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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
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
Cute Inequality
EthanWYX2009   0
8 minutes ago
Let $a_1,\ldots ,a_n\in\mathbb R\backslash\{0\},$ determine the minimum and maximum value of
\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]
0 replies
EthanWYX2009
8 minutes ago
0 replies
Interesting inequality
sealight2107   3
N 18 minutes ago by NguyenVanHoa29
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
3 replies
sealight2107
May 6, 2025
NguyenVanHoa29
18 minutes ago
Is this FE is solvable?
ItzsleepyXD   0
25 minutes ago
Source: Own , If not appear somewhere before
Find all function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in  \mathbb{R}$ . $$f(x+f(y))+f(x+y)=2x+f(y)+f(f(y))$$. Original
0 replies
ItzsleepyXD
25 minutes ago
0 replies
Symmetry in Circumcircle Intersection
Mimii08   2
N 37 minutes ago by mashumaro
Hi! Here's another geometry problem I'm thinking about, and I would appreciate any help with a proof. Thanks in advance!

Let AD and BE be the altitudes of an acute triangle ABC, with D on BC and E on AC. The line DE intersects the circumcircle of triangle ABC again at two points M and N. Prove that CM = CN.

Thanks for your time and help!
2 replies
Mimii08
3 hours ago
mashumaro
37 minutes ago
Blue chessboard
rcorreaa   10
N 40 minutes ago by Jaxman8
Source: 2022 Brazilian National Mathematical Olympiad - Problem 6
Some cells of a $10 \times 10$ are colored blue. A set of six cells is called gremista when the cells are the intersection of three rows and two columns, or two rows and three columns, and are painted blue. Determine the greatest value of $n$ for which it is possible to color $n$ chessboard cells blue such that there is not a gremista set.
10 replies
rcorreaa
Nov 22, 2022
Jaxman8
40 minutes ago
Substitution
JCE   3
N an hour ago by K124659
I've been working on this for about an hour or so, and I can't get this problem. I know the answer, but no idea on how to find it.
Please help?

2x-y^2=4
x^2+y=14
3 replies
JCE
May 27, 2006
K124659
an hour ago
Something nice
KhuongTrang   34
N 2 hours ago by TNKT
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
34 replies
KhuongTrang
Nov 1, 2023
TNKT
2 hours ago
Equation has no integer solution.
Learner94   34
N 2 hours ago by Ilikeminecraft
Source: INMO 2013
Let $a,b,c,d \in \mathbb{N}$ such that $a \ge b \ge c \ge d $. Show that the equation $x^4 - ax^3 - bx^2 - cx -d = 0$ has no integer solution.
34 replies
1 viewing
Learner94
Feb 3, 2013
Ilikeminecraft
2 hours ago
Polynomial of Degree n
Brut3Forc3   20
N 3 hours ago by Ilikeminecraft
Source: 1975 USAMO Problem 3
If $ P(x)$ denotes a polynomial of degree $ n$ such that $ P(k)=\frac{k}{k+1}$ for $ k=0,1,2,\ldots,n$, determine $ P(n+1)$.
20 replies
Brut3Forc3
Mar 15, 2010
Ilikeminecraft
3 hours ago
Really fun geometry problem
Sadigly   5
N 3 hours ago by GingerMan
Source: Azerbaijan Senior MO 2025 P6
In the acute triangle $ABC$ with $AB<AC$, the foot of altitudes from $A,B,C$ to the sides $BC,CA,AB$ are $D,E,F$, respectively. $H$ is the orthocenter. $M$ is the midpoint of segment $BC$. Lines $MH$ and $EF$ intersect at $K$. Let the tangents drawn to circumcircle $(ABC)$ from $B$ and $C$ intersect at $T$. Prove that $T;D;K$ are colinear
5 replies
Sadigly
Yesterday at 4:29 PM
GingerMan
3 hours ago
Polynomials and powers
rmtf1111   27
N 3 hours ago by bjump
Source: RMM 2018 Day 1 Problem 2
Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying
$$P(x)^{10}+P(x)^9 = Q(x)^{21}+Q(x)^{20}.$$
27 replies
rmtf1111
Feb 24, 2018
bjump
3 hours ago
Equilateral triangle formed by circle and Fermat point
Mimii08   0
4 hours ago
Source: Heard from a friend
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.

Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).

Prove that triangle A2B2C2 is equilateral.

0 replies
Mimii08
4 hours ago
0 replies
Problem 3 IMO 2005 (Day 1)
Valentin Vornicu   121
N 4 hours ago by Rayvhs
Let $x,y,z$ be three positive reals such that $xyz\geq 1$. Prove that
\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]
Hojoo Lee, Korea
121 replies
Valentin Vornicu
Jul 13, 2005
Rayvhs
4 hours ago
geo problem saved from graveyard
CrazyInMath   1
N 4 hours ago by Curious_Droid
Source: 3rd KYAC Math-A P5
Given triangle $ABC$ and orthocenter $H$. The foot from $H$ to $BC, CA, AB$ is $D, E, F$ respectively. A point $L$ satisfies that $\odot(LBA)$ and $\odot(LCA)$ are both tangent to $BC$. A circle passing through $B, E$ and tangent to $\odot(BHC)$ intesects $BC$ at another point $P$. $X$ is an arbitrary point on $\odot(PDE)$, and $Y$ is the second intesection point of $\odot(BXE)$ and $\odot(CXD)$.
Prove that $H, Y, L, C$ are concyclic.

Proposed by CrazyInMath.
1 reply
CrazyInMath
Feb 8, 2025
Curious_Droid
4 hours ago
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   51
N Apr 22, 2025 by Blast_S1
Source: IMO 2021 P5
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.

Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.
51 replies
popcorn1
Jul 20, 2021
Blast_S1
Apr 22, 2025
Bushy and Jumpy and the unhappy walnut reordering
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2021 P5
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JAnatolGT_00
559 posts
#41 • 1 Y
Y by cubres
Assume the opposite. Firstly paint each walnut white, and on $k-$th move paint $k-$th walnut black. By assumption each move swaps walnuts of same color, in particular it preserves parity of number of adjacent pairs of both black walnuts, which is $0$ at the beginning and $2021$ at the end, contradiction.
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PROA200
1748 posts
#42 • 3 Y
Y by sabkx, ike.chen, cubres
Wow! It has been a while since I first saw this problem, but for some reason, I never had the confidence to legitimately think about it.

Tag each walnut with a tag $t_i = 1$ at the beginning. The $1$s will represent the walnuts that have yet to be chosen as a value of $k$. Every move, we choose a walnut tagged with a $1$ and change the tag to a $0$, assuming for the sake of contradiction that the two walnuts adjacent to it are always both tagged with $1$s or $0$s (i.e., have either both not been chosen already or have both been chosen).

Of course, in each case, we have that the sum (indices mod $2021$)
\[\sum_i |t_i - t_{i+1}|\]changes by exactly $2$. But at the beginning, it's $0$, and at the end, it's $0$. This is a contradiction modulo $4$ -- after an odd number of moves, the sum should be $2\pmod 4$! This completes the proof.
This post has been edited 1 time. Last edited by PROA200, Mar 7, 2023, 6:43 AM
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ihatemath123
3446 posts
#43 • 1 Y
Y by cubres
Call a walnut that has been the center of a move a "changed" walnut.

For the sake of contradiction, assume that on every move, Jumpy swaps two walnuts that are either both less than the center one, or both greater. Then, Jumpy is either swapping two changed walnuts, or two yet-to-be-changed walnuts. Either way, one walnut becomes changed at each move, and the positions of the others remain constant.

Number each hole in clockwise order: $1, 2, \dots, 2021$. Because of what we established before, once a hole contains a changed walnut, it will always contain a changed walnut. Therefore, we may refer to each hole as "changed" or "unchanged" - each move is equivalent to picking a hole and changing it. Set, WLOG, walnut $1$ in hole $1$. Then, at some point, we must change hole $2$ - before we do that though, hole $3$ has to be changed. When we change hole $3$, holes $2$ and $4$ must both be unchanged, so we also must change hole $3$ before hole $4$. So when we change hole $4$, both holes $3$ and holes $5$ must be changed. This pattern continues, until we have that hole $2022$ must be changed after hole $2021$, which is a contradiction, since hole $2022$ is hole $1$, which we assumed was changed first.

Therefore, Jumpy must swap two walnuts $a$ and $b$ with $a < k < b$.
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huashiliao2020
1292 posts
#44 • 1 Y
Y by cubres
Assign a walnut a signature if it's been the center of a reordering the number 1, and 0 otherwise. AFTSOC Jumpy is swapping two walnuts that are both less or greater than the indice of the center walnut; in particular, Jumpy swaps two walnuts with the same signature, hence a hole never changes it's signature. WLOG set walnut 1 in hole 1 by shifting hole \#s; when we change hole 2, we must have that hole 3 had a signature of 1, since hole 1 has a signature of 1, whence hole 4 has a signature of 0 before changing hole 3, etc. We continue this way until we get that hole 2022=1 must have a signature of 0 before changing hole 2021, contradiction since we gave hole 1 a signature of 1 first. We conclude. :surf:

Remark. Signature seems like such a common word to assign now, I've seen it in several problems lol
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pi271828
3369 posts
#45 • 1 Y
Y by cubres
We assume the opposite for the sake of contradiction. Label a hole $k$ black on the kth move when it is chosen, and keep it black for the remaining of the process. Keep a hole white if it hasn't been chosen yet. Notice that on each move we should be swapping holes of the same color, and obviously we will turn the middle hole from white to black. Also note that once a slot holds a black hole, it must always hold a black hole otherwise the problem is proved. Now, simply consider an arc of the circle that is $B\underbrace{WW \dots WW}_{x}B$. The color of the endpoints of this arc can never be changed otherwise the problem is proved, so we shift our attention to the inner part. We prove by strong induction that if $2 \mid x$, then there must be a move that swaps two holes of opposite color. The base case of $x = 2$ is clear. Now assume that for an even $n$, that all evens less than $n$ follow the hypothesis. One of the inner $W$'s will be chosen and will be turned to a $B$. Now, we have $B\underbrace{W \dots W}_{a}B\underbrace{W \dots W}_{b}B$, where $a+b$ is odd, implying that one of $a$ and $b$ is even, so we have completed induction. Now, notice that after the first two moves, there must be an arc that will contain an even number of $W$'s that connects the two black holes because $2021$ is odd, so we are done.
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Shreyasharma
682 posts
#46 • 2 Y
Y by eggymath, cubres
Assume for the sake of contradiction otherwise. Rephrase the problem in terms of the color of a walnut. Color the $k$-th walnut red on the $k$-th turn. Note that as we swap two walnuts of the same color on the $k$-th term by assumption, the coloring remains the same.

Clearly we begin with $0$ pairs of adjacent red walnuts and end with $2021$ pairs of adjacent red walnuts. Now on any given turn a walnut that changes color has two neighbors of the same color, so the pairs of adjacent red walnuts either goes up by $2$ or by $0$, a contradiction.
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RGB
19 posts
#47 • 2 Y
Y by eggymath, cubres
Consider the process that happens after Jumpy rearranges the walnuts.

Initially, each walnut has a unique position around the circular pattern of holes. After Jumpy's sequence of moves, some walnuts might return to their original positions, while others might have moved around.

Now, let's imagine that none of the swaps involve a pair of walnuts where one walnut has a number smaller than $k$ and the other walnut has a number larger than $k$. This means that for every swap that occurs during Jumpy's sequence of moves, either both walnuts have numbers smaller than $k$, or both walnuts have numbers larger than $k$.

Let's consider the total displacement of all the walnuts that have numbers smaller than $k$ after Jumpy's moves. Similarly, let's consider the total displacement of all the walnuts that have numbers larger than $k$ after Jumpy's moves.

The sum of displacements of walnuts smaller than $k$ and walnuts larger than $k$ around the circle must be equal in magnitude but opposite in direction (due to the circular nature of the arrangement). This is because each swap results in a net displacement of $0$ for the overall circle, as each walnut returns to its original position after a full circle.

If there were no swap that involves a pair of walnuts where one walnut has a number smaller than $k$ and the other walnut has a number larger than $k$, then the sum of displacements of walnuts smaller than $k$ and the sum of displacements of walnuts larger than $k$ would have to be different, leading to a contradiction.

Therefore, there must be a swap involving a pair of walnuts $a$ and $b$ such that $a<k<b$ during Jumpy's sequence of moves.
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cj13609517288
1915 posts
#48 • 2 Y
Y by eggymath, cubres
Suppose that after the $m$-th move, walnut $m$'s value becomes $0$. If we can find a value of $k$ that works here, it must also work in the original problem as if $m<k$, then $0<k$, and if $m>k$, then $m$ hasn't turned into $0$ yet. Therefore, we now want to show that $0$ and something $\ne 0$ swapped during the process(call this a "forbidden move").

Suppose not. After the first move, there is a "contiguous chain" of $2020$ nonzero numbers. We can never have an end of a chain perform their swap unless the chain is of length $1$.

Claim. An even chain will always cause a forbidden move.
Proof. Let the chain be of length $2n$. We will strong induct on $n$. Base case $n=1$ is trivial, so assume that $1,2,\dots,n-1$ work. Then on the first move of that chain, the chain will be split into two parts of length $a$ and $2n-1-a$, one of which will have a smaller even length, which will cause a forbidden move.

Thus our chain of $2020$ will cause a forbidden move, contradiction. $\blacksquare$
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lnzhonglp
120 posts
#49 • 2 Y
Y by eggymath, cubres
Suppose otherwise. On the $k$th turn, color the $k$th walnut red. Each walnut we color red must be between two walnuts of the same color. The number of pairs of adjacent red walnuts changes by $2$ or $0$, but we must go from $0$ to $2021$. Therefore, this is impossible.
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OronSH
1738 posts
#50 • 4 Y
Y by megarnie, eggymath, ihatemath123, cubres
Consider the same operation, but after a swap you add $2021$ to the label of the walnut just operated on. Every swap now satisfies $k<a,b.$ We want to then show that there is some swap that swaps a number $\le 2021$ and a number $>2021.$ This is doable iff at every step, between every two consecutive $>2021$ walnuts there is an odd number of $\le 2021$ walnuts (by induction), but this is false after the first step, since there are $2020$ walnuts between the one with label $2022$ and itself.
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dolphinday
1324 posts
#51 • 2 Y
Y by eggymath, cubres
At the start, paint all walnuts white and then paint walnut $k$ on the $k$th turn black.
Then notice that the condition $a < k < b$ is equivalent to swapping two walnuts of differing colors. So then FTSOC assume that throughout, we always swap two walnuts of the same color.
However notice that the number of adjacent pairs of black walnuts is always $0 \pmod 2$ as we can only add a black walnut in between two black walnuts or in between two white walnuts. This is a contradiction as our final state with $2021$ black walnuts has an odd number, so we are done.
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HamstPan38825
8860 posts
#52 • 1 Y
Y by cubres
I found this very straightforward for IMO5? This is one of those ``quasi-invariant" problems where simplifying the structure inexplicably yields a neat solution.

Call an acorn large if we have already performed a move on it and small otherwise. Assume for the sake of contradiction that there exists an ordering for which the condition is not true. Let the dynamic variable $k$ denote the number of groups between two consecutive large acorns that have an even number of small acorns between them.

Claim. The parity of $k$ is constant.

Proof. In any move that does not satisfy the condition, one small acorn surrounded by either two large or two small acorns is converted into a large acorn. If the acorn were originally part of an odd group, the first case eliminates this odd group, and the second case splits the odd group into two even or two odd groups. If the acorn were originally part of an even group, the first case is impossible, and the second case splits the even group into one even and one odd group. In both cases, the parity of $k$ does not change. $\blacksquare$

But after the first move, $k = 1$ is odd, and after all moves are complete, all acorns are large, so $k=0$. This yields a contradiction.
This post has been edited 1 time. Last edited by HamstPan38825, Jun 15, 2024, 12:18 AM
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Martin2001
152 posts
#53 • 1 Y
Y by cubres
Rephrase the problem as so: Let there be $2021$ acorns arranged in a circle, and for each move we can color exactly one acorn red. We show that it is impossible to completely color all acorns red without coloring an acorn red where the immediately adjacent acorns are different colors.
We show that it is impossible for a string with an even number of acorns, where both endpoints of the string start as red. Note that this is an easy induction starting from where there are $4$ acorns. Because the original circle of $2021$ acorns can be arranged as a string with $2022$ acorns with both endpoints starting as red, we're done$.\blacksquare$
This post has been edited 1 time. Last edited by Martin2001, Aug 26, 2024, 1:04 AM
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Mathandski
757 posts
#54 • 1 Y
Y by cubres
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Blast_S1
358 posts
#55 • 1 Y
Y by cubres
Let $f(n)$ be $1$ if both neighbors of $n$ are greater, $-1$ if both are smaller, and $0$ otherwise. Since the sum of $f(n)$ over all $n$ is $0$, and since $2021$ is odd, it follows that there exist an odd number of $n$ such that $f(n) = 0$.

Consider the $k^\text{th}$ move. If $f(k) = 0$, then we're done. If $f(k) = 1$, then the move only affects $f(n)$ for $n > k$. If $f(k) = -1$, then the move only affects $f(n)$ for $n < k$. The parity of the number of $n > k$ for which $f(n) = 0$ is invariant in the latter two cases. The end.
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