competition-math

[hallow seal]

of the intermediate challenge?

[deft wraith]

Yes

Why do you ask?

[lusty tulip]

I have a mock of JEE Mains pattern exam on 30th someone pls help me
How can I prepare this stuff till then

Permutation and Combination - I can't even do a single question even after watching a 6 hours lecture
Binomial Theorem - I have to watch and practice 3 hours of more lecture
Parabola, Ellipse - I am kind of bad at this stuff I prefer not to watch lectures on it and just learn the formulae and its backstory and then just practice questions

Give more priority to binomial and parabola, ellipse as P&C came in my previous exam too and it will have less questions of that

[timber yacht]

Warm up ur brains a little:

Lim x --> e
f(x) = [(√(x - e^(x)⁴)) × (√(e^(x) - x)⁴)] / [( ln(x²) - x⁰)² + ( x⁰ - ln(x²) )]

[dry quartz]

cengage illustrations
should be enough

do the back ex. for parabola and ellipse

[sand hull]

Jee is easy dude how can you struggle with it if you ever did maths properly

[deft wraith]

That comment isn’t helpful

[rigid vault]

3

[leaden oyster]

Ok plz help

[leaden oyster]

a and b are real number prove that the roots of the equation …
are the vertices of an equilateral triangle in the complex plane.

[green thunder]

im curious to see the explanation im terrible at math

[ornate blade]

<@&268886789983436800>

[ornate blade]

maybe try subbing in $z \mapsto z - a/3$

[gilded halo]

south

[ornate blade]

ah it's perfect, like you just need z^3 = constant if you know complex numbers

[boreal sky]

How many positive integer solution to n^3+1=p^2 and how to prove it?

[hallow seal]

im gonna do it for the first time in a few days. do you have any tips

[split ore]

hi

i want help in this problem

[ornate blade]

can you factorise the top equation?

[split ore]

yeh

[ornate blade]

(x - y)(x - 2y) = 0 right? that implies x = y or x = 2y

so you have y^2 + y * y = 6 or (2y)^2 + (2y)y = 6

[split ore]

Thanka

[ornate blade]

np!

[split ore]

you're solve right

[ornate blade]

I can't read Arabic

the question asks you for all possible values of x, y that satisfy these two equations, right?

[split ore]

but how you factorise the top equation?

[ornate blade]

ah, okay so from x^2, we must have (x ....)(x ....) = 0

[split ore]

She is asking to solve the system of equations. )almost true(

[ornate blade]

and from the 2y^2 we could have (..... - 2y)(.... -y) or (..... + 2y)(... + y) or something

[ornate blade]

so yeah

[split ore]

she>it

[ornate blade]

then the middle term is negative so we guess that we need -2y and -y

putting everything together check that (x - 2y)(x - y) has middle term -3xy

alternatively, divide everything by y^2

$x^2 /y^2 - 3x/y + 2 = 0 \implies (x/y)^2 - 3(x/y) + 2 = 0$

sub $u = x/y$ and you just need to factorise $u^2 - 3u + 2 = 0$

[gilded halo]

south
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[split ore]

you're good person

[ornate blade]

thanks!

afwan brother (I only know a few phrases)

[split ore]

afwan brother ( so fun mix about arabic and english)

[quaint fog]

Does anybody know about the UKMT challenge?

[deft wraith]

You can find past papers on the UKMT website

If you do a few you should do well; you don’t need to know any theory

The questions are in order of difficulty so make sure to try the later questions, and when you get a solution wrong (and even if you don’t), there are additional related questions on the solutions which you should attempt

Good luck!

[deft wraith]

Refer to above

[quaint fog]

I have it 29 January but I have no idea how these tests are even conducted

Are these conducted in school or somewhere else?

and is this like a British version of Olympiad or something?

[deft wraith]

Is it the intermediate maths challenge?

If so, it’s in school

[neon island]

I had to guess the solutions:

(x,y)
(0,0)
(2,1)

The two solutions were

(x,x) = (0,0)
(2y,y) = (2,1)

Actully solution is this as well:

(sqrt(3), sqrt(3)) plus and minus

[hardy cipher]

Hi

I've been having some problem trying to solve this question. Can someone help me please

[pallid tundra]

!status

[scarlet root]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[hardy cipher]

1

[pallid tundra]

sums of two squares

think pythagorean triples?

[hardy cipher]

So.. is the answer just 2 ? (2+1)^2+4^2=(2+2)^2+3^2 ?

but how can I prove that this is the maximum of n?

[vague temple]

Try complex number maybe

[hardy cipher]

i've never use complex number for analysis how am I supposed to do it ?

[vague temple]

Ok then consider the consecutive difference of $y_{k+1}^2-y_k^2$

[gilded halo]

victor

[hardy cipher]

the consecutive difference between that is -2(x+k)+1 or the negative of consecutive odd number

then what should I do ?

[quaint fog]

oh alright. Thankyou!

[hallow seal]

all of them are in the school. they were for me. i did the junior one aswell

[quaint fog]

ok

are they hard?

[hallow seal]

well i got a silver in year 8 and a bronze in year 7

but ye some of the questions are hard but its multiple choice so if you dont know the answer just guess and you have a 20 percent chance of getting it right because there are 5 possible answers

[split ore]

Thank Bro

[lilac niche]

Woah

[hardy cipher]

Yeah this thing appeared on my school competition

[lilac niche]

School competition this looks like a higher level math problem

[hardy cipher]

Noo. it's from my school competition last year most of the problem is pretty easy but some of it is quite hard

[hallow seal]

is the pre uni chanel for sixth form?

[deft wraith]

Yes

[deft wraith]

Isn’t there negative marking

Or is that only smc

[hallow seal]

oh ye there is

forgot about that

[deft wraith]

Just practice question 15-25

If you can get silver you’re probably doing fine on the ones before that

[ivory ember]

You can Google this

https://artofproblemsolving.com/community/c6h1334377p7210502

Part d

tldr get a contradiction mod 4

[hardy cipher]

thanks

[final basin]

so it means that 1+1=3?

[gilded halo]

victor

[vague temple]

Did I make a mistake somewhere

[final basin]

idk

just asking

[vague temple]

Sorry, I don't understand your question

[final basin]

i want to know if 1+1=3

[vague temple]

Idk depends on your axioms

[final basin]

2+2=5, is that right?

[vague temple]

Wait my argument is wrong discard it

[final basin]

k

[vague temple]

You bash casework mod 4 I got confused at some point, it ends up being the same as the provided solution

[final basin]

math is hard, i dont know if 1+1=2 or 3

my brain is overheating

helelp

@vague temple

[hallow seal]

i just realised something

questions 1-15 are 5 marks each so thats a total of 75 marks and questions 16-25 are 6 marks each so thats a total of 60 marks each and questions 1-15 are also easier than questions 16-25

[deft wraith]

yes

but if you want gold/ qualify for olympiad you're going to need to do some of those 16-25 questions

(also best to do since u want to be better at maths in general)

[hallow seal]

whats the olympiad

[deft marsh]

\textbf{(i)} If (x, y) are positive real numbers, prove that:
[
\frac{4}{x + y} \leq \frac{1}{x} + \frac{1}{y}.
]

\textbf{(ii)} If (a, b, c, d) are positive real numbers, prove that:
[
\frac{2}{(a + b)(c + d) + (b + c)(a + d)}
\leq
\frac{1}{(a + c)(b + d) + 4ac}
+
\frac{1}{(a + c)(b + d) + 4bd}.
]
Also, determine when equality holds.

[gilded halo]

saintyzy

[hallow seal]

does anyone know what the olympiad is

[ivory ember]

i. Note that this is equivalent to $$\frac{4}{x+y} \leq \frac{x+y}{xy}$$ and thus $$4xy \leq (x+y)^2$$ I'll let you finish the rest.

[gilded halo]

Civil Service Pigeon

[ivory ember]

ii. This is a direct application of i.

[hallow seal]

does anyone know what the maths olympiad is

[deft wraith]

it’s the next round

Google cayley math Olympiad

What year are you in?

[tepid surge]

is this not just the GM-RMS ineq

[ivory ember]

Yeah but if I can keep it more elementary, then I might as well

[tepid surge]

Fair enough

[deft marsh]

Sokeone help

[tepid surge]

Someone literally did

[deft marsh]

Yeah i solved the first one

Sthe second one is the hard part

[placid robin]

i did it

pictures aren't clear at all

[obsidian lance]

Hi, I'm Satyam, author of 2 books [THE SILENT CLIMB] // [Learn Such That can't Forget Anything] .The Silent Climb is about how a person will become rich and and another one is all about rules, strategies on how to learn anything that you'll never forget that thing. If you want to buy them message me

[placid robin]

equality holds when ac=bd,i think it's easy

[mortal matrix]

how hard is the stanford math tournament (SMT) compared to the AIME?

pls ping

[obsidian lance]

@obsidian lance
i', new here

[deft marsh]

it is

find all values of and an integer n such that the number $n^2 + 10n + 160$ is a perfect square

[gilded halo]

saintyzy

[ivory ember]

$k^2=n^2+10n+160=(n+5)^2+135$

[gilded halo]

Civil Service Pigeon

[ivory ember]

Then, ||difference of squares||

[deft marsh]

how would anyone do that in the exam tho

cause it was for the jbmo tst

selection tst

[ivory ember]

.

Takes like 5 minutes tops once you get the idea

[deft marsh]

yeah but then its difficult finding the values

[hallow seal]

im in year 9

so im doing the intermediate challenge

[ivory ember]

Is it?

You’re just spamming casework

[hallow seal]

...

[ivory ember]

$(k-n-5)(k+n+5)=135$

[gilded halo]

Civil Service Pigeon

[ivory ember]

Then do cases over all of the ways to factor 135

[hallow seal]

thats kinda easy tbh

[deft marsh]

yeah true

u mean -135

[hallow seal]

why does this guy have a pfp of diddy💀

[deft marsh]

who?

[hallow seal]

you

[deft marsh]

thats kanye....

[hallow seal]

i think thats didddy

ooooohhhhhh ok

my bad

sorry

they kinda look the same

[deft marsh]

😭

its okay

[hallow seal]

im asian

[deft marsh]

oh nice

im greek

[hallow seal]

so i got that pass

nice

[deft wraith]

Yeah

[ivory ember]

?

$k^2=(n+5)^2+135$

[gilded halo]

Civil Service Pigeon

[ivory ember]

$k^2-(n+5)^2=135$

[gilded halo]

Civil Service Pigeon

[ivory ember]

$(k-n-5)(k+n+5)=135$

[gilded halo]

Civil Service Pigeon

[deft marsh]

oh yh thought u said (n+5)^2 - k^2

mb

[quaint fog]

How do I prepare for a competitive math exam on 29 Jan. I was selected for it on 25th and I have been trying my best but I still don't know how to fully prepare for it

[acoustic nova]

What is the name of the exam

[wet grove]

I find that it comes down to your mentality to stay concentrated and focused throughout the entire thing, the experience in solving and understanding the problems that'll be presented in front of you, and having inner-peace is crucial as well. A healthy brain and mindset clears out the nerves and the stress that you may otherwise face if you're a loose animal in an environment that is unfamiliar to the one you've originally developed in. This inner-peace mindset and the know how in understanding the problems takes months or even years to coalesce. Unfortunately asking 2 days prior isn't promising

[soft vigil]

Did any of you get the usamts email

[acoustic nova]

What email

[soft vigil]

Aime qual

If you get >=68 in total

[dusk urchin]

yo

anyone do amc this year?

[acoustic nova]

ooh nice. you got that?

[acoustic nova]

yeah

[dusk urchin]

how old r u

[soft vigil]

Yeah

Both thru amc and usamts

[acoustic nova]

ooh nice. i made it through amc. i shouldve done usamts more seriously i only had a good submission that i spent time on for round 3. but on that round i think i got perfect

[soft vigil]

I spent 1 day on round 2 lmao

[acoustic nova]

what did u get

[soft vigil]

24/25 on it

[dusk urchin]

amc8 i got 21 this year i think

[soft vigil]

Huh

We are not talking about silly amc8

[clever surge]

can i post any kind of competition math questions

[ornate blade]

ye

[obsidian lance]

https://youtu.be/8sGL5h0PN_c

[quaint fog]

I have been preparing for it from Saturday but I still couldn't cover all of it

What's the syllabus for it BTW?

[quaint fog]

And what will be the most effective way to prepare for it

[versed prism]

whos locked in for aime (not me)

[rough holly]

not me

[dreamy wadi]

any books for learning manipulations or out of the box/creative ideas to improve approach in problem solving? anything that's a good source for learning mathematical thinking

like often times you solve a difficult problem that involves "some out of the box idea" (which you can't think yourself)...

you are taught that method and you apply it in similar situations. HOWEVER we aren't taught to learn to "think" that way i.e the thought process behind that idea. How did it strike the solver? That's what I'd like to learn
And also some manipulation ideas, like those that get applied while solving integrals

[mellow summit]

This is my first year doing any math related competition, are there any tips I should know?

Also how would I compare 3⁹ and 4⁷ without a calculator

[pallid dragon]

beats me

343×343×343 vs 16k

uh

no 243

no 81

6400 * 80 > 16k

why is it still wrong

ah 27×27×27

too close to tell

yeah no idea

[sleek ivy]

3^9=81^23 4^7=64^24?

maybe js calculate em

should take a min

19683 vs 16384

3^9 is greater

[acoustic cairn]

Anyone did the UKMT IMC today

[leaden obsidian]

Four numbers √11, √12, √13, √14 are written on the board. In each step, we erase one number from the board and replace it with the product of some two of the remaining three numbers. Find out if it is possible to proceed so that after a few steps there are only whole numbers on the board.

[pallid dragon]

a,b,c,d
a,b,ab,d
a,b,ab,ab
...
aabb,aabb,ab,ab
...
aabb,aabb,4a4b,4a4b

[acoustic nova]

someone ban this guy <@&268886789983436800>

[compact glade]

No product of the numbers results in a perfect square, only whole numbers. So, not possible.

[pallid dragon]

i don't get it

[vague temple]

You can memorize prime log approximations

But yeah for this case it’s not too bad without a calculator to just compute

3^9 = 729(10-1)(3) = 19683
4^7 = 2^14 = 1024(16) = 16384
So 3^9>4^7

[jade widget]

That’s calculating

[fallen brook]

Any unsolved questions here?

[radiant jasper]

yes

17262x82y`7x3%=6x

[ornate blade]

Bad day to be a leopard

Leopards are not for life, it turns out

[fallen brook]

Ha lol he would love this server

[ornate blade]

I checked

leopard's already in the server somehow

@radiant jasper

[ornate blade]

oh yeah I remember now

he was in this channel asking about how to get better at UKMT

and then this jacked dude called LY came over cause he's a Cambridge student with a ton of UK comp math experience

so I bet he's complaining less now in the add maths channel

good times

@high goblet this is the guy btw

YOURE HERE

omgomg

[fallow elm]

idk why the term whole numbers are being used anymore but sure

let a, b, c, d be √11, √12, √13, √14 respectively.
notice that we require the square roots to be cancelled out - this may be done by simply squaring them

a, b, c, d is written on the board
a, ac, c, d
a, ac, c, ac
a, ac, a^2c^2, ac
a^2c^2, ac, a^2c^2, ac
a^2c^2, (a^2c^2)^2, a^2c^2, ac
a^2c^2, (a^2c^2)^2, a^2c^2, (a^2c^2)^2

this is just a common sensebash question actually

i thought it would be more difficult 😭

[fallow elm]

me

[fallen brook]

I don’t see him in the igcse server anymore

[ornate blade]

yeah

that's what I've noticed too, idk why he's not active here either

[mellow summit]

Wait why would we need them?

[radiant jasper]

hi

[high goblet]

why r u surprised lol, i haven't left

i need to finish off my problem sheet due in like 3hrs

[acoustic nova]

one use could be to find the the number of digits in a perfect power. Eg if u know log_10(3) u can find the number of digits in 3^2025

[sleek ivy]

What’s the way to find that without memorizing them?

[vague temple]

I had the same solution of (1-k/m)(m permute k) where k = 4 - 1 = 3. I had a feeling it would work but I have no idea at all how to prove that the probability of k/m is valid since the distribution is kind of weird, naive induction doesn’t seem to work. Can someone help

I was about to bash with Newton forward differences then realized this

[acoustic nova]

basically u need to be given it. For example an AMC 12 problem this year asked about it, but also gave the log approximation in the problem

[sleek ivy]

Ah ok

[deft marsh]

We color each of the 49 cells of a
7×7 square grid either red or blue so that the following two conditions are satisfied simultaneously:
(i) There are exactly 4 rows in which the number of blue cells is greater than the number of red cells.
(ii) There are exactly 4 columns in which the number of red cells is greater than the number of blue cells.
Among all possible such colorings, what is the maximum possible side length of a monochromatic square that can be formed in the grid?
(A square in the grid is called monochromatic if all its cells have the same color. For example, in the partially colored figure on the right, there are two 2×2 monochromatic squares at the bottom left, colored red.)

[gilded halo]

victor

[radiant jasper]

cuz im focused

for igcses

[radiant jasper]

im preparing for m/j igcse

and those servers wasted too much of my time

[mortal matrix]

Hi, I'll be taking the AIME II

Will I get to see the AIME I problems before my AIME II?

When is the embargo usually until?

[reef condor]

Yeah

[reef condor]

Depends, aime I has no internationals so it could be sooner

But usually 8 am est the following day

[random laurel]

Guess the grade

[meager sail]

find all solutions for
p^2-p-1=n^3 when n is an integer (n≥1) and p is a prime number

I can't seem to solve this can anyone help

[solid cloud]

P(p-1) = n^3+1

Factor n^3+1

Then case solve it

[meager sail]

that doesn't work

[solid cloud]

Case 1 p divide n+1

Case 2 p divide n^2-n+1

[deft marsh]

Someone help

[meager sail]

you can prove that p is the divisor of n^2-n+1 but i dont know what to do next

[meager sail]

yea

[sacred ether]

Hello, I was wondering for people who are studying/planning on doing maths in uni, how are ya'll preparing for it? I'm in 11th grade (going to 12) and I'm worried about my entrance exams. I have one year to prepare and I lowkey have no idea what to do.

[civic leaf]

Are u from India

[sacred ether]

Yes

[civic leaf]

Me too

[civic leaf]

Are u preparing for jee advanced

[sacred ether]

Nope

ISI

[civic leaf]

Me too

Bro can u give me some tips on how can I crack it coz I gotta k ow About it yesterday

[sacred ether]

That's what I came here for 😭😭

I have no clue what to do 😭

[blazing dagger]

Hi ^-^
Once you're done showing p | n² - n + 1, ||let n² - n + 1 = pm, for m ≥ 1. Then, p = (n + 1)m + 1 => n² - n + 1 = m[(n + 1)m + 1]||
||=> n² - n(1 + m²) - (m² + m - 1) = 0||
||D = (m^4 + 6m² + 4m - 3)||. ||Show that this is perfect square only for certain values of m. Conclude from there.||

[meager sail]

Thanks!

[grand cape]

Just two indian dudes understanding the situation exactly @civic leaf @sacred ether lol, that's all sorry for the ping

Also good luck guys

[coarse shuttle]

when doing induction, its fine to do k-1 => k instead of k => k+1, right?

[acoustic nova]

yes

[austere sky]

does anyone have advice for JEE math

im in 10th going to 11th, ill be joining a coaching

[jagged hearth]

yeah its basically the same thing if you replace k-1 with n to get n=>n+1

[jagged hearth]

me too

[queen moss]

i need some advice, i'm going to study elementary number theory for an upcoming math competition. what are the crucial topics i should study?

[jagged hearth]

depends upon what's the level of the competition like if its proof writing based then you should look into congruence theory (a≡b (mod n) stuff), diophantine equations, divisibility, and maybe bit of arthematical function (such as floor, ceiling, phi(n), d(n) etc.)

that's ug pretty much it for elementary number theory

[vague temple]

Yes, you can recall the principle of induction as falling dominoes: the first domino falls, and the previous domino makes the next domino fall, so all dominoes will fall. E.g. here, f(k) = k+1 is exactly the same function as f(k-1) = k. Mathematically, as long as you have a sequence that generates the full set of natural numbers from the base case, it will work.

[high goblet]

this is generally a bad way of thinking about induction

example:

[coarse shuttle]

i know how to do induction

but notationally

it would have been a pain in the ass

to use k+1

[high goblet]

Theorem: every graph on n vertices with positive degree is connected
Proof: by induction. The case n=1 is clear. Now assume true for n=k. We shall prove it's true for n=k+1. Sps we add a new vertex to our connected graph with k vertices. Then clearly this vertex is connected to some vertex in our subgraph of k vertices. But by the induction hypothesis, every vertex in our subgraph of k vertices is connected, so every vertex is connected.

but this is obviously a very bogus proof

why?

[coarse shuttle]

well it only applies to some k+1

you have to show its true for all the k+1

[high goblet]

yh

but if you go k => k+1, it is so easy to make mistakes like these

it is much better to work from k -> smaller k

[coarse shuttle]

wdym k -> smaller k

i just try to prove it for k+1

and then use k somewhere

i dont build off of k

[high goblet]

as in like the problem here is we started off with a graph with k, and then we build from k to k+1

you should instead start with a graph with k, then remove vertices to get to i.e. k-1 or k/2 or sqrt(k) etc.

[coarse shuttle]

yeah

[high goblet]

as an example, 7^n - 3^n is divisible by 4 for all n

wait idk how to do a bad induction proof with this

anyway in general you should start with 7^n - 3^n and rearrange to get something in terms of i.e. 7^(n-1) - 3^(n-1)

because ur arrows only go =>

for these kinds of qs, technically the arrows are all <=> but it is still a very bad habit to fall into

like if you look at a lot of people's proofs, they start with i.e. 7^n - 3^n and then rearrange to get 7^(n+1) - 3^(n+1)

[vague temple]

Oh that’s interesting I haven’t really considered that, thanks for the explanation LY

[high goblet]

nw!

[coarse shuttle]

also can u have a base case other than 1

[acoustic nova]

of course

[acoustic nova]

7^n ≡ 3^n (mod 4)

[sleek ivy]

Make it so it’s 3(7^n-3^n) +4*7^n

[acoustic nova]

n ∈ ℕ

[sleek ivy]

Not counting 0

[acoustic nova]

well it’s debatable whether 0 is a natural number

[sleek ivy]

Fair enough

[reef condor]

Induction should just be a general proof technique

Whenever you work over n things just treat the result for n-1 things as trivial / can be built off of

[tepid surge]

it is

i mean induction is a well-known and powerful technique to demonstrate a property over some subset of the integers

[vague temple]

n=0 still works though

0 is divisible by 4

[honest falcon]

Mostly when we deal with natural numbers in math books, 0 is not a natural number. But in the philosophy of mathematics, it is a debate whether 0 is a natural number or not

[chrome glacier]

The length from origin to a point through which an ellipse passes(general horizontal ellipse), where the tangent from the point is equally inclined to each axes. HOW DO YOU SOLVE THIS?

[atomic agate]

Guys, any advice on what to study to prepare for a pascal test (from university of waterloo)?

[ornate blade]

https://cemc.uwaterloo.ca/contests/pcf

try practicing past contests from the CEMC

you might also find AMC8 very useful as it's around the same level as the Pascal contest

I found a list online actually

Algebra
Arithmetic
Counting
Elementary Geometry
Logic
Number Theory
Probability
Estimation
Proportional Reasoning
Reading and interpreting Graphs and Tables
Pythagorean Theorem
Spatial Visualization
Linear Functions and Equations
Quadratic Functions and Equations
Coordinate Geometry

a lot of these you should be learning in class

and it's hard to get a sense of what these really mean without looking at past questions

[quick beacon]

Can somebody solve this

Or give a hint

[sleek ivy]

just do mocks its pretty easy to get a distinction

[vague temple]

Are you able to draw a diagram as a first step

[knotty tusk]

how do i get into/start competition math as a sophomore highschooler with no previous experience?

[quick beacon]

Ya basically z1z2z3...z10 lie on a unit cirvle

Circle

And same with w1w2w3..w9

And we have w9-z1 mod value as 2sinpi/9

So we have a definite configuration of these complex numbers on argands plane

On drawing i feel i and iii are correct but i want a rigorous method to solve

[lone saffron]

You have the numbers 1-n and want to make a string of these numbers. There are 3 rules. Each number must be used exactly once. A number i cannot be the ith number. Going left to right, the numbers increase in value until reaching n, where they start decreasing in value. How many possible strings of numbers are there for a value of n, and what are these strings? (This second question matters less to me to answer.)

[stuck orchid]

Does any of yall been to USAD in 2022-2023v

[lone saffron]

never been good at combinatorics, all i can see currnetly with this is that for an odd n, you cant have the string be strictly increasing or decreasing, because the middle value will be in its own position.

[quick beacon]

Bro

But like

Suppose u have 1 2 3 4 5

Not necessary u use all 5 of them

[still geode]

you have to use all

[quick beacon]

Q says it may be used exactly once

[lone saffron]

mb, u do tho

[quick beacon]

Ohok

[vague temple]

Where are you stuck, show your progress

[still geode]

ok so i feel like you have to fix a position for n and then try find the number of ways to arrange the other numbers

[lone saffron]

see i tried kinda doing that

[still geode]

but idk the not being at its own pos thing is weird

[lone saffron]

but i couldnt figure it out

except for the ends

bc those are easy

[still geode]

like even with n at the very start how would you do it

[lone saffron]

just numbers decreasing

unless n is odd, in which u cant place it at the start

[still geode]

yeah but thats 1 way to do it but didn't u want to find all the ways

[lone saffron]

yees

[still geode]

oh wait nvm it has to decreas

because of the thing

right

[still geode]

well, the beginning

n can't go at the end

[lone saffron]

yes it can

it has to increase before n and decrease after n

if n is even jsut make it 1 2 3 4 5 ... n

[quick beacon]

In the end its like we substitute omega and omega square in the function

Cube roots of unity basically

[still geode]

yeah but then every number is in its own place

which is exactly what you don't want

[lone saffron]

oopsy poopsie

[still geode]

💀

[lone saffron]

so n doesnt go at the end

[still geode]

i guess if n goes at #2 then its (n-2) options for #1 and the rest is fixed?

[vague temple]

Have you tried recursion

[lone saffron]

wdym recursion

[still geode]

he means induction

[vague temple]

Do you know what recursion is

[vague temple]

No I mean recursion

[still geode]

ok, explain

[quick beacon]

I feel ith position rule and ascending order rule contradict each other just for a feel can u think of any order whcih would be the part of answer

[lone saffron]

i know what recursion is, yes...

[lone saffron]

n=3, 231

[still geode]

2, 3, 1

[vague temple]

Have you considered the casework on inserting n+1 in a valid string for n

[lone saffron]

n=2, 21

[still geode]

or n = anything, [2, 3, ... n, 1] works

[quick beacon]

Wait what

[still geode]

yes thats called induction

[quick beacon]

Mbmb

[lone saffron]

no hes talking about something else i think

[vague temple]

I’m trying to find a recursive relation

[still geode]

also how do you know that every arrangment for n + 1 can be made by inserting n+1 into a valid string of n

[vague temple]

?

Have you studied recursion

[quick beacon]

@vague temple

[vague temple]

I’m not saying it will work but have you tried it, it seems useful

[lone saffron]

yeah, so you could insert n

+1 directly before or after n

[vague temple]

It almost always works for string problems

[still geode]

explain what you mean by recursion

because i swear you're describing induction

[lone saffron]

oop, im wrong. if you have 21, 321 wouldnt work

[vague temple]

Yes, you have to be careful in the casework

Induction is a form of recursion

[still geode]

you said yourself that just going backwards doesn't work for odd numbers

[lone saffron]

yes, so my answeer to his question is wrong

i think for n>3 my thin works tho

[still geode]

i mean i guess the n+1 has to go right after the n right

or right before

ok this is promising

only 2 cases

[vague temple]

Given that they lie on a circle, you tried unfolding the definition of |z_i-z_j| = 2sin pi/10

[still geode]

oh

[vague temple]

E.g. let z = z_0, z_1 = …

[still geode]

its just [2, 3, ... n, 1]

wait HOLD UP

[vague temple]

No

2,4,5,3,1

[quick beacon]

I tried it u dont get the answer algebraically it becomes too long

[still geode]

4 is in 4th

but ur right because 4, 3, 2, 1 exists

[quick beacon]

We have to proceed geometrically

[still geode]

which is not made by inserting a 4 into a n = 3 solution

does 2, 5, 4, 3, 1 not work

yeah it does bruh

3, 5, 4, 2, 1

as well

[vague temple]

I don’t think you can necessarily infer that Im(z_1) = 0

Geometrically we have inscriptions in a units circle of a rotated of a regular decagon and a rotated nonagon

You can use algebra

It works, 1 and 3

[vague temple]

Recursion works, f(n) = f(n-1) + 2f(n-2)

[lone saffron]

yes, we found that out a few mintues ago

i just wonder why

any ideas for how i could prove or at least reason why it is that pattern?

[vague temple]

Well just exhaustively construct the n-th string from string n-1 and string n-2 (equivalently, delete elements until each case works)

[lone saffron]

2341
3421
4321

4

23451
24531
25431
34521
35421

5

234561
235641
236541
245631
246531
345621
346521
456321
465321
564321
654321

6

ok lets look

[vague temple]

Say you have a working string n, then obviously deleting n still works if there is no position clash. However, if there is a position clash, then it means you can remove n-1 and it will fix.

Notice that in a position clash n-1 and n can be interchanged, so there are 2 ways

[lone saffron]

how do you know that second part?

[radiant jasper]

how do i prep for USAMO as a 9th grader with barley any experience in comp math

[vague temple]

Break into cases: n-1 valid or n-1 not valid, you will see there are 2 times n-2 valid

[radiant jasper]

im currently studying AOPS Alg 2 with the book no tutor

[lone saffron]

solve past problems

[lone saffron]

khan academy is very useful

[radiant jasper]

ty

[vague temple]

Start with Sohil Rathi’s mastering AMC 10/12 book, very useful

Then AoPS books can be useful but mostly handouts and problem sets

[vague temple]

From OEIS:

a(n) is also the number of derangements in S_{n+1} with empty peak set.

That’s the closest thing I could find

[lone saffron]

Im not exactly sure what that means.

[vague temple]

Derangement means that number j is not in position j

I presume S_{n+1} refers to a group of some kind

[still geode]

what are you doing

we literally kinda solved it in the other channel

[lone saffron]

trying to find why its the jacobsthall

we know it is

[still geode]

right

[lone saffron]

i wanna be able to prove it is or at least give a good reason it is

ok whatever im just gonna stop this im done fr

[vague temple]

I told you that you can derive the n-th string by casework after removing n
I’ll write it out for you:
If we remove n and it works, then we get f(n-1).
If it doesn't work, first notice that n-1 and n must be consecutive to maintain the mountain shape, either permutation giving the same result after the pair's removal. Then it must be that the number right after the pair n-1, n is clashing, since the sequence is decreasing while the position index is increasing. Formally, WLOG consider a1a2a3...(a{k-2}=n-1)(a{k-1}=n)aka{k+1}...an. Then, we have a1<a2<a3<...<a{k-1}<n-1>ak>a{k+1}>...>an.The condition that n removing it doesn't work implies that ai=i+1 for at least one i>=k. We want to show that it's impossible to have aj=j+2 for any j>i. But the condition j>i implies that aj<ai = i+1 < j+2. Hence, we are done: if removing n breaks it, then removing n-1 afterward will fix it. From the 2 permutations of removing n and n-1, we get 2f(n-2).
Hence, we obtain our desired result:
f(n) = f(n-1) + 2f(n-2)

These 2 cases are exhaustive so you will find the formula

[soft vigil]

First prep for aime

[quick beacon]

Ya this i drew for feel

[quick beacon]

The answer?

[half thicket]

Question from JEE advanced 2024 (paper 1)

[pallid tundra]

!status

[scarlet root]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pallid tundra]

also this is not really competition math, #precalculus is a better place for this question

[digital hedge]

Prove we can cut a rectangle R1 in such a way that the pieces can be put together to form a rectangle R2 with same area as R1

[grand valley]

iit bombay olympiad qs

[reef condor]

Draw a graph

[radiant jasper]

ok thanks

[radiant jasper]

ok

[median moon]

im trying to study for amc 10/12

any suggestions?

like textbooks or courses?

i have volume 1 the basics and volume 2 beyond in my possession right now

by aops

[soft vigil]

isnt that in a year

[median moon]

yea but i just want to get my hands on the correct material beforehand

[coarse shuttle]

part of the solution for a problem i was looking at says that the number of digits in floor(2.5^x) is ceil(log(2.5^x))

but shouldn't it be ceil(log(floor(2.5^x)))

[untold thunder]

Two parallel segments of length 2 are one unit apart. A line intersects these two segments at $A$ and $B$. What is the expected length of $AB$? If I wanted to solve this could I take the average of the maximum and minimum of $AB$? I don't have any way to intuitively explain it but it seems right.

[gilded halo]

Fatherous

[untold thunder]

You can read the basics (I don't know much about it) I wouldn't recommend reading beyond because that builds off non-beginner topics. All the other AoPS books are good for the AMC except prealgebra (In my opinion). Reading those books won't completely cut it though so doing the actual AMC problems after you've learned the material is good. Also, AoPS Alcumus is good for a nice variety.

[still geode]

this feels like the kind of problem where the answer depends on very subtle points, like "how is the line randomly chosen"

[untold thunder]

You can read the basics (I don't know much about it) I wouldn't recommend reading beyond because that builds off non-beginner topics. All the other AoPS books are good for the AMC except prealgebra (In my opinion). Reading those books won't completely cut it though so doing the actual AMC problems after you've learned the material is good. Also, AoPS Alcumus is good for a nice variety.

I made it but as I said I don't really know how to solve it

I guess a good way to define "random" is choosing two points on each segment and making a line from there

[still geode]

ok well then surely it would just be

the expected value of $\sqrt{1+(a-b)^2}$ for $a,b\in[0,2]$

[gilded halo]

DiamondPanda16

[still geode]

integration yay

[pallid dragon]

if you take a random A and a random angle

wolfram said it can't figure it out, but desmos can

i'm not sure if it gets averaged or i need to divide something

[still geode]

random A and random angle is different to random A and random B

-# i totally didn't just choose the easier interpretation

[still geode]

divide the integral by the range

which is 2

to find the average

[pallid dragon]

i think it's already correct

it only counts upwards, so ends up as half

and then i don't divide by 2

[still geode]

what do you mean only counts upwards

[pallid dragon]

like this

[still geode]

also what if it doesn't intersect

with the other segment

[pallid dragon]

it feels right

i'll make a simulation

[vague temple]

Yes, it depends on your probability distribution, but assuming a and b are uniformly distributed on [0,2] (the difference will be triangular) then the integration evaluates to (sqrt5+1)/6 + 1/2 ln(2+sqrt5)

[still geode]

is this what $\int_0^2\int_0^2\sqrt{1+(a-b)^2}\dd{a}\dd{b}$ evaluates to

[gilded halo]

DiamondPanda16

[vague temple]

Should be if no mistake

[vague temple]

Divide by 4

[still geode]

lets just

oh righ

[vague temple]

It’s expectation

[still geode]

i was gonna wolfram alpha but ur right already

nice

[vague temple]

This doesn’t seem like comp math unless it’s like uni comp math not high school

Feels like probability theory question

[still geode]

it doesn't seem like comp math because it has integration

[pallid dragon]

simulation says mine is wrong :\

[untold thunder]

I was making a question for a kid that just qualified for aime bro

Imma ask the mo server if they can get a non-calculus solution or if this one is wrong

But thanks for the work on my problem anyways

[radiant jasper]

I feel like the level gap between Cayley and Fermat is quite significant, am I the only person who feels this? (Waterloo math contest)

Like obv both are nowhere close to amc level, but I expected both contests to be at same difficulty, but i dont think they are

[ornate blade]

how about Pascal and Cayley? does it feel closer to you?

[radiant jasper]

I haven’t looked at pascal for a long time

Oh I guess it’s meant to have that level gap

[ornate blade]

yeah I mean I'd say the AMC gap is even wider

between AMC 10 and 12, well it's a 2 year gap of course

but I feel that AMC 12 questions are ridiculously long / technical for contests internationally, especially q20 to q25

[vague temple]

Classic Bertrand’s paradox

[radiant jasper]

I thought for amc u would only need to solve 1-15 and leave all the questions blank

Then ur safe for aime

Or that’s at least what my friends have told me

[ornate blade]

yes that's correct

[sleek ivy]

fermat seems easier to me this year than pascal seemed last year

maybe i js got better

[reef condor]

Yeah amc 12 gets hard

Recently the final 5 gets hard

[versed prism]

whos locked in on aime (not me)

[swift imp]

forgot it existed

[still geode]

in the wrong country

[torn gate]

I don't have aime tomorrow due to early dismissal in my school (winter storm)

[heavy ruin]

That's crazy dog

[drifting moon]

Hi

Anyone active here

[still geode]

somewhat

[radiant jasper]

Has anyone taken aime

[radiant jasper]

aime was easy

anyone did the mathcounts chapter

[versed prism]

yo

i think i cooked up a storm on the aime

[reef condor]

Pretty sure no discussion is allowed

[versed prism]

well for the questions themselves yeah

[little lintel]

aime was tolerable

[radiant jasper]

aime was so weird thou

like it was easy

but like still

[vestal yacht]

wtf my aime ended early

Did we have to finish before 5:30 est?

[radiant jasper]

idk

its 3 hrs

[vestal yacht]

I thought we had to start before 5:30

My freaking teaxher got it wrong 😑

[radiant jasper]

lmao

ya i think its start before 3:30

[vestal yacht]

icl ts pmo

Bruh are u being so freaking serious rn bro

[radiant jasper]

i meant 5:30

[vestal yacht]

Oh

Then why did my test end an hour early exactly at 5:30 🤔

Hopefully they messed up the coding or something

And everyone can take aime 2

[radiant jasper]

idk

i think its prob time dif

idk

[heavy ruin]

Hello

[radiant jasper]

hello sir

[raw cave]

yo

who took aime

and Mathcounts chapter

[thorny solar]

I had an interest in numbers division , thinking took me away can we have divisible numbers for primary numbers

like 7/2 and give us a whole number

I need your help in that

[plush wharf]

are you asking if dividing two prime numbers can give a whole number?

if so, well 2/2=1 certainly is one

in general if two numbers a/b divide to get c, then a=b*c. (you can show that if a,b are prime numbers, then c has to be 1)

[balmy lintel]

Proved everything other than the last part where right hand side implies for any continuous function that identity holds, no solutions online either.

[acoustic nova]

for two distinct primes, no. think about the definition of divisibility and the definition of primes

[knotty egret]

site name?

sites for comp math

[hoary radish]

If someone can implement math formulas as a ready function to generate mouse movements in Python, dm me (paid task)

[mystic crane]

cool project, good luck with that, hope you find someone

[jade widget]

i like 2024 AIME 2😭

[vague temple]

o3-mini consistently gets up to 13 problems correct on 2025 AIME I

https://matharena.ai/

[radiant jasper]

wait did u take

it

[reef condor]

Not surprising

I used the publictly available free ChatGPT to check my answer to p4 after the contest

And it double counted 0,0 but otherwise was correct

[queen surge]

last time I checked gpt was really bad with math

[wintry glacier]

o3 really upped its game

[ornate blade]

only the publicly accessible versions are

cause you think a bunch of STEM-obsessed edgelords would be content with their machination failing maths?

no! they've secretly wanted it to get good at maths, especially contest maths cause of its reputation, and that's why they've been working at it behind the scenes

[dapper magnet]

god i hate ai lords 😭

[ornate blade]

or rather, hiring code monkeys to accomplish this for them

[fallow elm]

who's taking aime 2 fr

[eternal valley]

For each positive integer $n$ not divisible by 3, let $r(n)$ denote the number of pairs of solutions $(a, b)$ of the Diophantine equation
\begin{eqnarray}
a^{2} + 2b^{2} = 3n
\end{eqnarray}
with positive integers $a$, $b$.
Prove that $r(n)$ is odd if and only if there is a positive integer $m$ not divisible by 3 for which $n = m^{2}$ or $n = 2m^{2}$

[gilded halo]

Elly

[eternal valley]

this is from a german math olympiad

for 10th grade

has been bothering me since like last year lol

(i forgot about it and just recently remembered it again)

and

i dont really know how to

approach this?

its be awesome if smn could

give hints or

a direction

maybe some sources i could check out to get an idea

also this is translated so there might be some issues so if something is unclear its prob translation error

[still geode]

hmm

[eternal valley]

a and b cant be divisible by 3

cus if they were

then

n would have to be divisible by 3

which contradicts the statement

hm

but like

individually they can be

probably

[still geode]

yeah you need a = b != 0 mod 3

a^2 + b^2 is annoying though like

maybe try some small cases for n?

[eternal valley]

if n = m² then we get following relation
2(b+m)(b-m) = (m+a)(m-a)

for n = 2m² we get
(a-2m)(a+2m) = 2(m-b)(m+b)

might be useful

[eternal valley]

ill try

i gtg now

ill try later

[eternal valley]

this suggests that b>m and a<m or b<m and m<a

or b = a = m

maybe its

what if theres always an even number of solutions but because in this specific case a = b = m is a solution that would account for 1 more solution instead of two making r(n) odd

thats the idea im getting rn

[eternal valley]

if a>m then a^2 + 2b^2 < 3a^2 <-> 2b^2 < 2a^2 <-> b < a since a,b > 0

okay ill thik about this later

and write it down on paper instead of flooding this channel lol

oopsie

and try out some easy cases first

[eternal valley]

i thought about using complex numbers but that got me nowhere mostly cus im not that experienced yet

also the cases gave me nothing

[vague temple]

Depending on how much theory you know about Gaussian integers the problem can be very simple using Z[sqrt(-2)] but we will avoid it for grade 10
A perfect square can only be 0,1 mod 3.
Also, notice that a^2+2b^2 is divisible by 3 but not by 9.
So mod 3, a^2==b^2==1. (Implying neither a,b, is divisible by 3)
Either a-b ==0 or a+b == 0 mod 3. Either way, we get something like:
a-b=3x,a+2b=3y
or
a+b=3x,a-2b=3y
where x,y are coprime to 3.

to get 2x^2+y^2=n.
Motivated by this, we see that for case 1,
let a=2x+y and b=x-y.
While in case 2,
let a=2x+y and b=y-x.
such that we define an involution (x,y) mapping to (x,-y) in case 1 or (-x,y) in case 2. Verify that positivity holds based on constraints so solutions are valid
But some of these must be duplicates, then. When? This must happen if x=-x, or x = 0, meaning n = y^2, or y = 0, meaning n = 2x^2.

[soft vigil]

me fr

i only need 9 for jmo and 11 for amo 🙏

[vague temple]

Good luck, wish you the best

[eternal valley]

okayyy i need to reread this a few times to understand it

but thanks for the help :D

[vague temple]

The key argument is of symmetry: I think you found (a+b)(a-b) == 0 earlier. The core of the argument lies in that for each solution (x,y) this can be paired with a solution (x,-y) if case 1 or (-x,y) if case 2. But there will be fixed points when -a=a or -b=b.
This is a common way to prove multiplicity of solutions, you pair each solution with a distinct one through some mapping (for evenness typically an involution)

I think I messed up the signs somewhere but if you fix it it should be alright

[eternal valley]

2 questions
how did you get a^2 == b^2 == 1 (mod 3)
how does that imply a-b == 0 or a+b == 0 mod 3?

sorry im pretty bad at math

ohh

i think i get it now

how did you know what to set a and b equal to though?

ohh wait

i think i get that too

[vague temple]

Taking the equation mod 3, you get
a^2==b^2
Perfect squares can only be 0,1 mod 3. But if a^2==b^2==0, then it is divisible by 9. Hence it must be ==1

[vague temple]

Solving the linear system

[eternal valley]

ohh that makes sense

had a brain fart moment there

[vague temple]

No worries
The second part is just factoring
a^2-b^2==0
(a-b)(a+b)==0
Since there can only be 1 factor of 3 it has to come from one of them

[eternal valley]

mhm

[eternal valley]

why does that work?

[vague temple]

What do you mean

[eternal valley]

ohhh

i see

that makes sense

okay i understand it now

:D

tysm

[rotund cloak]

Give me your problem. I will help you as much as I can.

[knotty egret]

hello guys i am very good at math i i would like to do comp math problems where can i do that?

@everyone

@eternal valley

[eternal valley]

id recommend google

[knotty egret]

?

what it mean

[eternal valley]

try googling like

[knotty egret]

google is a very big

[eternal valley]

comp problems

like

[competition name] problems

[knotty egret]

yea i dont find some

where do you do that

@eternal valley

do u have site that you use

[eternal valley]

the competitions i compete in have their own websites

alternatively you can ask your teacher/professor

[acoustic nova]

How good is very good

[knotty egret]

Yes I am very good

[knotty egret]

What the website name

[karmic marten]

aops or mathdash

[knotty egret]

What is better?

[knotty egret]

.

[karmic marten]

aops for learning concepts and reading solutions to problems

mathdash is good for practicing problems and it is fun too

[vague temple]

Search for "olympiad problems" like olympiad handouts, past olympiads, etc.

Those are very challenging

You can refer to AoPS, it's useful for finding these
https://artofproblemsolving.com/wiki/index.php/List_of_mathematics_competitions
https://artofproblemsolving.com/wiki/index.php/Category:Mathematics_competitions

[torn gate]

Me

[scarlet root]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[ivory ember]

I don't have a lot of time but I remember this question being so bs 💀

1. Notice that the denominators are perfect squares

2. ||Write the denominators as dot products||

It reduces to a fairly easy ||geometric probability|| question

[knotty egret]

Site name?

@torn gate

[hoary radish]

No one could do that?

[knotty egret]

Guys do you know what site he use for this problem @everyone @hoary radish do you know?

[neat mural]

hello , i was looking at the first problem of IMO 1979 (https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_1)

and i got stuck at the part where they say that (qr) / s belong to the relatives number

does anybody know why it is true and could explain it to me?

[ivory ember]

what is "relatives number"? 😕

[neat mural]

isn't that how they're called in english?

integers i think

[torn gate]

https://cdn.artofproblemsolving.com/attachments/1/3/24e38f5e414fcfc6645bd94800eaed22d45aa3.pdf

Test itself: https://artofproblemsolving.com/community/c594864h3225558p29529398

[neat mural]

i meant integers

[ivory ember]

s doesn’t divide 1979 -> no cancellation when multiplying 1979 and 1/s -> both 1979 and qr/s must be integers

[neat mural]

that's what i don't understand

[neat mural]

how does there being no cancellation when multiplying 1979 and 1/s implies that both 1979 and qr/s must be integers?

[ivory ember]

Because for 1979(qr/s) to be an integer, {qr/s} is some integer multiple of 1/1979

But s can’t divide 1979

And hence {qr/s}=0 -> qr/s is an integer

[neat mural]

??????

what's that {} notation?

and i also somehow don't understand how any of that works

[neat mural]

could you explain what this mean?

please

oh wait

i understand

thanks

[knotty egret]

$[
S = \sum_{n=1}^{\infty} \sum_{k=0}^{\lfloor n/2 \rfloor} \sum_{m=0}^{\lceil k/3 \rceil} \frac{(-1)^{n+k+m} \prod_{p=1}^{\lfloor \sqrt{n} \rfloor} \binom{3k}{m} \left(4^{\lceil n/4 \rceil} - \frac{(2p)!!}{p!}\right)}{\binom{\lfloor 3n/2 \rfloor}{k} \cdot (n+k+m)!} \cdot \frac{(2n)!!}{(2k)!!(2m)!!} \cdot x^{\lfloor n - k/2 + m \rfloor}
]$

[gilded halo]

Ori
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty egret]

Evaluate this summation

@torn gate your doing aime 2025 or all the aime right there is calculs in aime?

[reef condor]

there is not suppose to be calc for aime

[reef condor]

fractional part of x
{x} = x - floor(x)

[jade widget]

any prediction for usajmo cutoff?

[tepid surge]

Philadelphia Eagles winning ong

[jade widget]

so true bestie

[reef condor]

230/235 on the a and b

225/230 low end

[deft marsh]

Find all pairs of positive inegrrs numbers (p,x) p prime such that $$x! - p^x = px$$

[gilded halo]

saintyzy

[deft marsh]

Need a full solution

[molten ridge]

i can see that (2,4) is a solution

[vague temple]

Modulo x, we have p^x == 0, meaning x divides p^x. Hence, let x=p^k for k<=p^k. By inspection, (p,k) = (2,2) is a solution. In addition, notice that isolating (p^k)! on the LHS, it grows faster than the RHS of p^p^k + p^(k+1) = p^(p^k+k+1) when increasing p or k for k > 2. (e.g. supposing from the base case that (p^k)! = p^(p^k+k+1) for (p,k)=(2,2), if you increment k, then (p^(k+1))!/(p^k)! > (p^k)(p^k)… = (p^k)^(p^(k+1)-p^k)=(p^k)^(p^k(p-1)) whereas p^(p^(k+1)+k+2)/p^(p^k+k+1) = p^(p^k(p-1)+1) < (p^2)^(p^k) < (p^k)^(p^k(p-1)). I’ll let you check for incrementing p, you can also check using either x>4, p>2). Hence, it’s the only solution. (checking values less than 2 doesn’t work)

[deft marsh]

Yh i know also x divides p^x so x is a power of p

[median kindle]

has any1 ever participated in the IJSO or JBMO

(IMO works too, but thats a distant goal)

[median kindle]

oh and usajmo is a selection exam for what

[high goblet]

i participated in balkan MO, but didn't quite make IMO team for my country

[median kindle]

BMO or JBMO

[high goblet]

BMO

[median kindle]

ahh nice

what country

[high goblet]

uk

[median kindle]

ohh guests?

[high goblet]

yeah

[median kindle]

nice

do you happen to know if UK participates in the JBMO

[high goblet]

i think it's something like historically the UK used to own cyprus or smth smth lol

[high goblet]

nah we don't

[median kindle]

ahh ye

[median kindle]

ohh alr

im preparing for the JBMO this year, so i was looking for someone w past experience to give me some pointers

anyway do you have any tips for MO prep in general

[high goblet]

hmm what sort of difficulty level is JBMO?

[high goblet]

u might have better luck in MODS

[high goblet]

in general, u just gotta grind problems

[median kindle]

idrk difficulty scale

[median kindle]

thats what im doing rn, and i kinda seem to be inconsistent

[high goblet]

when u start getting to the point where u can comfortably solve like IMO 1s/4s u could start looking into deeper theory

do you have a copy of euclidean geo in maths olys?

[high goblet]

yeah dw, it always feels like that

maths olys have high variance so it can be quite easy to feel like ur not really progressing

[median kindle]

bc one day i solve an imo problem in like 20 mins, and the other im struggling on a 1st/2nd rounder

(serbia has A LOT of qual rounds)

city round
district round
national round
junior SMO / SMO (i can participate in both bc of age)
if participated in SMO, then selection exam for IMO

way too much for my liking

[median kindle]

yee

and not to mention that last year i fucked up math AND physics competitions

so basically this year is now or never

[high goblet]

i think it's also worth saying to not stress too much, like ok obviously rn making i.e. IMO will be a high priority

but just bear in mind that there's still more to maths after maths olys

it can be quite easy to stress too much in a competition setting and like not do well

[median kindle]

for me not really, i am a year 9 student, and i have time on my hands

[high goblet]

so if u have the attitude that it's ultimately just gonna be for fun, then that helps

[high goblet]

oh lol u have ages

[median kindle]

for the IMO

jbmo and ijso is this year or never

[median kindle]

ik, but math olys are important rn(not only math, im preparing math and physics simultaneously)

and i kinda need good olys to get into a good uni

basically every uni in serbia is shit

and im tryna escape

[high goblet]

yh i just meant don't stress too much

like obviously in the exam it can be quite easy to stress out and not do as well

so obviously easier said than done but yeah

[median kindle]

yeah that happened last year for me

[median kindle]

yeah

[high goblet]

i think if you think of it as oh well it's just for fun it's easier to not stress out

if you think of it as i have to make it to the next round it can be easier to stress out

anyway gl on ur mathematical journey!

[soft vigil]

i know someone who probably went

[median kindle]

ahh nice

[soft vigil]

need to ask him but theres a very high chance he went

[median kindle]

ty ❤️

[median kindle]

on what

ijso or jbmo

[soft vigil]

jbmo

[median kindle]

ohh nicee

what country

[soft vigil]

wiat a minute

is jbmo for <9th grade or smth

[median kindle]

<= 9

[soft vigil]

oh

[high goblet]

jbmo is junior balkans maths olympiad

[soft vigil]

he went to b(alkan)mo not jbmo

[median kindle]

9 if u have age

ohh nvm still

im preparing for those, but im more than glad to talk w people that have mo experience

anyway i gtg cya guys

[soft vigil]

ive taken usajmo if that means anything

[median kindle]

ahh ye

lets talk tmrw

[shadow niche]

sorry if this is off topic, but is there anywhere i could read on IJSO stuff online? i just qualified for the first few rounds and cant seem to find anything in my country

[slender saddle]

I’d say the math problems there are pretty much like round I POSN here, although there are some IJSO problem books as online pdfs. In my year there’s no math for the 2nd round IJSO.

[hidden rose]

India is next level bruv

becuz of the population the competition is insane

We have IOQM the RMO then INMO then IMOTC and then top 5 out of like 250k go for IMO

In my class only 4 people qualified for RMO and all 4 of them failed that

[burnt trail]

find ab, knowing that $\\quad a,b \in \mathbb{Z}^{+}\$ $\\frac{a^3 + b^3}{ab + 4} = 2024$

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[worldly lynx]

anyone doing the olympiades de maths in france?

[ivory ember]

https://math.stackexchange.com/questions/3669889/find-the-all-positive-integer-solutions-a-b-to-fraca3b3ab4-2020?noredirect=1

[burnt trail]

how to apply it to mines

[burnt trail]

I was looking but didn't found, good, ty

[burnt trail]

good catch, thanks

[shadow niche]

theirs still no math in the second round, by the way. I feel like the bound of the content in round 2 is just round 1 but harder, and i probably barely passed the cutoff

[slender saddle]

Yeah, round 2 was really tedious, although I’d say biology was the hardest part.

[latent copper]

Random question, I took the Putnam for the first time this year and I heard that the scores come out this month, is there an estimate to when this month it'll be coming out?

[burnt trail]

how do I apply it to mines

[radiant jasper]

what is the answer?

[vernal axle]

1011, 1013

[ivory ember]

It’s literally the exact same logic

[shadow spruce]

😭

[radiant jasper]

Thanks~!

[radiant jasper]

It should be smth like this ig...

[ivory ember]

This isn’t always exhaustive

Here’s an example of this logic failing

,w cos x + sin x tan x - 2 sin x = -1, 0 \leq x \leq 2pi

[gilded halo]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[radiant jasper]

Ah right

[radiant jasper]

if sinx- cos x = -1 (that we assumed)
then, tan x - 1 should be equal to 1..no?

[radiant jasper]

I was thinking of this thing..can be wrong dk

[ivory ember]

The point is that you can’t just match up the factors

Ex. In the example above, doing so yields the wrong answers

(Note that all of the other steps are correct)

As a simpler example, note that 24=(4)(6)=(3)(8)

By your logic, 4=3 and 6=8

Also small remark (not sure where to put this) - note that multiplication is commutative

So even if ab=cd -> a=c and b=d

We could’ve rewritten cd as dc to get a=d and b=c

I assume that you can see the issue (namely symmetry with the multiplication order)

[radiant jasper]

mhm got it

what am I supposed to do at the place of that step?

[ivory ember]

.

You kinda just got rlly lucky lol

These are what a rigorous solution would be along the lines of

[radiant jasper]

I think so too.

[radiant jasper]

Alright I am not familiar with few concepts used here..no wonder I couldn't get it.
Anyways, thanks for pointing out that mistake
won't do it in future.

[median kindle]

ik that

i mean like 1.5b people

[median kindle]

wait

international junior science oly

problems in general physics by I E Irodov is good

for physics

idk about chem and bio

[hidden rose]

bro India is not full of 1.5 billion teenagers 😭

[median kindle]

ik but still

a lot of people

[hidden rose]

yea

[acoustic nova]

in the last 3 years on the imo they haven’t done so well given their population

[radiant jasper]

Yepp can't agree more ... IMO used to be fun... Indians are naturally good at mathematics

[deft wraith]

What’s your point here, seems like the opposite of hockeydude85’s

[radiant jasper]

Nope I was referring to the post to which this guy replied

yea maths is not my forte why am i here

im js tryna get better fr

[deft wraith]

Right okay

[acoustic nova]

why would the last part be true (of anyone actually)

[hidden rose]

Some guy called Aditya tripathi from my school got full marks in RMO

1st time in history

Genius

[radiant jasper]

Even I cleared RMO back in day

[hidden rose]

damn

i failed ioqm

[radiant jasper]

Did you take any coaching and which class are you in

[hidden rose]

9th 😭

[radiant jasper]

Like 8-11 fall in same category

[hidden rose]

and yes

i did take

im in sri chaitanya

techno

[radiant jasper]

It's fine go for prashant jain

[hidden rose]

so we have coaching along with school