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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
trigonometric inequality
MATH1945   9
N 33 minutes ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
9 replies
MATH1945
May 26, 2016
sqing
33 minutes ago
Wrong Answer on a Street Math Challenge
miguel00   2
N an hour ago by miguel00
Hello AoPS Community,

I was just watching this video link (those of you that are Korean, you should watch it!) but I came across a pretty hard vector geometry problem (keep in mind contestants have to solve this problem in 5 minutes). No one got this problem (no surprise there) but I am posting because I actually think the answer he gave is wrong.

So the problem goes like this: Referencing the diagram attached, there are three externally tangent circles $C_1, C_2, C_3$ on a plane with centers $O_1, O_2, O_3$, respectively. $H$ is feet of the perpendicular from $O_1$ to $O_2O_3$ and $A$ and $B$ are intersections of line $O_1H$ with circle $C_1$. Points $P$ and $Q$ can move around the circle $C_2$ and $C_3$, respectively. Find the maximum possible value of $|\overrightarrow{AQ}+\overrightarrow{PB}|$.


I got my answer but the video said their 1st answer but they later corrected it on the comments to their 2nd answer. I'll let you guys attempt the problem and will give my solution shortly after. Thanks in advance!

-miguel00
2 replies
miguel00
an hour ago
miguel00
an hour ago
Tangents inducing isogonals
nikolapavlovic   56
N an hour ago by Ilikeminecraft
Source: Serbian MO 2017 6
Let $k$ be the circumcircle of $\triangle ABC$ and let $k_a$ be A-excircle .Let the two common tangents of $k,k_a$ cut $BC$ in $P,Q$.Prove that $\measuredangle PAB=\measuredangle CAQ$.
56 replies
nikolapavlovic
Apr 2, 2017
Ilikeminecraft
an hour ago
Elementary Problems Compilation
Saucepan_man02   25
N an hour ago by trangbui
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
25 replies
Saucepan_man02
Monday at 1:44 PM
trangbui
an hour ago
partitioning 1 to p-1 into several a+b=c (mod p)
capoouo   5
N an hour ago by NerdyNashville
Source: own
Given a prime number $p$, a set is said to be $p$-good if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod{p}$.
Find all prime number $p$ such that $\{ 1, 2, \cdots, p-1 \}$ can be partitioned into several $p$-good sets.

Proposed by capoouo
5 replies
capoouo
Apr 21, 2024
NerdyNashville
an hour ago
Not homogenous, messy inequality
Kimchiks926   11
N an hour ago by Learning11
Source: Latvian TST for Baltic Way 2019 Problem 1
Prove that for all positive real numbers $a, b, c$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ the following inequality holds:
$$3(ab+bc+ca)+\frac{9}{a+b+c} \le \frac{9abc}{a+b+c} + 2(a^2+b^2+c^2)+1$$
11 replies
Kimchiks926
May 29, 2020
Learning11
an hour ago
Problem 5
blug   2
N an hour ago by Jjesus
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
2 replies
+1 w
blug
May 19, 2025
Jjesus
an hour ago
D,E,F are collinear.
TUAN2k8   0
an hour ago
Source: Own
Help me with this:
0 replies
TUAN2k8
an hour ago
0 replies
Introducing a math summer program for middle school students
harry133   0
2 hours ago
Introducing IITSP, an online math summer program designed for middle school students over summer.

The program is designed by Professor Shubhrangshu Dasgupta from the Department of Physics at the Indian Institute of Technology Ropar (IIT Ropar).

Please check out the webpage if you are interested in.

https://www.imc-impea.org/IMC/bbs/content.php?co_id=iitsp
0 replies
harry133
2 hours ago
0 replies
NT Game in Iran TST
M11100111001Y1R   6
N 2 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 2
Suppose \( p \) is a prime number. We have a number of cards, each of which has a number written on it such that each of the numbers \(1, \dots, p-1 \) appears at most once and $0$ exactly once. To design a game, for each pair of cards \( x \) and \( y \), we want to determine which card wins over the other. The following conditions must be satisfied:

$a)$ If card \( x \) wins over card \( y \), and card \( y \) wins over card \( z \), then card \( x \) must also win over card \( z \).

$b)$ If card \( x \) does not win over card \( y \), and card \( y \) does not win over card \( z \), then for any card \( t \), card \( x + z \) must not win over card \( y + t \).

What is the maximum number of cards such that the game can be designed (i.e., one card does not defeat another unless the victory is symmetric or consistent)?
6 replies
M11100111001Y1R
Yesterday at 6:19 AM
sami1618
2 hours ago
Brilliant Problem
M11100111001Y1R   1
N 2 hours ago by aaravdodhia
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
1 reply
M11100111001Y1R
Yesterday at 7:28 AM
aaravdodhia
2 hours ago
Problem 3
blug   2
N 3 hours ago by LeYohan
Source: Polish Junior Math Olympiad Finals 2025
Find all primes $(p, q, r)$ such that
$$pq+4=r^4.$$
2 replies
blug
Mar 15, 2025
LeYohan
3 hours ago
2 headed arrows usage
mathprodigy2011   1
N 3 hours ago by alcumusftwgrind
Source: 2003 USAMO 4
I can't upload the file but I was working with someone on 2003 USAMO p4. When we saw "if and only if" I thought it meant we have to prove it both directions. However, when we looked at Evan Chen's solution after writing it out; Evan Chen used double headed arrows and left it at that. My question is, how did he use them and how do I know when I can or can not use them?
1 reply
mathprodigy2011
3 hours ago
alcumusftwgrind
3 hours ago
4th grader qual JMO
HCM2001   42
N Yesterday at 6:58 PM by BS2012
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
42 replies
+1 w
HCM2001
May 22, 2025
BS2012
Yesterday at 6:58 PM
sussy baka stop intersecting in my lattice points
Spectator   24
N Apr 29, 2025 by ilikemath247365
Source: 2022 AMC 10A #25
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

IMAGE

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
24 replies
Spectator
Nov 11, 2022
ilikemath247365
Apr 29, 2025
sussy baka stop intersecting in my lattice points
G H J
Source: 2022 AMC 10A #25
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Spectator
657 posts
#1 • 1 Y
Y by megarnie
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

[asy]
//kaaaaaaaaaante314
size(8cm);
import olympiad;
label(scale(.8)*"$y$", (0,60), N);
label(scale(.8)*"$x$", (60,0), E);
filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white);
label(scale(1.3)*"$R$", (55/2,55/2));
filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white);
label(scale(1.3)*"$S$",(-14,14));
filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white);
label(scale(1.3)*"$T$",(3.5,25/2));
draw((0,-10)--(0,60),EndArrow(TeXHead));
draw((-34,0)--(60,0),EndArrow(TeXHead));[/asy]

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
This post has been edited 3 times. Last edited by Spectator, Dec 24, 2022, 12:54 PM
Reason: Asy
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peelybonehead
6290 posts
#3
Y by
You guys had time to solve P25?!!?
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iamhungry
149 posts
#4 • 22 Y
Y by Paul10, balllightning37, bobthebuilder1234, andrewwang2623, bestzack66, michaelwenquan, Ritwin, ivyshine13, eibc, aidan0626, Lamboreghini, plang2008, mahaler, mathboy100, megarnie, mathmax12, spiritshine1234, the_mathmagician, EpicBird08, aidensharp, akliu, Sedro
I looked at this problem, saw colored squares and the word "lattice points" and immediately went back to the previous problems
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Snowyowl2005
57 posts
#5 • 8 Y
Y by Spectator, wamofan, iamhungry, Mogmog8, Lamboreghini, cosmicgenius, Aopamy, centslordm
Solution at the end of https://randommath.app.box.com/s/fnajq15ew2gvx0bmt8ucri8karksulyo.

Suppose that the square $R$ has $r^2$ lattice points (i.e. has coordinates ranging from $(0, 0)$ to $(r - 1, r - 1)$). Similarly, suppose that the square $S$ has $s^2$ lattice point (i.e. has coordinates ranging from $(0, 0)$ to $(1 - s, s - 1)$).
Then $\frac{r^2}{s^2} = \frac{9}{4}$, and so $r = 3x$ and $s = 2x$ for some positive integer $x$.
Furthermore, the union of $R$ and $S$ contains $r^2 + s^2 - s$ lattice points since they share $s$ lattice points of the form $(0, 0), (0, 1), (0, 2), \ldots, (0, s - 1)$. Therefore:
\[t^2 = \frac{r^2 + s^2 - s}{4} = \frac{13x^2 - 2x}{4}\]and so:
\[13x^2 - 2x = 4t^2\]Since the right hand side is even, $13x^2$ is even, and so we can write $x = 2y$ for some integer $y$. Thus:
\[52y^2 - 4y = 4t^2\]\[13y^2 - y = y(13y - 1) = t^2\]But by the Euclidean algorithm, $y$ and $13y - 1$ are relatively prime, and so both $y$ and $13y - 1$ must be perfect squares. Write:
\[y = m^2\]\[13y - 1 = 13m^2 - 1 = n^2\]Suppose that the top left corner of $T$ is given by $(1 - k, 0)$. Then the top right corner must be given by $(t - k, 0)$. We have:
\[\frac{|S \cap T|}{|S|} = 27 \cdot \frac{|R \cap T|}{|R|}\]where $|A|$ denotes the number of lattice points in a region $A$. Plugging in the relevant values, we get:
\[\frac{k \cdot t}{s^2} = 27 \cdot \frac{(t - k + 1) \cdot t}{r^2}\]Cross-multiplying, we get:
\[k \cdot t \cdot r^2 = 27 \cdot (t - k + 1) \cdot t \cdot s^2\]Simplifying and using $r^2 = \frac{9}{4}s^2$, we get:
\[9 \cdot k = (t - k + 1) \cdot 4 \cdot 27\]\[12 \cdot (t - k + 1) = k\]\[12 \cdot (t + 1) = 13k\]Therefore, for any $t \equiv 12 \pmod{13}$, we have a solution for $k$. But:
\[y(13y - 1) \equiv -y \equiv t^2 \equiv 1 \pmod{13}\]and so $y \equiv 12 \pmod{13}$. Since $y = m^2$, we have:
\[m^2 + 1 \equiv 0 \pmod{13}\]and that $13m^2 - 1$ is a perfect square. The smallest $m$ satisfying both conditions is $m = 5$. Plugging this in, we get:
\[y = m^2 = 25\]\[x = 2y = 50\]and so $r = 3x = 150$, $s = 2x = 100$, and $t^2 = y(13y - 1) = 25 \cdot 324$, so $t = 5 \cdot 18 = 90$, and:
\[(r - 1) + (s - 1) + (t - 1) = 150 + 100 + 90 - 3 = 340 - 3 = 337\]so the answer is B.
This post has been edited 2 times. Last edited by Snowyowl2005, Nov 11, 2022, 6:09 PM
Reason: The problem is asking for the sum of the edge lengths, which are $r - 1, s - 1, t - 1$, instead of $r + s + t$. It has been fixed above.
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mathboy100
675 posts
#6
Y by
Can we use Pick's theorem on this?
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bryanguo
1032 posts
#7
Y by
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$
This post has been edited 1 time. Last edited by bryanguo, Nov 11, 2022, 5:35 PM
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ihatemath123
3449 posts
#8
Y by
Uhhh what would you need picks for here
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Spectator
657 posts
#9
Y by
can someone include asy code for this
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Snowyowl2005
57 posts
#10 • 1 Y
Y by bryanguo
bryanguo wrote:
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$

The only error seems to be the answer extraction - it has been fixed above.
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HamstPan38825
8868 posts
#11
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This problem is essentially how quickly you can do number theory.

Let the side length of $R$ be $x$, so the number of lattice points in $R$ is $(x+1)^2$, the number of lattice points in $S$ is $\frac 49(x+1)^2$. It follows that the side length of $S$ is $\frac 23 x - \frac 13$.

Now, let $a$ and $b$ be the lengths of the portions of the sides of $T$ in $S$ and $R$, respectively. We have $$\frac{(a+1)(k+1)}{\frac 49(x+1)^2} = 27 \cdot \frac{(b+1)(k+1)}{(x+1)^2}$$by the third condition, so $a = 12b+11$, and the side length $k$ of $T$ is of the form $13b+11$ for some $b$.

Now, using the second condition, $$\frac{(k+1)^2}{\frac {13}9(x+1)^2 - \frac 23(x+1)} = \frac 14.$$Note Then $$36(13b+12)^2 = (x+1)(13x+7).$$Notice that $\gcd(x+1, 13x+7) = \gcd(x+1, 6) = 6$, so actually
\begin{align*}
x+1 &= 6r_1^2 \\
13x+ 7 &= 6r_2^2
\end{align*}for some positive integers $r_1, r_2$. These $r_1, r_2$ satisfy $$13r_1^2 - r_2^2 = 1.$$At this point, there isn't really a better way than to test out values for $r_1$, as we know by the answer choices that the answers cannot be too big. Indeed, $(r_1, r_2) = (5, 18)$ is a solution. Then, $x = 149$, so $$(k+1)^2 = \frac{150 \cdot 1944}{36} = 324 \cdot 25,$$which yields $k = 89$. Finally, the last side length is $99$, so the answer is $89+99+149 = \boxed{337}$.
This post has been edited 2 times. Last edited by HamstPan38825, Nov 11, 2022, 7:32 PM
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Somersett
71 posts
#12
Y by
peelybonehead wrote:
You guys had time to solve P25?!!?

Bruh this test went by so fast i didn’t even get to look at this problem
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Math_Shisa
158 posts
#13
Y by
All the solutions look so long I don’t think I would have time to even answer this one.
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saturnrocket
1306 posts
#14
Y by
I didn't even get to 15
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Dino76
707 posts
#15 • 1 Y
Y by Mango247
That's a lot better than me. I didn't even bother to take it...
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qwerty123456asdfgzxcvb
1087 posts
#16
Y by
me when not 340
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mannshah1211
652 posts
#17
Y by
if There is one more Baka in AMC Titles, i will Riot.
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Bluesoul
899 posts
#19
Y by
Not necessary to solve exact R,S,T, just compute R+S+T mod 13

Denote the side lengths of $R,S,T$ as $r,s,t$ respectively. According to the first condition, we have $\frac{9}{4}(s+1)^2=(r+1)^2, 3s+1=2r$

Second condition yields $4(t+1)^2=(s+1)^2+(r+1)^2-(s+1)$. Plug $r=\frac{3s+1}{2}$, we get $16(t+1)^2=13(s+1)^2-4(s+1)=(s+1)(13s+9)$

Let the length of $T$ in negative x-axis as $x$, the third condition implies $\frac{(x+1)(t+1)}{(s+1)^2}=27\frac{(t+1)(t-x+1)}{(r+1)^2}$ simplify this to $x+1=12(t-x+1), 12t+11=13x$. From here, we can attain $t\equiv -2\pmod{13}$. Consider equation $16(t+1)^2=(s+1)(13s+9)$, we have $(s+1)(13s+9)\equiv 9(s+1)\equiv 3\pmod{13}$

Thus, we have $s\equiv 8\pmod{13}$

Finally, we have $r=\frac{3s+1}{2}$ which yields $r\equiv 6\pmod{13}$. Consequently, $r+s+t\equiv 8+6-2\equiv 12\pmod{13}$ while only $\boxed{B}$ satisfies.
This post has been edited 1 time. Last edited by Bluesoul, Oct 25, 2023, 12:05 AM
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MathyMathMan
250 posts
#20
Y by
mannshah1211 wrote:
if There is one more Baka in AMC Titles, i will Riot.

Bouta say, but what is that title? :huuh:
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Amkan2022
2023 posts
#21
Y by
.........
This post has been edited 1 time. Last edited by Amkan2022, Apr 29, 2025, 4:07 AM
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IbrahimNadeem
888 posts
#22
Y by
aops being wild today
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IbrahimNadeem
888 posts
#23
Y by
why am I getting an irrational side for T?
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ESAOPS
263 posts
#26
Y by
oops
sol
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xHypotenuse
787 posts
#27
Y by
This feels harder than average aime #12s and I lwk have 0 clue on how to approach it
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MC_ADe
183 posts
#28
Y by
write out each condition and turn it into different forms, since you know each is an integer you can use divisibility,modulo, euclidean algorithm and other methods to find the values
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ilikemath247365
261 posts
#29
Y by
I was actually able to solve this problem......except it took me like 30 minutes. :rotfl:
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