , convenient meaning the one that behaves well inside logarithmic functions. A very natural one is
$$(d+1)\max_{0\le i\le d}\{a_i\}x^d\ge P(x)\qquad\text{when $x\ge 1$}.$$Call the messy constant in front of $x^d$ as $K$. Recall that $9\lceil \log_{10}P(99\cdots 9)\rceil\ge P(9\ell)$, so the bound simplifies as,
\begin{align*}
 9d\underbrace{\lceil \log_{10}99\cdots 9\rceil}_{\text{$\ell$}}+ 9\lceil \log_{10}K\rceil & \ge 9\lceil \log_{10}P(10^{\ell}-1)\rceil\\
& \ge \sum_{0\le i\le d}a_i(9\ell)^i.
\end{align*}It's not hard to see why the leading coefficient of $P$ must positive so taking sufficiently large $\ell$, the above does not hold when $d>1$ and hence the claim.

Now we are only left with the cases when $d=0,1$. The case when $d=0$ is trivial, so we are only interested in the other case. Using similar ideas as before, we know that $9\lceil \log_{10}2n\cdot\max\{a_0,a_1\}\rceil\ge S(n)a_1+a_0$. Again we put $n=99\cdots 9$ and we have,
\begin{align*}
9\lceil \log_{10}2\times 99\cdots9\rceil + 9\lceil \log_{10}\cdot\max\{a_0,a_1\}\rceil &=9\ell+9\lceil \log_{10}\cdot\max\{a_0,a_1\}\rceil \\
&\ge 9\ell a_1+a_0.
\end{align*}If $a_1\ne 1$ we have a contradiction by taking sufficiently large $\ell$, therefore $a_1=1$. Somehow we need to force that $a_0=0$. Suppose $a_0>0$ (the case when $a_0<0$ is easy to rule out), as $$S(n+a_0)=S(n)+a_0,$$take $n$ such that $n+a_0$ is an extremely large power of $10$, meanwhile making $S(n)$ arbitrarily large. But this seems a bit off as $S(n+a_0)=1$ throughout the process whereas $S(n)+a_0$ is just getting bigger and bigger unless $a_0=0$ and we are done. $\square$
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cursed_tangent1434
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#49
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First observe that the leading coefficient of $P$ must be positive since if $P$ is negative for all sufficiently large positive integers we have a contradiction by considering $n$ such that $S(n)$ is sufficiently large. Consider $n=10^k-1$ for sufficiently large positive integers $k$. Then,
\[P(S(n))=P(S(10^k-1))=P(9k)\]Since $P$ has a positive leading coefficient, for all sufficiently large $k$ we have $P(10^k-1)<P(10^k)$. Let $D(X)$ denote the number of digits of $X$ in base-ten (excluding leading zeros). Thus,
\[D(P(n))\le D(P(10^k)) = D(a_d10^{kd}+a_{d-1}10^{k(d-1)}+\dots + 10^ka_1 + a_0)\]Hence,
\begin{align*}
    S(P(n)) & \le 9D(P(n))\\
    &\le 9D(a_d10^{kd}+a_{d-1}10^{k(d-1)}+\dots + 10^ka_1 + a_0)\\
    &\le  9D(a_d)+9kd
\end{align*}for all sufficiently large positive integers $k$. Let $a_m$ denote the minimal coefficient of $P$. We have,
\begin{align*}
    P(9k) &= a_d(9k)^{d} + \dots + 9ka_1+a_0\\
    & \ge a_d - \left[|a_{d-1}|(9k)^{d-1}+\dots + |a_1|(9k)+|a_0|\right]\\
    & \ge a_d(9k)^d - |a_m|\left(\frac{(9k)^d-1}{9k-1}\right)
\end{align*}We wish to show,
\begin{align*}
    a_d(9k)^d & \ge |a_m|\left(\frac{(9k)^d-1}{9k-1}\right) + 9D(a_d)+9kd \\
    a_d(9k)^{d+1} &\ge a_d(9k)^d + |a_m|(9k)^d - |a_m| +  9D(a_d)+9kd
\end{align*}But this is clear since,
\begin{align*}
    a_d(9k)^{d+1}&= a_d(9k-1)(9k)^d + a_d(9k)^d \\
    & > (a_d+|a_m|)(9k)^d + a_d(9k)^d
\end{align*}for all sufficiently large positive integers $k$. Further, for all $d>1$
\begin{align*}
    a_d(9k)^d & > (d+1)(9k)a_d\\
    &= (9kd+9k)a_d\\
    & \ge 9kd + 9a_d\\
    & \ge 9kd+9D(a_d)
\end{align*}which proves the result. Hence, we must have $P$ being linear or constant. If $P(x) \equiv c$ for some constant $c$ we must have,
\[S(c)=c\]which implies that $1\le c<10$. Further, if $P(x) \equiv ax+b$ for $a>0$ we may consider $n=10^k$ for sufficiently large positive integers $k$. Then,
\[a+b = P(1)=P(S(10^k))=S(P(10^k))=S(\overline{a0\dots 0b})=S(a)+S(b)\]However, $S(a)+S(b) \le a+b$ with equality if and only if $0 \le a,b \le 9$. We are now left with a finite case check.
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