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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Plz help
Bet667   3
N 20 minutes ago by K1mchi_
f:R-->R for any integer x,y
f(yf(x)+f(xy))=(x+f(x))f(y)
find all function f
(im not good at english)
3 replies
Bet667
Jan 28, 2024
K1mchi_
20 minutes ago
2019 SMT Team Round - Stanford Math Tournament
parmenides51   17
N 30 minutes ago by Rombo
p1. Given $x + y = 7$, find the value of x that minimizes $4x^2 + 12xy + 9y^2$.


p2. There are real numbers $b$ and $c$ such that the only $x$-intercept of $8y = x^2 + bx + c$ equals its $y$-intercept. Compute $b + c$.



p3. Consider the set of $5$ digit numbers $ABCDE$ (with $A \ne 0$) such that $A+B = C$, $B+C = D$, and $C + D = E$. What’s the size of this set?


p4. Let $D$ be the midpoint of $BC$ in $\vartriangle ABC$. A line perpendicular to D intersects $AB$ at $E$. If the area of $\vartriangle ABC$ is four times that of the area of $\vartriangle BDE$, what is $\angle ACB$ in degrees?


p5. Define the sequence $c_0, c_1, ...$ with $c_0 = 2$ and $c_k = 8c_{k-1} + 5$ for $k > 0$. Find $\lim_{k \to \infty} \frac{c_k}{8^k}$.


p6. Find the maximum possible value of $|\sqrt{n^2 + 4n + 5} - \sqrt{n^2 + 2n + 5}|$.


p7. Let $f(x) = \sin^8 (x) + \cos^8(x) + \frac38 \sin^4 (2x)$. Let $f^{(n)}$ (x) be the $n$th derivative of $f$. What is the largest integer $a$ such that $2^a$ divides $f^{(2020)}(15^o)$?


p8. Let $R^n$ be the set of vectors $(x_1, x_2, ..., x_n)$ where $x_1, x_2,..., x_n$ are all real numbers. Let $||(x_1, . . . , x_n)||$ denote $\sqrt{x^2_1 +... + x^2_n}$. Let $S$ be the set in $R^9$ given by $$S = \{(x, y, z) : x, y, z \in R^3 , 1 = ||x|| = ||y - x|| = ||z -y||\}.$$If a point $(x, y, z)$ is uniformly at random from $S$, what is $E[||z||^2]$?


p9. Let $f(x)$ be the unique integer between $0$ and $x - 1$, inclusive, that is equivalent modulo $x$ to $\left( \sum^2_{i=0} {{x-1} \choose i} ((x - 1 - i)! + i!) \right)$. Let $S$ be the set of primes between $3$ and $30$, inclusive. Find $\sum_{x\in S}^{f(x)}$.


p10. In the Cartesian plane, consider a box with vertices $(0, 0)$,$\left( \frac{22}{7}, 0\right)$,$(0, 24)$,$\left( \frac{22}{7}, 4\right)$. We pick an integer $a$ between $1$ and $24$, inclusive, uniformly at random. We shoot a puck from $(0, 0)$ in the direction of $\left( \frac{22}{7}, a\right)$ and the puck bounces perfectly around the box (angle in equals angle out, no friction) until it hits one of the four vertices of the box. What is the expected number of times it will hit an edge or vertex of the box, including both when it starts at $(0, 0)$ and when it ends at some vertex of the box?


p11. Sarah is buying school supplies and she has $\$2019$. She can only buy full packs of each of the following items. A pack of pens is $\$4$, a pack of pencils is $\$3$, and any type of notebook or stapler is $\$1$. Sarah buys at least $1$ pack of pencils. She will either buy $1$ stapler or no stapler. She will buy at most $3$ college-ruled notebooks and at most $2$ graph paper notebooks. How many ways can she buy school supplies?


p12. Let $O$ be the center of the circumcircle of right triangle $ABC$ with $\angle ACB = 90^o$. Let $M$ be the midpoint of minor arc $AC$ and let $N$ be a point on line $BC$ such that $MN \perp BC$. Let $P$ be the intersection of line $AN$ and the Circle $O$ and let $Q$ be the intersection of line $BP$ and $MN$. If $QN = 2$ and $BN = 8$, compute the radius of the Circle $O$.


p13. Reduce the following expression to a simplified rational $$\frac{1}{1 - \cos \frac{\pi}{9}}+\frac{1}{1 - \cos \frac{5 \pi}{9}}+\frac{1}{1 - \cos \frac{7 \pi}{9}}$$

p14. Compute the following integral $\int_0^{\infty} \log (1 + e^{-t})dt$.


p15. Define $f(n)$ to be the maximum possible least-common-multiple of any sequence of positive integers which sum to $n$. Find the sum of all possible odd $f(n)$


PS. You should use hide for answers. Collected here.
17 replies
parmenides51
Feb 6, 2022
Rombo
30 minutes ago
orthocenter on sus circle
DVDTSB   1
N an hour ago by Double07
Source: Romania TST 2025 Day 2 P1
Let \( ABC \) be an acute triangle with \( AB < AC \), and let \( O \) be the center of its circumcircle. Let \( A' \) be the reflection of \( A \) with respect to \( BC \). The line through \( O \) parallel to \( BC \) intersects \( AC \) at \( F \), and the tangent at \( F \) to the circle \( \odot(BFC) \) intersects the line through \( A' \) parallel to \( BC \) at point \( M \). Let \( K \) be a point on the ray \( AB \), starting at \( A \), such that \( AK = 4AB \).
Show that the orthocenter of triangle \( ABC \) lies on the circle with diameter \( KM \).

Proposed by Radu Lecoiu

1 reply
+1 w
DVDTSB
Today at 12:18 PM
Double07
an hour ago
Incircle triangles inequality
MathMystic33   0
an hour ago
Source: 2025 Macedonian Team Selection Test P5
Let $\triangle ABC$ be a triangle with side‐lengths $a,b,c$, incenter $I$, and circumradius $R$. Denote by $P$ the area of $\triangle ABC$, and let $P_1,\;P_2,\;P_3$ be the areas of triangles $\triangle ABI$, $\triangle BCI$, and $\triangle CAI$, respectively. Prove that
\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]
0 replies
MathMystic33
an hour ago
0 replies
Collinearity of intersection points in a triangle
MathMystic33   0
an hour ago
Source: 2025 Macedonian Team Selection Test P1
On the sides of the triangle \(\triangle ABC\) lie the following points: \(K\) and \(L\) on \(AB\), \(M\) on \(BC\), and \(N\) on \(CA\). Let
\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]and let the line \(CS\) meet \(AB\) at \(Q\). Prove that the points \(P\), \(Q\), and \(R\) are collinear.
0 replies
MathMystic33
an hour ago
0 replies
Brazilian Locus
kraDracsO   15
N an hour ago by Ilikeminecraft
Source: IberoAmerican, Day 2, P4
Let $B$ and $C$ be two fixed points in the plane. For each point $A$ of the plane, outside of the line $BC$, let $G$ be the barycenter of the triangle $ABC$. Determine the locus of points $A$ such that $\angle BAC + \angle BGC = 180^{\circ}$.

Note: The locus is the set of all points of the plane that satisfies the property.
15 replies
kraDracsO
Sep 9, 2023
Ilikeminecraft
an hour ago
Circumcircle of MUV tangent to two circles at once
MathMystic33   0
2 hours ago
Source: Macedonian Mathematical Olympiad 2025 Problem 1
Given is an acute triangle \( \triangle ABC \) with \( AB < AC \). Let \( M \) be the midpoint of side \( BC \), and let \( X \) and \( Y \) be points on segments \( BM \) and \( CM \), respectively, such that \( BX = CY \). Let \( \omega_1 \) be the circumcircle of \( \triangle ABX \), and \( \omega_2 \) the circumcircle of \( \triangle ACY \). The common tangent \( t \) to \( \omega_1 \) and \( \omega_2 \), which lies closer to point \( A \), touches \( \omega_1 \) and \( \omega_2 \) at points \( P \) and \( Q \), respectively. Let the line \( MP \) intersect \( \omega_1 \) again at \( U \), and the line \( MQ \) intersect \( \omega_2 \) again at \( V \). Prove that the circumcircle of triangle \( \triangle MUV \) is tangent to both \( \omega_1 \) and \( \omega_2 \).
0 replies
MathMystic33
2 hours ago
0 replies
Roots of unity
Henryfamz   0
3 hours ago
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
0 replies
Henryfamz
3 hours ago
0 replies
A problem of collinearity.
Raul_S_Baz   0
3 hours ago
Î am the author.
IMAGE
P.S: How can I verify that it is an original problem? Thanks!
0 replies
Raul_S_Baz
3 hours ago
0 replies
Inequalities
nhathhuyyp5c   2
N 3 hours ago by alexheinis
Let $x,y$ be positive reals such that $3x-2xy\leq 1$. Find $\min$ \[
M = \frac{1 - x^2}{x^2} + 2y^2 + 3x + \frac{24}{y} + 2025.
\]

2 replies
nhathhuyyp5c
4 hours ago
alexheinis
3 hours ago
Aime 2005a #15
4everwise   22
N 3 hours ago by Ilikeminecraft
Source: Aime 2005a #15
Triangle $ABC$ has $BC=20$. The incircle of the triangle evenly trisects the median $AD$. If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n$.
22 replies
4everwise
Nov 10, 2005
Ilikeminecraft
3 hours ago
Nice geometry...
Sadigly   1
N 4 hours ago by aaravdodhia
Source: Azerbaijan Senior NMO 2020
Let $ABC$ be a scalene triangle, and let $I$ be its incenter. A point $D$ is chosen on line $BC$, such that the circumcircle of triangle $BID$ intersects $AB$ at $E\neq B$, and the circumcircle of triangle $CID$ intersects $AC$ at $F\neq C$. Circumcircle of triangle $EDF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. Lines $FD$ and $IC$ intersect at $Q$, and lines $ED$ and $BI$ intersect at $P$. Prove that $EN\parallel MF\parallel PQ$.
1 reply
Sadigly
Sunday at 10:17 PM
aaravdodhia
4 hours ago
AM=CN in Russia
mathuz   25
N 4 hours ago by Ilikeminecraft
Source: AllRussian-2014, Grade 11, day1, P4
Given a triangle $ABC$ with $AB>BC$, $ \Omega $ is circumcircle. Let $M$, $N$ are lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. $K(.)=MN\cap AC$ and $P$ is incenter of the triangle $AMK$, $Q$ is K-excenter of the triangle $CNK$ (opposite to $K$ and tangents to $CN$). If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.

M. Kungodjin
25 replies
mathuz
Apr 29, 2014
Ilikeminecraft
4 hours ago
Simson lines on OH circle
DVDTSB   2
N 4 hours ago by SomeonesPenguin
Source: Romania TST 2025 Day 2 P4
Let \( ABC \) and \( DEF \) be two triangles inscribed in the same circle, centered at \( O \), and sharing the same orthocenter \( H \ne O \). The Simson lines of the points \( D, E, F \) with respect to triangle \( ABC \) form a non-degenerate triangle \( \Delta \).
Prove that the orthocenter of \( \Delta \) lies on the circle with diameter \( OH \).

Note. Assume that the points \( A, F, B, D, C, E \) lie in this order on the circle and form a convex, non-degenerate hexagon.

Proposed by Andrei Chiriță
2 replies
DVDTSB
Today at 12:10 PM
SomeonesPenguin
4 hours ago
Hard number theory
td12345   7
N Apr 10, 2025 by td12345
Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
7 replies
td12345
Apr 9, 2025
td12345
Apr 10, 2025
Hard number theory
G H J
G H BBookmark kLocked kLocked NReply
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td12345
233 posts
#1
Y by
Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
This post has been edited 1 time. Last edited by td12345, Apr 14, 2025, 6:17 PM
Reason: variation of the problem
Z K Y
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mathprodigy2011
334 posts
#2
Y by
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
Z K Y
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td12345
233 posts
#3
Y by
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 1:57 AM
Reason: typo
Z K Y
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mathprodigy2011
334 posts
#4
Y by
td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned
Z K Y
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td12345
233 posts
#5
Y by
mathprodigy2011 wrote:
td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned

Yes, they lead to the same one when computing the square root, but the set counts the \(x\) not the set of the square root expression no? I got a different number from yours and I don't know where I went wrong, I will lay down the \(x\) I found as it goes over your claimed result and maybe one of us can figure out:

For \( M_2\):
- \(x\) can be \( (2^{2025-k}-2^{k-1})^2 \) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be \( -(2^{2 \cdot 2025-2k}+2^{2k-2}+2^{2025})\) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be $2^{2 \cdot 2025-3}+2-2^{2025},  -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. This leads to \(2\) extra values for \(x\) distinct from the above.

For \(M_{2027}\):
- \(x\) can be \( \frac{2027^k+2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2025\) distinct non-zero values of \(x\).
- \(x\) can be \( \frac{-2027^k-2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2026\) distinct non-zero values of \(x\).
I double checked with W|A to see if they lead to perfect squares under the square root and they do, so unless I double counted, the answer should be $2026+2+2025 + 2026$?


Edit: ok there is a miscount for \( M_2\) because when \(k=1013\) it leads to \(0\) for the very first family and \(x\) can't be \(0\). So for \(M_2\) we shall have $1012+1013+2=2027$. But for \(M_{2027}\) count seems solid? I got $2024+2+2025+2026$
This post has been edited 8 times. Last edited by td12345, Apr 10, 2025, 8:34 PM
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td12345
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Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?
This post has been edited 2 times. Last edited by td12345, Apr 10, 2025, 8:40 PM
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mathprodigy2011
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td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong
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mathprodigy2011 wrote:
td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong

Yes it's integers that are not 0. $\mathbb{N}$ is natural numbers (positive integers) and \( \mathbb{N}^*\) is natural numbers without \(0\). Would you like to give it another try just to see if it's really $8100$?
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 9:30 PM
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