competition-math

[modest marten]

are you talking abt the USAJMO?

[deft marsh]

Yeah

[modest marten]

i think you need to be a sophmore or below for that

[severe eagle]

If youre in singapore i recommend coco education

[modest marten]

oh

[deft marsh]

Oh

[modest marten]

im not in singapore

[modest marten]

you can still try for USAMO though

[deft marsh]

Yh ik

[modest marten]

type shi

nonchalant dreadhead

[modest marten]

im in the US

[deft marsh]

Someone help me with this

[ornate coyote]

my goal is to make jmo this year

(probably not)

[deft marsh]

The national jmo?

[feral canyon]

Why is one of the first problems in Marions classical mechanics so fucking difficult istg

Spherical coordinates make my brain deteriorate

[slow tree]

i do better on AMC A tests than B tests; are A tests easier or is it just a coincidence

[pallid tundra]

idk you tell me

[ornate coyote]

They are built to be the same level but maa keeps messing it up but generally they are the same

[high goblet]

where is the problem from?

i'll have a look at your problem soon

[deft marsh]

Greek junior math olympiad 2024

I sokved it with pells equation but i think u can do it with vieta jumping too

But idrk how to hse bieta jumping

[high goblet]

are you aware of AOPS?

if u just want solns you can see solns on AOPS, but i can have a look at ur problem soon if u want me to

[deft marsh]

There is a solution posted but j dont rlly understad it

From the greek mo site

[deft marsh]

also i think that aops doesnt have the solutions for jmo

for greece atleast

it would be great if u could solve it

and helpful

[high goblet]

lmao for this q doesn't analysing the case x=1 just work

anyway i haven't spotted the pell's eqn soln but the vieta jumping soln i found was

[deft marsh]

Ifu want ill tell u the pell equation solution

[high goblet]

set x=1, then we want to show y^2 + z^2 + 1 + y + z - 5yz = 0 has infinitely many solns

[high goblet]

sure

[high goblet]

are you familiar with vieta jumping?

[deft marsh]

Not rlly

[high goblet]

ah k

well the idea is that we can spot that y=1, z=1 is a soln

then you think of z as a constant

[high goblet]

for z=1, there should be 2 values of y which solve this equation

let's call them y_0 and y_1

(say y_0 = 1)

let's also assume y_0 <= z

then by vieta's formula, y_0 + y_1 = 5z - 1, which tells us that y_1 > z is an integer

so (y_1, z) is another pair which works

then repeat this process

[deft marsh]

Ohh

Alr

[high goblet]

https://artofproblemsolving.com/community/c6h3268111p30059272

[deft marsh]

oh thanks

[deft marsh]

u can also let x=1, y=a-b ,z = a+b a>b so after some calculations u get the equation $$7b^2 = 3a^2 - 2a - 1$$. multiply by 3 and after some more calculations u get $$(3a-1)^2 - 21b^2 = 4$$ so let a = 2A+1 , b= B so u get the equation $$(2A+1)^2 - 21B^2=1$$ so u get the pell equation m^2 - 21n^2 = 1

[gilded halo]

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

[deft marsh]

21 is not a perfect square so the equation has infinite amount of solutions (Mk,Nk) with the smallest one being (M1,N1)=(55,12)

so M_k+1 = 55M_k + 21 (12 N_k)

[deft marsh]

b= 2B

mb

[worldly bough]

hey, has anyone here ever done the International Maths Challenge? this one https://international-maths-challenge.com/

[acoustic nova]

I think that’s like a scam thing I’ve heard <@&268886789983436800> ?

[rose hazel]

what

it's not a scam lol

[worldly bough]

thats what im actually curious about

ive found both stuff that makes me think its a scam and stuff that doesnt make sense to me if its a scam

but i havent been able to find many (or, any, really) first-hand accounts of actually taking the challenge, so id like to know if anyone here had

[radiant jasper]

same bruh

[burnt trail]

Find the sum of all positive integers such that their expression in base $7$ digits is the reverse of their expression in base $16$ digits. Express your answer in base $10$.

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[burnt trail]

hints

[ornate coyote]

play around with the numbers

narrow it down to the max number of digits in the base10

form

both the base 7 and base 16 representation should have the same length

[ornate coyote]

i have never heard of it

but its problems seem too easy to have so many people participate

[worldly bough]

yeah im fairly sure it lies about the numbers on its front page

especially because if you check the wayback machine they were listed as higher in the past

[radiant jasper]

what

there's no way that many people do this

the sample questions are way too easy

[worldly bough]

yeah

honestly part of my interest in looking into it is just, making a fraudulent maths competition is so niche it feels almost silly. though a scam's a scam

though part of what confuses me about it is the stuff that seems like it isnt a scam, like all the video testimonials from kids and their parents

[radiant jasper]

how would you solve this (from the 2017 australian maths comp)

[high goblet]

number of digits of n ~ log(n)

[warped anvil]

yo can anyonse send their usamts r2 proofs?

my p4 proof is not cooking

[radiant jasper]

find how mnay base 7 numbers even havw the same number of digits as base 16 numbers

tkae diffent cases for number length in these bases

and solve

[burnt trail]

how

[radiant jasper]

I mean

At some point the number of digits in a base 7 number will have way more digits in that of a base 16 number

I think I saw a problem something like this at an ARML practice

Most it will ikely go. up to is 2 or 3 digits

[digital hedge]

Prove that given any set of six point in the plane in general position has a subset of three point that form the vertices of a triangle with its smallest angle not exceeding 30°

[deft wraith]

just looks like a money grab

not a scam

[acoustic nova]

isn’t this related

[radiant jasper]

why is he saying it's bad

[opaque orchid]

Hi what do y’all recommend to learn for kangaroo math competition ? And is there a book that will help

[radiant jasper]

the amt one?

[hexed oak]

so basically you want to prove you can find an angle less than 30*

let us first make a polygon

that will have angle sum

um

6*180-360

=4*180

=720

and the number of angles is

6 * (6-2)

=

24

720/24

=30

so by PHP

atleast one of the angles is <= 30

so we are done

@digital hedge

[hexed oak]

DOE = DOB - EOB = DAB/2 - EAB/2 = A/4 - (90-A)/2 = 3A/4 - 45

AOD = 180-A

CA = AO/(cos(A)) (sorry for the bash)

AD = 2AOcos(A/2)

=>

cos(A/2)cos(A) = 1/4

[pulsar terrace]

i dont think thats correct

[hexed oak]

why?

[pulsar terrace]

nvm

[hexed oak]

i have a solution after this

but it is a trig bash

(as in i have to do trig)

are you dine with that?

[pulsar terrace]

im okay

but this came in a exam without calculator

so im trying to find geometric solutions

[hexed oak]

no

calculatorr required

let y = a/2

cos(y)cos(2y)-1/4=0

cos(y)(2cos^2(y)-1)-1/4=0

2cos^3(y)-cos(y)-1/4=0

notice if cos(y) = -1/2 this would be true

factorising out 2cos(y)+1

(2cos(y)+1)(cos^2(y)-cos(y)/2-1/4)=0

since 0<=y<=45, cos(y)=-1/2 is not a soln

solving the quadratic eqn

we get

[pulsar terrace]

yeah got it

understood

thank you

[hexed oak]

sry for the bashiness

[digital hedge]

But how if the convex hull is not a hexagon, for example like a Pentagon or 4-gon etc?

I've found the solution which is different with this one

But I'm interested to find other solution aswell

[hexed oak]

i dont need a convex polygon, i only need a non self intersecting one

[digital hedge]

Oh so what you're doing is constructing a polygon that contains all those 6 points?

[hexed oak]

what is yours?

[digital hedge]

Let the point with highest y coordinate be v1 and consider the line l through v1 which parallel to x-axis. Let P be a point in l and P≠v1. We labelled the other points v2, v3, v4, v5, v6 such that <Pv1vi is increasing as i increases. Then dividing the cases when <v2v1v6 ≤120 or ≥120.
• if it's ≥120, then at least one of <v1v2v6 or <v1v6v2 is not exceeding 30
• if it's < 120, then
<v2v1v3 + <v3v1v4 + v4v1v5 + v5v1v6 < 120 which means at least one of them must not exceed 30.

[hexed oak]

yes

[acoustic nova]

https://math.stackexchange.com/questions/1011305/proof-of-eulers-totient-theorem

for completeness can someone explain the proof for when n is a power of a prime?

[mortal cedar]

ϕ(p^k) = p^k - p^(k-1)
a^ϕ(p^k) = a^(p^k - p^(k-1))
=(a^(p-1))^(p^(k-1))
And a^(p-1) ≡ 1 mod p

[acoustic nova]

but that’s mod p not mod p^k right?

[mortal cedar]

Oh, right. Sorry about that

[acoustic nova]

do u see how to do it otherwise?

[acoustic nova]

Oh someone showed it to me so I got it

[deft marsh]

find all pairs of positive integers x,y,z with z odd such that $$2018^x = 100^y + 1018^z$$

[gilded halo]

saintyzy

[deft marsh]

i need a bit help

[jolly solstice]

just started looking but a constraint I see is the power of 2 dividing 100^y + 1018^z must be min(2y,z) since 2y is even and z is odd. To make the power on 2018 match, it forces x=min(2y,z).

in fancier language, this is an example of the 2-adic absolute value's ultrametric inequality's strong property.

of course there's more to do, but nice to be able to kick out a constraint like this without too much effort to get started

[gilded halo]

christmasisoversus

[high kelp]

^ this question was very fun to solve and very gratifying once i got it; i daresay this is the nicest looking question to involve a fractional power

[sturdy panther]

i guess i'll try the problem :D

[mortal cedar]

Boils down to a single factorisation over the integers

[swift imp]

cute

[sturdy panther]

i think i'll try smth else...

we know x-16 and x+43 are square numbers with a difference of 59

so we can say that there will be a maximum case for which there may be x, because the difference between consecutive squares always increases with the value of the squared number. and that minimum case will be 2 consecutive squares.

(n+1) ²- n ²=59

2n + 1 = 59

n=29

so we have 29^2 and 30^2 as the two numbers under the radical.

so x = 900-43=857

is this correct?

it is!

[deft marsh]

im trying to understand a solution to an easy problem but i cant

the problem is :

if $x,y,z$ are positive real numbers prove that $(3x+y)(3y+z)(3z+x)\ge64xyz$

[gilded halo]

saintyzy

[deft marsh]

nvm im dumb

[marsh merlin]

Can someone help me with the following problem

14 - Reaching 2025 (Coefficient 14)

We can construct a sequence of integers by adding, at each step, twice the sum of the digits of the number. For example, starting with 1000, we get:

1st step: 1000
2nd step: 1000 + 2 × (1 + 0 + 0 + 0) = 1002
3rd step: 1002 + 2 × (1 + 0 + 0 + 2) = 1008
4th step: 1008 + 2 × (1 + 0 + 0 + 8) = 1026, and so on.

How many numbers can we generate, starting from an initial number strictly less than 2025, that allow us to reach the number 2025?

Thanks for your help

[inner ermine]

diabolical question

[marsh merlin]

You don’t have any idea to solve this problem ?

If someone has an idea, it will really really helps me

[inner ermine]

not a fan of these typa comp qns

[marsh merlin]

No problem

Anyone else ?

[marsh merlin]

Not a big help 😭

[gilded halo]

Vanellope von Schmugz

Vanellope von Schmugz

[marsh merlin]

Thanks.
I noticed some things like that before but I can’t find the final solution

[gilded halo]

Vanellope von Schmugz

[latent bough]

wsg guys

[marsh merlin]

Thank you !!!!
I’m French so I didn’t understand everything at the end. So how many numbers are working ?

[ember plank]

Little warm up for middle schooler

[marsh merlin]

what is the question with this function ?

[next sorrel]

Mod 9 digit strats can lower the possibilities significantly

[ember plank]

Small warm up. This is good for 6th or 7th graders so that they loose any faith in humanity

[marsh merlin]

Yea but what do we have to do ? Derivate it ?

[next sorrel]

It doesn’t specify what to do with the function…

[ember plank]

Uhhh simplify it

Find x

[marsh merlin]

That’s not the same things

Find x ?? There isn’t any x in this function

[ember plank]

Find f(x)

By simplyfing the expression at the right

[marsh merlin]

We already have f(x)

[ember plank]

I mean simplify the expression at the right of the "="

[marsh merlin]

Ahh ok

[ember plank]

I found this on tiktok and there´s an more fast answer

[marsh merlin]

What is it ?

[ember plank]

It is 0

[marsh merlin]

Why ?

[marsh merlin]

That’s the derivative

[ember plank]

(It´s in french so idk the english of it) I think that´s because at the bottom there´s an constant i didn´t watch all video

[marsh merlin]

Je suis de

Fr

Explique vite fait

[radiant jasper]

Stop you're giving me nightmares

[high goblet]

this is just a random splurge of symbols

[high goblet]

there is no answer because you've written nonsense

it's like "solve $\sin(magic(x)) = \exp(\ln(-\pi)) + \sum_{x=\pi}^{lmao} gigafactorial(x)$

[gilded halo]

LY

[high goblet]

like...no there is no answer because you've not written an actual problem

[ember plank]

This is not me but an populat french math tiktoker who resolves problems and he said it was 0

[gilded halo]

زَكَـرِيَّـا | ⵣⴰⴽⴰⵔⵉⵢⴰ

[radiant jasper]

129

[patent thistle]

https://artofproblemsolving.com/community/c123h3455653_3624_indexed_alternating_sums

[deft marsh]

Yeah i knew i shouldve used am gm

But got confused cause o thought 3x+y>= 2√3xy

😭

But its four terms

[lapis cloak]

Can someone help me please

[radiant jasper]

Nice

But use the givens

[lapis cloak]

I am, how is your answer coming A?

It's D in my opinion

[radiant jasper]

Why?

[lapis cloak]

As it's less than 2

None of the option is matching

[radiant jasper]

Well like say they were asking for the max but didint have gamma or delta in absolute values

What would your answer be?

[lapis cloak]

A

[radiant jasper]

Why?

[lapis cloak]

I doubt if |f(x)|<=1 h then how did max|f(x)|=9 come

[radiant jasper]

?

Based on the problem statement Do you know if f(1) is <1 or > 1?

[verbal kayak]

2=1+i2

[verbal kayak]

lol whos that realy stupid guy

[lapis cloak]

No

[radiant jasper]

Is f(x)<=|f(x)|?

[lapis cloak]

<=1

[radiant jasper]

Yeah both are less than or equal to 1

Now what if I ask about the max

If |f(x)|<=1 for all x in -1 to 1 then what do we know about the maximum value of f?

[lapis cloak]

It must less than or equal to 1

[radiant jasper]

Yeah not sure

My default is to say none of these because there doesnt seem to be enough information to get bounds for finding a maximum

[lapis cloak]

Means it's correct?,

[radiant jasper]

Well no you have enough information to say it cant be the other answers right

Break it up into three cases,

gamma,delta >0
gamma,delta<0
gamma > 0 delta <0

If you have case 1 then with the assumptions of the problem you automatically have its less than 1

[lapis cloak]

Yeah did it once bro

[radiant jasper]

Case 2 I think you can use reverse triangle inequality on f(-1)

Same with case 3 too

So like

Mah

I dont see it

Wait nvm

1/delta is less than 1 since |f(0)|<1

[lapis cloak]

Max is 2 ig?

Or 1?

[radiant jasper]

I think gamma/delta might also be less than 1 depending on assumptions

But this seems messy I wont lie

[lapis cloak]

Yeah that's why

Is this correct?

[radiant jasper]

I dont know where the x^6 comes from

That entire expression is mysterious to me

X^6+x^4 and so on

This is unclear to me

Where are you getting these first equations from?

Like you are saying cauchy schwartz?

But I dont see how that fits

Oh like you are dot producting them

I sorta see it then nvm

Its not too helpful for getting bounds tho

Oh

Maybe you can take f(D)-f(-D)?

We know both are less than 1 individually

So like |f(D)-f(-D)|=|2D(DB+1)|<2
implies |D(DB+1)|<1

Not too helpful

(f(1)-f(0) )- (f(-1) - f(0) )gives something else

[lapis cloak]

Yeah

[vernal axle]

If they had asked about the supremum instead of the maximum, the answer would have been 7, since we could take polynomials arbitrarily close to 4x^3-3x. But as they ask exactly max the answer should be (D). I think the answer 10 is ruled out, although I can't prove it rigorously.

[crystal bison]

How were u and v chosen here ? And at last,where I have marked,how it came?

[vernal axle]

any root of the equation z^n=1 has this form. You can just substitute and check. As for cosine, just draw the unit circle and mark points on it for which cos(alpha)>=-1/2.

[vernal axle]

Though I think that the phrase 'there are 2*673 possibilities of m for each k' is a wrong statement. So the final answer is not correct.

[vernal axle]

That's what I've said above. So, the correct answer should be (D)

[crystal bison]

Then what is the correct answer?

[vernal axle]

1682/3027

[hexed oak]

can someone recc me a nice combigeo?

[radiant jasper]

$$ 2024 + 1 = Goodbye 2024. $

[radiant jasper]

I dont get this argument regarding supremum

[vernal axle]

It looks like you can make this max as close to 7 as you want, but you can never reach it with some specific non-zero alpha, beta, gamma, delta.

[empty talon]

Anyone here who knows about Fubini Integrals?

[radiant jasper]

4x^3-3x doesnt get close to 7 though?

Oh you mean for the max though

Yeah I see that

[surreal willow]

Chat why is the amc 10b aime cutoff 10.5 points higher than the 10as ???

[high goblet]

i think i did this problem a long time ago, but i think this was a fairly decent combigeo

this is somewhat combigeo

[high goblet]

what exactly do you mean by this?

also this would belong in #real-complex-analysis or #advanced-analysis depending on what exactly your asking

[hexed oak]

Thanks

[high goblet]

i've tried searching for fubini integrals but i can't find anything

did you perhaps mean fubini's theorem, which is about integrals?

or were you more just asking about generalising integration?

[empty talon]

oh thanks a lot

[empty talon]

yes- so those specific integrals are called fubini integrals

[high goblet]

i don't think i've ever heard anyone refer to them as fubini integrals

fubini's theorem is about integrating over product measures and is often applied in the case where we have the lesbesgue measure i.e. the lebesgue integral

[high goblet]

also what are you asking about?

[empty talon]

yeah I mean they call that iterated integrals I guess

[high goblet]

yeah i think that's more common

[high goblet]

^

!da2a

[scarlet root]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[empty talon]

yeah I had a doubt with a particular question-

[high goblet]

if it's about measure theory, post it in #advanced-analysis

[empty talon]

alr, thanks a lot

[radiant jasper]

guys

i just came across a suspicious server

that seems to be leaking answers to AIME

i left only seconds after i joined

it's called "anti-maa"

there is something seriously shady about it

i think i should just join back and report

how do i report it my fears are confirmed

code for server

someone help me report it

[radiant jasper]

please

[warped anvil]

broski

just report it

how did they even get aime answers??

i would highly doubt it is real but still report it

also i would recommend

not to put the code

you are just giving more people the opportunity to join

[radiant jasper]

idk how to 😭

[warped anvil]

you can report the guy to maa staff

https://support.discord.com/hc/en-us/requests/new?ticket_form_id=360000029731

you can make a ticket here

[radiant jasper]

bro

i found out theres an entre network of these crooks

[warped anvil]

💀

there always will be

i would recommend the best thing you can do is not tell regular people about it

you could report to maa staff, and hope that all comp math servers such as ommc and tops oj insta delete any ads

or you could report this guy to them beforehand

[pulsar terrace]

i have tried menelaus theorem

but cant proceed

[ivory ember]

Hint: There’s a cyclic quad. Find it.

[pulsar terrace]

okay

can you like ||the hints||

[ivory ember]

Hint 2: ||Power of a point|| tells you that ||(KM)(KN)=(KO1)(KO2)||.

[radiant jasper]

Hi

Apnara bangla cotha bolte paare?

[pulsar terrace]

lol

[sand hull]

Ha bolte paare

XD

[sand hull]

What?

[sand hull]

It's wrong

[torn gate]

Two real numbers, x and y are chosen at random over the interval [0,1]. What is the average of the value ceil(10x-y) - (10x-y)?

[warped anvil]

the expected value over a uniform range is a+b/2 right?

so wouldnt x and y expect to be 0.5

so wouldnt it be 0.5?

or am i wrong @torn gate

[torn gate]

Not all values of 10x-y are equally likely

[warped anvil]

oh yea ur right

[high goblet]

<@&268886789983436800> advertising

[warped anvil]

why were the moderators advertising

[ornate blade]

You ping mods if someone is advertising

[blissful hemlock]

hi guys! does anyone know any sources to learn olympiad geometry from scratch?
i mean i do know like id say 60% of the theorems and stuff but my mind goes blank the second i see a geometry question

[surreal willow]

How does a server get their hands on AIME answers 😭 😭 😭 😭

[surreal willow]

The euclidean geometry in mathematical olypmiad book is really good

its free to

except it does use some more advanced stuff and techiniques

you can also try grinding alcumus aops

[ivory ember]

? This follows directly from the fact that $O_1 O_2 MN$ is cyclic

[gilded halo]

Civil Service Pigeon

[reef condor]

No it’s not (unless you pirate it)

[surreal willow]

I found a website that gave the url out

[sleek ravine]

probably pirating lol

[blissful hemlock]

yeah i got a pdf from a friend, thanks again!

[surreal willow]

Np

[surreal willow]

Nah website seemed legit

[sleek ravine]

o hm interesting

[surreal willow]

Ye and it didn’t force me to download anything either

It redirected me to the pdf

[hearty otter]

Is K a random point ?

[radiant jasper]

someone signed up as a proctor, got the answer key a day early, and sold the key to kids in the server

the guy even sent a picture as proof

[surreal willow]

😭

How much did it cost??

[radiant jasper]

idk and idc

it's just plain wrong

[swift imp]

Yea it’s pirated but

Who doesn’t pirate

Pirate to see if it’s good then buy it

[jade widget]

A group of 10 people visited a bookstore. It is known that:
1. Each person bought exactly 3 books.
2. For every pair of individuals, there is at least one book that both have purchased.

What is the minimum number of people who could have bought the most-purchased book?

I need a better way to solve this!!

[hexed oak]

and you draw a graph from books to ppl who bought the book

so you get degree of each person is 3

and that any two ppl are at a distance of 2

and you have 10 ppl

hm

so let us say a person get 3books

and those books are shared with all 9 others

so we have

that

(by PHP)

there exists a books which is bought by 4 ppl

now the only construction which works for this

is if all books are bought by 4 ppl

but you have 30 buyings

and

4 doesnt divide 30

so

some book must be bought by 5 ppl

notice 5*6 = 30

so just have 6 books

and have any person by 3 of those books, this is possible to do with overlap in any two ppl, since there are 6C3 = 20 ways of buying 3 books, and overlap doesnt exist only for 2 buyings, so there are 10 pairs of buyings in which overlap soesnt exits, and we can have each person buy from a different pair, and so all buying will have overlap

so we are fone

[verbal kayak]

sigma boy bad

[jade widget]

Ahhhh

Thanks

[teal wave]

im absolutely cooked but can u guys help me

[deft marsh]

can someone explain to me how to solve exponential diophantine equations

for example 3^x + 4^y = 5^z which is a rlly simple one

[reef condor]

No easy way

Just bashing mods and residues is really the only way

[hexed oak]

hm

3^x=5^z-4^y

if x =0, we are done

3^0+4^1=5^1

z is even

3^x+4^y=5^2c

if y = 0, then there are no solutions

x is also even

3^2a+4^y=5^2c

if z is 0, there is no solutiom

2^2y=(5^c)^2-(3^a)^2

2^2y=(5^c-3^a)(5^c+3^a)

both of these are (non 1)powers of 2

2^r and 2^j suppose

2^j-2^r = 2*3^a

so

(2^j-1)-(2^r-1)=3^a

so

r = 1

so

5^c-3^a=2

and

5^c+3^a=2^j

so

2^(j-1)+1 = 5^c

if we are allowed to use catalm this is trivial

j-1 = 1 or c = 1

only

2^2+1=5 works

j = 3

r = 1

2y=j+r=4

y=2

so

3^2+4^2=5^2

@reef condor

[deft marsh]

why is z even

[hexed oak]

since

mod 3

[deft marsh]

oh because $$(1)^y &\equiv (-1)^z\mod{3}\$$

[hexed oak]

yep

[gilded halo]

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

[deft marsh]

and x is even because u took mod 4

[deft marsh]

$$ 2025^x = 5^y + 2000 ^z$$

find all solutions to the equation $2025^x = 5^y + 2000 ^z , x,y,z \in \mathbb Z$

[gilded halo]

saintyzy

[reef condor]

Consider the powers of 5 on each side for a start

(The greatest power of 5 dividing LHS and RHS should be equal)

[lilac niche]

Please?

Does anyone know how to solve this ?

<@&286206848099549185>

[hexed oak]

do you know how to figure out the number of discontinuities of this function?

try the values were [x+n] or [x] changes

and see if those points are discontinuities

rgar way you can see which value of n has how many discontinuities

[low fiber]

are there any good tools to make a prep plan for next years math olympiad?

[fast coral]

study what you like and when the exams are nearby mock the sample questions of the oly you’re taking

[radiant jasper]

"Guess what? I haven’t showered since last year!"🤣

[inner bison]

do you have the a_0, a_1, a_2? If you calculate some examples it might help

[plain jacinth]

So I barely qualifed for Aime

And I am a sophomore

And I was wondering if getting into usamo even matters senior year for college apps

[swift imp]

It can

[plain jacinth]

how?

Aren’t college apps due before usamo resulte come out

[swift imp]

Wdym

Ur question is asking about

Doing usamo senior year

Or doing usamo for college apps?

[plain jacinth]

If you do good enough on the aime senior year and qualify for usamo will you be able to put it on your college application or will it be too late

[radiant jasper]

Brooo what grade lvl

[lilac niche]

JEE MAINS level

[radiant jasper]

Oh it’s jee

[lilac niche]

Yea

[radiant jasper]

What does mains level mean

??

[deft marsh]

could someone drop a solution?

[tame burrow]

Do you need full solution?

[swift imp]

For college apps yea

[lilac niche]

It should be understandable how you got further steps

@radiant jasper you are blackpink fan a BLINK💜!!

[tame burrow]

ok

Lemme see

so each discontinuity occurs at x = k-n, where k is an integer. To make sure that x falls into the interval, we set the boundary -2<k-n<1, so n-2<k<n+1. The number of integers that satisfy this inequality is 3. So, for each n, there are 3 points of discontinuity. Since there are 2026 n values, there are a total of 6078 discontinuities.

[lilac niche]

There are 2025 n values and the options are 4 3 4052 and 2027
Btw even I got till getting the three inequalities but after that finding the summation I am stuck

[fast coral]

floor of (x+n) = floor of (x) + n n-> integer

[tame burrow]

isn't it 0 to 2025? so there are 2026 values of n

[fast coral]

so you justt need to care about x

not n

i think

[tame burrow]

No but there are 3 discontinuities for each value of n

so its just number of ns * 3

[fast coral]

yh

i just mean number of discontinuities is irrelevant to what the value of n is

[tame burrow]

Yeah

there's a summation with bounds

I just used that

Am I dum?

[radiant jasper]

Yeahh

[gaunt mural]

There are exactly 2 discontinuities for each n, because there are two integers in the range (-2,1). So the answer is 2*2026=4052.

[tame burrow]

Oh yeah

That's true...

[deft marsh]

In chess, a king threatens another king if, and only if, they are on neighboring squares, whether horizontally, vertically, or diagonally . Find the greatest amount of kings that can be placed on a $12 \times 12$ board such that each king threatens just another king. Here, we are not considering part colors, that is, consider that the king are all, say, white, and that kings of the same color can threaten each other.

[gilded halo]

saintyzy

[deft marsh]

someone drop a solution ples

[shadow spruce]

$Happy:\left(\sum_{i=1}^9i\right)^2$

[gilded halo]

Aestusy

[warped anvil]

treat each two pieces as a 3x3 grid

one of them is in the center and the other is in the edge

we want to minimize the number of areas a king gets so we put him in the edge and corners

so we get 12/3 * 12/3 = 16 * 2 for each 3x3 grid which should be 32?

@deft marsh is that right

or can we do better

[deft marsh]

ill check it out tomorrow cause im going to sleep rn

[warped anvil]

okay

no wait we should be able to do better

[deft marsh]

but the answer is not right

[warped anvil]

im thinking it is 40

[deft marsh]

if u want i can send it to u

[warped anvil]

yeah can u send

[deft marsh]

56

[warped anvil]

oh aight

hmmm

[deft marsh]

im not sure yet im trying to get a solution but ill check it out again tomorrow

[warped anvil]

okay

ill try to figure out one

[deft marsh]

alrr

lmk if u do

[warped anvil]

yeah aight

ok i found a way to get 48

[radiant jasper]

Will anyone comment on my videos if i upload some math videos on yt🥺

[deft marsh]

whats the name of ur channel

[radiant jasper]

Havent made one yet

[deft marsh]

oh alr

[shadow spruce]

I will

[hexed oak]

36

with proof

take the 36 2x2 squares

each square can have atmost one king

so atmost 36kings

now just put each king at points which can be marked (even,even)

clearly no king can attack another king

and this config has exactly 36 kings

so

atleast 36 kings

so there must be 36 kings

[deft marsh]

the answer ive seen was 56

[deft marsh]

We have a standard chessboard with dimensions ( n! \times n! ), where ( n \geq 2 ). We remove the top-right and bottom-left squares. Can we cover the remaining chessboard with ( n \times n ) dominoes? Each domino can be placed either horizontally or vertically, covering two squares of the chessboard.

[gilded halo]

saintyzy

[shadow spruce]

When does this bot know he needs to render LaTex

Hello $x$

[gilded halo]

Aestusy

[craggy drift]

eyy 56 is possible. no idea how to get the solution though

[craggy drift]

^

[radiant jasper]

Is anyone here taking the waterloo math contest?

I keep getting the same scores on the practice, and the score is like 2-3 questions away from making it to the honor roll

How can I break this barrier?

[ebon escarp]

No. Just shooting of the top of my head. n = 1 yes. because you said nxn right and you removed from the top corner, so you can't ahve 1x2(n). So your grid is going to collapse when you remove those corners, everything is going to shift, you will need 1 x some number. I Think.

[hexed oak]

in these squares, some kings attack each ither?

[hexed oak]

what do you mean by nxn domnioes?

oh

i misread the qn

i though it means no king threatens another king

sry

the answer should be

hm

so

let us say there are n kings

then adding the 4 diagonal adjecencies and the 4 normal adjecenies, we get there are n attacks

[hexed oak]

we know that in a 3x3 square there are atmost 4 kings

so we get an upperbound of 4*16 = 64

hm

we can use that in the 3x3 surrounding a king there are exactly 2 kings

and combing there arenas

we get

14

14/2 = 7

14^2/7

=

28

hm

how many arenas can share a point

upto 4

so

so

[deft marsh]

the problem says $n \ge 2$

[gilded halo]

saintyzy

[shadow spruce]

Hi

[plain jacinth]

So I barely qualified for aime and I was wondering how difficult the Sumac and Promys entrance exams are compared to aime

[silk spindle]

Did u take the amc 12?

[burnt trail]

which book is good for preparing for cemc cayley

[radiant jasper]

Oh ur taking it too so am i

D - 54

[plain jacinth]

No amc 10

I got 115.5 on the A

[burnt trail]

which resources are you using to prepare for it

[radiant jasper]

Just practice tests

U?

[reef condor]

Harder but more time

Definitely harder though

[burnt trail]

just doing past problems, doing what I can

[radiant jasper]

Ah same here

What do u usually get on the practice tests??

I can’t do 24 or 25

23 if the question is my type

[cyan forge]

Who wants? 1650 questions in total (50-100 questions each topic)

[upbeat jay]

woah I would like to see

[dapper magnet]

is that x^3 + 3x + 29

[sick tapir]

na it's +2 , there's a comma after tht

also you can try this if you want it is good but u might do it

( More than one correct )

If need solution for this DM me

[sullen crypt]

can anyone solve this

[sick tapir]

this is ez

[sullen crypt]

r u sure

[sick tapir]

yes

wait

just doin

[sullen crypt]

u r from which country

[sick tapir]

there you go and imma go play basketball have fun

[sullen crypt]

thanks

[sick tapir]

actually i take it back

[sullen crypt]

@sick tapir

[sick tapir]

it was good i mean challenging

[sullen crypt]

i have a way more tougher question wanna see

[sick tapir]

Korea

[chilly pilot]

Bro i answered your qn in pre calculus

[sick tapir]

i'll pass wanna go basketball

[sick tapir]

i saw it u r right

its ans are 16 and 666

cya boys in 2 hrs

[sick tapir]

nah bro matrix is way too ez

i only love calculus and trigo

and geometry

i mean i love MATH

cy

[sullen crypt]

u know its questions from world toughest exam

2nd toughest

it is held in our country

[chilly pilot]

Its easy

Matrix is the easiest in jee adv ig

[sullen crypt]

bro but its tough for me

[chilly pilot]

definite integration and seq series 💀☠️

[sullen crypt]

bruh i am in 8th

[chilly pilot]

Why u solving jee qn's then lol

[sullen crypt]

because till 11th i have completed

now 12th started in winter

break

[chilly pilot]

Bro ☠️💀

You are Future AIR 1 then if it's true

[sullen crypt]

from 5th i have only one goal

360/360

1st person to do so

but thing is that

i only study maths and physics

[chilly pilot]

That's all you need

Jee chemistry is ncert based

Its most scoring in the exam

[sullen crypt]

ur class

[chilly pilot]

12th

[sullen crypt]

did u participated in ioqm

@chilly pilot

[chilly pilot]

Nah focusing on jee rn

[sullen crypt]

its better than it

[chilly pilot]

And the questions asked In the exam are way too easy .. it's just about the speed

[sullen crypt]

have heard same line

[sullen crypt]

its good to watch anime with study

[thin perch]

Could someone help me with P4a

[gusty verge]

Split the sum into odd and even terms

This will allow you to get rid of the floor and hence use |xy|<1

[chilly pilot]

Is that greatest integer function? In the exponent

[thin perch]

thanks

[thin perch]

you mean this [x]?

[stable path]

any jee student

?

[chilly pilot]

Me

[thin perch]

it returns the integer part

ex: [3,7] = 3

[sick tapir]

ik every exam is world in which mathematics is there

cause my favourite subect is MATH and i cant afford to lose even 1 single mark in MATH

[chilly pilot]

Wth 😂

[sick tapir]

and your JEE ik it very well its hell too ez for us and the math which comes in it is very low level

Come to MIT then we'll have a chat

[chilly pilot]

☠️💀☠️💀☠️💀

I strongly disagree with you here

[sick tapir]

some1 from indian named Chirag is at MIT i m his friend he showed me his question and they were average

but not too good that can;t be done

[chilly pilot]

Just so you know jee is given at the age of 17

[sick tapir]

but i deeply respect you all for giving it at an age of 17

[chilly pilot]

Yeah bro i think this level of mathematics is good enough for 17 y/o people

[sick tapir]

for a 17yo i agree

[chilly pilot]

Bro have you ever tried the book "50 challenging problems in probability"

[sick tapir]

no

[thin perch]

sorry for bothering but could you show me the next step after that. I got two sequence and used geometric series but it seems did not work

[sick tapir]

but now definetly will

thanks for the book btw

[chilly pilot]

Yeah bro its difficulty level is extreme for me

You will definitely enjoy it if you like solving difficult problems

[gusty verge]

Uh I haven't done it but it might be easier to set |xy|=r

[sick tapir]

yeah ido

wait tell me is this the one

[sick tapir]

which one of these 2 iis it

[chilly pilot]

The second one

Bro are u straight away buying it 😂

[sick tapir]

and can any1 try my own made question

[sick tapir]

ya

[chilly pilot]

Ahh i hate conics

[sick tapir]

try pls

it took 3 weeks to create a tough one

[chilly pilot]

I don't even remember the formulae

Let me take out my notes

[sick tapir]

wait

i'll help you

its easy

just look for foci and asymptotes

[chilly pilot]

Bro i dont remember shit in hyperbola

Where the f is its asymptote ... Is it its directrix

[sick tapir]

still try any1 ?

nah

see the image

hey

wanna see just the result of the area what it came out ?

or i should send the detailed solution

[chilly pilot]

Yeah send it maybe I'll revise hyperbola Tommorow and then try the question

[stable path]

check dm

[sick tapir]

hu ?

[chilly pilot]

I found the equation of tangent by derivative but its different

[sick tapir]

send the pic

[chilly pilot]

Oh sorry calculation error

[sick tapir]

nvm

its tough

[chilly pilot]

I didnt even differentiate the constant 💀

[chilly pilot]

T=0 better fr

[sick tapir]

ha ?

[chilly pilot]

The method by which you derived the eqn of tangent

We say it "T=0 method "

[sick tapir]

ohhhok

well its ok

you can try your methods

but just remember to mark the steps pls

[sick tapir]

yeah ik T=0

i just couldnt connect to it cause u randomly said T=0 better

mb

it can be applied but it will go long

[chilly pilot]

Yeah i am bad geometry

I am good at calculus and probability

[sick tapir]

want it ?

[chilly pilot]

Question ?

Sure

[sick tapir]

tell me topic ?

[chilly pilot]

Give me probability

Avoid pells theorem cuz it isn't taught in india

[sick tapir]

alr

you got this

[chilly pilot]

(39!)^2/(52)!(26)!

Multiplied by 4 too

[sick tapir]

why dont you do its full solution and give the numerical value

[chilly pilot]

Bro 💀 factorials

[sick tapir]

is 4 in N or D

[chilly pilot]

N

[sick tapir]

well it's right

good

[sullen crypt]

wanna do 1v1

[sullen crypt]

r u in mit currently

[sullen crypt]

thats my spirit

and in jee it includes pcm all 3 subjects in limited

and its way tougher than mit entrance exam

INDIA WILL ALWAYS LEAD MATHS

and i will prove IIT>MIT

[sturdy panther]

as an indian, ek baat kahu?

it's seldom that we see an indian winning in a prestigious competition such as IMO, THE math olympiad.

the type of questions that are asked in iit-jee math and ESPECIALLY mcq's require rote memorization of formulas; while it can be really useful, doesn't truly test mathematical and logical thinking to the fullest extent.

a more well-rounded and smart 12th-grade-end advanced math paper would be smth like HSC math extension 2.

[split quail]

bro stop texting N study

[sturdy panther]

texting?

you are an undergrad. should care a little more about yourself then to tell a 13 year old what to do.

[split quail]

ik i care abt myself

[sturdy panther]

but like isn't undergrad supposed to be hard

like maybe you should study a little more?

because i'm still young and i dont need to study too much

[split quail]

doesnt mean ill study 16 hrs a day

[sturdy panther]

so you expect me to do so?

[split quail]

na even if u studied that much u will fail

enjoy

[sturdy panther]

how can you make that assumption?

give me your basis man

[split quail]

coz ur dumb

[sturdy panther]

im not going to argue, i just want to know why you made a handful of snide remarks right now.

[sturdy panther]

in what way?

[split quail]

every

[sturdy panther]

explain bro.

well, first of all, you have some french in your bio.

do you know how to speak french?

im an indian but i can.

[split quail]

kinda

[split quail]

oh

[sturdy panther]

vous pouvez comprendre tout ce que je parle?

[split quail]

lemme get translator wait

[sturdy panther]

and even then you call me dumb?

please don't be rude. this is a math forum not an argument club.

[split quail]

idh much time to focus on other langs like u

[sturdy panther]

study 16 hours a day then

[split quail]

learning lang js requires practice not skills

[split quail]

ur dumb if u will study that much ur brain wont get rest

[sturdy panther]

skills are earned through practice.

just like in math.

[split quail]

learning lang isnt skill

[sturdy panther]

language is a skill.

of conveying things elegantly

of speaking and drawing words easily

[split quail]

ur dumb

fr

[sturdy panther]

i understand that nothing constructive can be done with you in this state

i will stop talking over here. do as you please man

[split quail]

times up kiddo

[sturdy panther]

dont be rude, peace out

[split quail]

piss off

[sturdy panther]

<@&268886789983436800>

idek why he decided to attack me :(

[blazing osprey]

very troubled individual

[sturdy panther]

💀

[blazing osprey]

@split quail you know better than to say anything you've said here. come back when you can be nice to others. or don't.

[fossil adder]

Battle of virgins

[sturdy panther]

😭🗣️‼️

[fossil adder]

My bad yall are 13

Continue

[sturdy panther]

nah i dont care dw

[sturdy panther]

hey man, you're in 8th grade right?

i see that u completed ur 11th grade syllabus as well :o

[prime raft]

eww noo

[fossil adder]

@sturdy panther btw I agree with what you said

The other guy seem immature

[prime raft]

is this bullying

[fossil adder]

Made some good points on competitions

[sturdy panther]

def

[prime raft]

well he could have at least @ them

ur taking too long to type @fossil adder

see, problem solved

[fossil adder]

Bullying is way better than high testosterone virgin fights in mathematics

[prime raft]

wtf

what r u talking about now

[fossil adder]

Iykyk

[prime raft]

idont

it has nothing todo with me

[fossil adder]

People in math are very passive aggressive shall I say

Bullying is better

[prime raft]

why better

[fossil adder]

Plus calling someone immature is not fucking bully

[prime raft]

and as far as i can tell no ones passive aggressive

[fossil adder]

Get it right

[prime raft]

that just means you dont like context

sorry

[fossil adder]

I’m not implying anyone here is

[prime raft]

what are u an english person

[fossil adder]

I’m saying in general, people in Math

[prime raft]

no clue why u say that but ok

[fossil adder]

Go to college then you will find out

[prime raft]

i dont have money but sure maybe one day

i literally dont get what ur going on about high school stereotypes

[fossil adder]

I’m not talking about HS

[prime raft]

yeah well it sounds like it

[fossil adder]

where

[prime raft]

^

[fossil adder]

Maybe communication is not your strong suit

[prime raft]

wha why does it matter

i keep getting this feedback

but idk why ppl say that

https://tenor.com/view/asteroid-in-love-anime-mai-inose-crying-gif-9272778323128881761

[fossil adder]

Rather not get into it, let people continue their debate on math competition

[prime raft]

ok

still dont know why were talking about this

[fossil adder]

You should go to college btw, if you’re interested in mathematics

[prime raft]

none of ur business

[fossil adder]

Just a suggestion

[prime raft]

https://tenor.com/view/j-train-gif-23675589

i see very funny

[fossil adder]

Very sincere suggestion

[prime raft]

why do you even wanna talk to me

im questioning ur intentions

[fossil adder]

Cuz internet is not real

[prime raft]

well tht's a debate on it's own

[deft wraith]

is this competition maths

[prime raft]

no

[fossil adder]

Here comes user95040

I’ll leave, 95040

[deft wraith]

ok

[gaunt mural]

x~/~x+n/k since n goes to infinity

[chilly pilot]

True

[plain jacinth]

How do I learn the material on Promys and Sumac

[reef condor]

Wtf

[reef condor]

Grind comp math

[radiant jasper]

pumac susus

[reef condor]

avg pumac a8 impossible question

[sick tapir]

good question

[stable path]

ye jee ka que h ?

guys anyone who is preparing for jee so badly ?

[sick tapir]

there you go have fun

Btw indian boys

i m learning hindi also

i learned NAMASTE BAI KHAISE WHO MERA NAME RUDYII HAI

[sullen crypt]

its namaste bhai kaise ho mera nam rudyii hai

[sick tapir]

OK

[sullen crypt]

r u in mit

@sick tapir

[sick tapir]

yes

i m on holidays thats why i joined this server , in 3 days i wont be online anymore

[sullen crypt]

ohk

[sick tapir]

what you wanted somethin from me ?

[sturdy panther]

hmm

when n is even, we can instead rewrite all instances of n with 2n, thus simplifying some fractions in trig.

so pi *n /24 becomes pi *n/12

we could do some more...

there's a certain point at which cos(pi * n/12) = cos (pi *m/12) although m and n are distinct integers

taking cos(pi * 0)=cos(2pi)= (cos(24pi/12))

So the denominator actually repeats itself!

every 24 numbers.

we'll calculate the same for odd summation.

we replace n with 2n-1

so cos pi *n/24 in any case becomes

cos (π(2n-1)/24)

so we now take

cos(π(1/24)) = cos(π(1/24)+2π)

cos(π(49/24))

49 is 2n-1 where n is 50.