competition-math

[olive haven]

And I am talking even the Silicon Valley

[raven oyster]

oh

[olive haven]

There’s different towns and regions and unless you’re like near Stanford or in Cupertino, you’re not getting the best in a public school

[raven oyster]

lol

i mean my school really forces u into this ciricullum where PCM(physics chem math) is so easy but geo history bio and ur second language r tough

so i hv to focus on psychology bc its rly hard and im also studying for history bc our syllabus is 3 yrs long

[olive haven]

That might also depend on you. For some people the history may be super easy but the math may be hard

You are just more interested in the scientific aspects and have self-studied there

[raven oyster]

ey but ur restoring my faith in my edu system ty : )

[olive haven]

Lol no problem

[raven oyster]

yk sometimes the problem is tht the principal doesnt know how the teachers teach, she js gets somewhat qualified teachers

but i can blame her

bc like

my physics teacher got rly good in JEE physics section

but shes not the best at astrophysics which is a large part of our portions

but what can u do

[olive haven]

Lmao it all depends on the principal

[raven oyster]

but the principal teaches eco so our eco teachers are COMPLETELY locked in

we hv fire eco classes

[olive haven]

Anyway, we kind of got off topic. This is the competition-math channel

[raven oyster]

lol good point

what grade r u in?

[olive haven]

Going into 9th

[raven oyster]

oh same

did u take amc 8

i js want to know like

what portions were in there

[olive haven]

Nope. Never taken AMC. Just done past papers

And I’ve only really looked at AMC 10 papers

[raven oyster]

oh

my friends say that its really combi based

like how much tho

30%?

[olive haven]

AMC 10??

[raven oyster]

yea

[olive haven]

It depends on the year Ig

Last paper I did and looked through had like 4 combi questions I think. Might be wrong tbh

And those were the ones that got me stuck because I kind of suck at combinatorics

[raven oyster]

lol sameee

when i took my 'mock' amc10 test my problem solving had to pull through bc i had no idea how to solve it

[olive haven]

So combi is around 20% in papers

[raven oyster]

ah

thts not aa lot

[olive haven]

But it really depends on the year

[olive haven]

Real

Anywayy, pretty late for me, so I am going to sleep. Bye 👋🏻

[raven oyster]

byeee

[granite zenith]

Can anyone suggest a beginner- friendly textbook or any resource on The Invariant Principle?

[near hill]

Is that even a standard name of anything? It sounds much too nondescript to be the name of a single thing.

[radiant shoal]

maybe one day ill join comp math

where can i practice besides aops

[inland totem]

hmmm

[severe eagle]

its a method in combinatorics

[severe eagle]

Rushil Mathur combi

theres a chapter on invariants and monovariants

[thorny tusk]

how do yall know all these books

[fast atlas]

Listening to recommendations of people who probably never read them

[scenic remnant]

Can vouch for him

[opaque heart]

hello guys, i have just started 10th grade this year, however i have covered syllabus uptill 11th, does anyone know any good competitions or any olympiads around the corner that i could register for?, im also tryna work on my college profile and it could reeally help, (P.S. - dont judge the bland profile, i have only started discord today), i'm a fast learner so even if it is a little advanced i can truck through
Thank you!!

[molten lotus]

does anyone have a resource I can use to practice linearity of expectation problems

[thin pumice]

amc12 would be good if hr in north america

[upper jackal]

Is there a systematic approach to test if a large number is prime or not during competitions?

[dark meadow]

take the sqrt

and check all primes lower than the sqrt

[upper jackal]

damn looks like im gonna be spending the next 30 mins checking 1-130

[flint dagger]

help

[molten lotus]

Not really u can first check if it is divisible by 2 then skip all even integers until 130

[upper jackal]

yea i didnt mean 1-130 all numbers inclusively i meant all the prime numbers in it

[molten lotus]

Oh ye

[upper jackal]

Still takes a while to do 15959/every prime number up to 130 haha 😭

[quiet belfry]

open a ticket

[inland totem]

hmmm

[vernal axle]

||1. integrate x^2(1+x/n)^n over [0,1]
2. use definition for exp(x)||

[molten lotus]

What contest is this for

[flint dagger]

advance

jee advance

[radiant shoal]

is that tetrated C_k

[upper jackal]

whats the fastest way to solve a quadratic like this on an amc test (no calc)

[near hill]

Plug into the quadratic formula.

[upper jackal]

Alr but it would consume a lot of time calculating the discriminant i was just wondering if there were any tricks

[frank hornet]

I guess you can use rrt and try if some roots work?

[upper jackal]

Oh perfect thank you so much it already works at -1/2

[quiet belfry]

anyone planning to do los angeles math tournament

[sturdy knoll]

Which quesyion resulted in tjis equation if i may ask

[upper jackal]

It was a trig question with law of cosines where i had side x and y:
x^2=452-448cos(θ) y^2=281-160cos(θ)
and x-y=7

I let cos(θ) = a which is where i got a

But dw i already solved it

[stuck tartan]

LOC?

oh oops nvm

ignore that

[upper jackal]

yea

[sturdy knoll]

Which yewr

And what question

[viral sierra]

No, i live at the opposite side of the country

[sturdy knoll]

Complete the square maybe

[upper jackal]

Its not on any actual competitions but its part of a competition-math problem set my teacher gave me

I solved it though dw

[sturdy knoll]

Oh

[fallen apex]

Guess the solution

[thorny tusk]

isnt that... just proving the quadratic formula with the values already plugged in?

[uncut pond]

factorising?

[sturdy knoll]

It’s quicker yhan computing the values by hand

Imo

[opaque heart]

im acutally in india

[opaque heart]

u can also try middle term split

quadritic i feel would be too long

[bold oar]

IOQM for olympiad

[opaque heart]

thank yu

[short sand]

hello could anyone in france share JBMO selection tests, i would really appreciate it

i am willing to share my countries national imo selection test

[near hill]

The point of the quadratic formula is that it describes exactly the same calculations you'd do when completing the square, just without writing down the intermediate equations -- and with some overall simplifications that postpone working with fractions until the very end, in the case where the coefficients are integers. Those simplifications are there to make the calculations easier than completing the square directly would be.
The main benefit of completing the square vs plugging into the formula is that completing the square can be easier to remember because the intermediate steps make individual sense. But if you've memorized the formula anyway, using it will generally be a win.

[opaque heart]

dayumm, intelligence quotiant as high as the name

[inland totem]

hmm

[lament bluff]

Any tips on how to score high on the Putnam?

[viral sierra]

become dotted

but in general, i think youll prolly have better luck in the olympiad server, because most of us here are far too young to do putnam

[inland totem]

and far too norz

[sturdy knoll]

Guys

How would I figure out all possible integer divisors of n!

[inland totem]

uhh if it's small enough i js write it out

if n is large then i use legendre

[sturdy knoll]

It’s 12

I’m noy tryna write it out

What’s legendre

[inland totem]

legendre's formula

search it up

it's pretty useful

[sturdy knoll]

Ts look so complicated

Ain’t there an easier way

[inland totem]

uhhhh

idrk

[sturdy knoll]

How would this work for 12!

Is it likr

[inland totem]

like what?

[sturdy knoll]

Nvm

I found an easy method

That has to do with prime factorization

How would I find the nunber of perfect square divisors tho

[inland totem]

hmmm

[zinc ridge]

u do the same but instead count the choices where u do 0 2 4 6... up to n

[sturdy knoll]

What

Do the same as what

[zinc ridge]

nevermind

i have to go

[sturdy knoll]

Hmmm

So like

What are we thinking

[viral sierra]

it isn't too bad once you get used to it. the idea is pretty simple.

[sturdy knoll]

I found an easier method

I think I like it more

[inland totem]

ooh

[sturdy knoll]

It’s with like prime factorization

[zinc ridge]

why do you keep saying that

legendres is literally prime factorization

[sturdy knoll]

It looked to complicated

[zinc ridge]

whatever your method is is likely some variation of legendres

[sturdy knoll]

I js did smth with the exponeents

Probably

[radiant shoal]

67

[proper sigil]

IM NOT TAKING MY SNEAJERS OFF I AM SNEAJERS O'TOOLÉ

Can someone help me enroll in sneajer math tournanent?

LOL SNEAJER

OI SNEAJER

[remote phoenix]

you've already been warned not to shitpost in this channel.

[zealous reef]

Does anyone have a handout on Burnsides lemma

[severe eagle]

??

[near hill]

A serial troll or group of trolls seem to have picked this channel to raid every so often with "sneajer" as their warcry. It's probably not worthwhile to attempt to figure out why.

[carmine junco]

is there any calculus olympiad or something, i hate geometry and number theory

[severe eagle]

aw

[radiant shoal]

probably

[viral sierra]

You could probably look into past competitions such as bmt which have a calculus round

[round crater]

heya

[pallid dragon]

looks like PIE

like it can have 1 to 5 pairs that add up to 21

and you know how many total

etc.

yeah it should work

no okay it's easier

you just have to pick one number from every pair, so 2^10

[carmine junco]

alr

[round crater]

yea how does that work tho

[pallid dragon]

there are 10 pairs

you can't take both numbers, so at most 1

but you need to pick 10 so at least one

why can't i get the same answer with pie

got it

like suppose you choose 4 numbers from 1-10
now you need to pick 6 from 11-20
but there are 6 numbers left there, the other 4 would pair

so 10c0 +10c1 +10c2... +10c10 = 2^10

[echo ocean]

from 1 to 20, if you pick a number then there will be another one such that their sum is 21

Now you decide to choose one of them

that's 2 possible ways each

[round crater]

ohhhh

nice

yall shud give CAT for the fuck of it

[carmine junco]

has anyone done psat here?

[echo jacinth]

https://www.xplusa.com/m/openproblem/be05e2e6-077c-42d1-b7ce-2645bedc6d97/

Can anyone help?

[near hill]

Minimum? Surely 250.

[kindred wraith]

distinct integers

[near hill]

Whoops, I need to read properly.

0+40+320+640=1000 gives 640 -- but that might not be optimal.

[twin raft]

50 + 200 + 250 + 500 gives only 500

you can lower it further by using numbers with large gcd

which might be optimal actually

[twin raft]

oh wait no i cant count, this gives 1000 as lcm

[twin raft]

hmm, do we allow 0?

since the set of common multiples would just be {0}

but also lcm is the least positive multiple, which is kind of an issue if there are none

[near hill]

Oh foo, you're right.

[twin raft]

i think i figured it out

[near hill]

Then -400+200+400+800 gives 800.

[twin raft]

write a, b, c, d using the lcm (and introduce some new variables), if you dont understand what I mean,
||Let lcm = L, then a = L/x, b = L/y, c = L/z and d = L/w||

oh i forgot about negatives, nvm, doesnt break my proof

[near hill]

Oh, or 8+32+320+640=1000 for 640 still.

That looks convincing.

[twin raft]

ill delete it and leave just the hint, hope the op can figure it out from there

[inland totem]

hmm might check this out later

[kindred wraith]

the fact that it says "be the first one to provide a solution" and that its an open problem that is easily solvable by computer search is just a huge turnoff for me

like, why do i want to spend my time working on a problem that currently feels like brute force, no guarantee that a solution offers any real insight, seems too tedious for any real substantial shortcuts, if there isnt at least some kind of indication by the problem writer themselves that this problem is worth working on?

the "be the first" feels like they're just trying to manipulate and bait me. i would have respected the problem statement more had they just flat presented the problem with zero fluff

[near hill]

MathIsAlwaysRight found a pretty nice solution (though now deleted).

[kindred wraith]

even "i thought this could be interesting, curious what others think" is more inviting than "be the first to solve this open problem"

[echo jacinth]

The catch is, you need to provide a solution, not just the answer.
Brute force search using some program is not called a solution.

[kindred wraith]

it technically is a solution, but i get what you mean, but it doesn't make it any better

in fact it makes it worse

[near hill]

I'd say that definitely is a solution, just not a particularly enlightening one.

[echo jacinth]

What is definitely a solution?

[near hill]

Brute force search.

[kindred wraith]

because worst case scenario i just convert my program into handwritten out complete cases and that is now qualifying as a solution

except it still sucks

[echo jacinth]

Well it is a matter of definition and to some extent is subjective. But when you say that solve the following equation, would you accept a brute force search as a solution?

[near hill]

I don't think there's anyone here who are a natural audience for that criticism, though.
The reason why they write "be the first" may be that (other than embedding automation like Discord's) they're not even letting anyone see the problem without passing through a "create an account" spamwall.

[echo jacinth]

Or could be, there is no solution for this at present. Your solution will be the first one.

[near hill]

Well, only if you interpret that as "no solution that anyone has jumped through the hoops to submit to that particular website".

[kindred wraith]

if the brute force search provably catches all the possible cases, yes

[echo jacinth]

Don't worry too much about registering. Do you have a proper mathematical solution? If so, post it hear. I will enter the solution under my login.

[kindred wraith]

usually it does not because there are infinitely many options, but if somehow that infinitude can be eliminated as possibilities and the remaining finite possibilities are brute forced, yes that is a solution

ultimately i just struggle to care about this math problem and the person who wrote that page made it less appealing, not more

[near hill]

Again, MathIsAlwaysRight already posted a nice solution -- and then deleted it again, but left a perfectly good hint for it still there.

[kindred wraith]

thats all im saying, but ive said enough, especially being negative so shrug

[echo jacinth]

I don't think brute force searches count as mathematical solutions. But anyways, that acceptance is subjective.
And providing a solution by brute forcing ~1000C4 combinations is at the bare minimum stupid, if not impossible.

[near hill]

Even MIAR's strategy wasn't one that immediately occurred to me, in retrospect I think I'll say it doesn't take any particular genius to get that idea -- and it was only a few lines -- so it is not realistic to assume it is an open problem.

[echo jacinth]

Fine give the full solution here, I will post it wherever is required.

[near hill]

I'm going to respect MIAR's decision to leave it as a hint.

[spark scaffold]

Hi I got stuck here

If I see like 1/(x-1) * 1/(x-1) will u group it

[echo jacinth]

I am not sure what you mean by "not realistic to assume it is an open problem".
Anyone can post a problem without a solution as an open problem. You can also post a regular problem with a solution. So, not sure what you meant by that.

Again, give the solution here itself, if you already have it.

[near hill]

Claiming it is an open problem doesn't make it so, especially one that (as witnessed by MIAR's quick solution) is as easy as this one.

[echo jacinth]

What is the solution? And what is the answer?

[spark scaffold]

Asking me?

[near hill]

Again, again, again: MathIsAlwaysRight posted a perfectly good hint for a pretty simple solution, and I'm going to respect their decision to delete the fully digested writeup afterwards.

[echo jacinth]

Ok... don't post it here. DM me the solution. Let's see what answer you have.

[near hill]

For about the fourth time, I'm going to respect MathIsAlwaysRight's decision to leave the full solution unposted.

[echo jacinth]

Or why don't you simply say that you don't have either the solution or the answer.

Leave your bs for some other day.

[near hill]

Because that's not true.
If you had spent the same amout of time actually looking at their hint, as you've spent claiming that the problem is open, you would probably have gotten it too now.

[blazing comet]

is anyone doing purple comet

[twin raft]

i can guide you through the sol, if you want

[twin raft]

Well, its probably not a hw so ill just send the full sol. I'll spoiler it in parts so that you can use it as a series of hints. If you just wanna see it, ig thats fine with me, but you will probably learn nothing.

||Let L = lcm(a,b,c,d). Now notice that we can write a = L/x, b = L/y, c = L/z, d = L/w for some integers x,y,z,w||
||The condition simplifies to L/x + L/y + L/z + L/w = 1000, which can be rewritten as L(1/x + 1/y + 1/z + 1/w) = 1000||
||Clearly, if we want to minimize L, we need to maximize the sum 1/x + 1/y + 1/z + 1/w||
||The maximum is achieved at 1/1 + 1/2 + 1/3 + 1/4, with (x,y,z,w) = (1,2,3,4) (+ permutations)||
||1/1 + 1/2 + 1/3 + 1/4 = 25/12. So L = 1000 / (25/12) = 480, we can also compute (a,b,c,d) = (480, 240, 160, 120) ||

[twin raft]

And this is not very nice btw

[echo jacinth]

Beautiful. Thanks bro. With your permission can I post the solution?

[echo jacinth]

Sorry about that. But I think posting the solution using spoiler was a good idea.

[twin raft]

Np, you can

[echo jacinth]

Frankly speaking the L/n part was there with me also, but your minimization and maximization steps are very elegant. Far from being brute force.

[twin raft]

As helpers, we arent supposed to give full solutions. Full solutions give the OP essentially nothing in most cases. If OP is trying to learn, then full solution doesnt help. If OP is trying to do HW / exam, then full solution is academic dishonesty.

So thats why I was reluctant to give you the full sol - I dont think it can help you

[echo jacinth]

I think many people will post problems here after giving them sufficient due diligence.

Giving the sol under the spoiler is always a good idea, those who want will see it, those who don't will not. Their choice.

[twin raft]

anyway, my point was more about rudeness. Troposphere just politely told you that I have the solution, you could've pinged me and asked about it, that'd be much better than being unnecessary rude

[twin raft]

usually the approach we take is giving hints and guiding the helpee through the problem

after the first hint, the helpee usually has some new ideas - and they should be given the time to explore them

[echo jacinth]

I thought he said he also has the soln, that's why I asked him to dm it if he doesn't want to post it here.

[twin raft]

he did see it, yeah. But he told you that they'll not share it with you, because I deleted it. At that point, you shouldve just asked me

[echo jacinth]

Let's leave it. It's a good soln. Hopefully I will have some time this weekend, and will post it.

[tiny eagle]

Wow that pretty much insta solves

[novel karma]

I am when is it

My team literally said we doing it and gave us no info

[sturdy knoll]

Yeah

[sturdy knoll]

I take it Thursday

[neon summit]

LHS=5x(9x^2-16)/(3x^2-4x+3x-4))((4x-4)/(x(3x-4)))
I'm pretty sure you can factor it out from there, and then just multiply the reciprocal of RHS to the LHS and then figure it out from there and make sure there are no null solutions or whatever that was

[neon summit]

first off, it's not (x-4)(x+4)
it's 9x^2-16 which is (3x-4)(3x+4) after removing 5x,and 3x-4+x=4x-4 which is 4*(x-1)

[neon summit]

(2a+1)(162a-157)=0
you can note that if this is factorable(which it surprisingly is) you can assume that the values for the coefficients of the factors would have to be rather small since the middle of the equation is small
do also note that after dividing 324 by 2, you get 162 which is exactly 5 away from 157(which is also 5 away from 152) which is how ur gonna form the thing I reached above
IF factorable, from there you can get values of the coefficients and constants in the factors and solve

[upper jackal]

Tysm

[neon summit]

that's the non-quadratic way because honestly who memorizes square roots above 1089

[severe eagle]

Intpoly is so skibidi

usatstst 2020 p7 is nice

[blazing comet]

well i did mine just now

@ruby quest @quick stag how was purple comet!

[atomic tulip]

Can you all give me some resources to use

[keen oriole]

me too

[vague basin]

hint pls topic is invariants

[ivory ember]

Did I not answer this earlier? Consider the total perimeter (aka total boundary length) of the black region. Specifically, how does it change as squares change color?

[vague basin]

yea i solved it sry i wasn't getting any help for a bit and i also needed a furthur hint so i decided to post it here

[novel karma]

Was it hard

[blazing comet]

well ye

[novel karma]

Ok

[inland totem]

hmm

[radiant shoal]

what is mu alpha theta

and what are the questions like

[inland totem]

i only know that competition from the aops books

[zealous reef]

i think mu alpha theta is like an old amc 12

[novel karma]

It’s a club

[radiant shoal]

interesting

my home high school has it

[radiant shoal]

how hard is that

[neon summit]

mm minecraft

[thorny tusk]

o this mc50 hw p9 right

[static basin]

hello... this my alt dont tell anyone

[zealous reef]

im not sure how difficult it is but some problems from the aops books are from there

[radiant shoal]

oh yeah i looked

amc 10 looked easy so amc 12 prolly wont be too hard

[spark scaffold]

Hey when I cross multiply 2

I get 4 x (1/3) x (2/3) x (2/5) x (3/5) x (3/7) x …

Is that true

[vernal axle]

why? Looks like you've lost a number 2 factors in the numerator

[untold urchin]

hi

people

can someone help me yo

[severe eagle]

send the question

😭 😭

[untold urchin]

A line of equation 3x+5 has a y-intercept which is a vertices of an equilateral triangle, and the line 3x+5 also counts as a line of the triangle. find all the equation of the lines that make up these 2 equilateral triangles, as well as the coordinate of the vertices

[severe eagle]

gl

[untold urchin]

???

are you not gonna trie

try

[severe eagle]

nah i got no time

i just finished polynonial lesson

[untold urchin]

is it easy tho

[severe eagle]

is what easy

[untold urchin]

the question

its like aime level style

[severe eagle]

idk?

[untold urchin]

im not there yet

[severe eagle]

i havent tried

[untold urchin]

okok

[severe eagle]

i carelessed 6 questiona for aime lmao

should have gotten 10

[untold urchin]

wowie

aime contestant

ill get there next year

for sure

[severe eagle]

anyway i have 6 more years

[untold urchin]

and ill get into usamo

[severe eagle]

gl

[untold urchin]

you sweat

[severe eagle]

im not US so i cant do

[untold urchin]

how much hours do u study a day

????

[severe eagle]

too little

[untold urchin]

whatttt

like how much

[untold urchin]

are you 13

[severe eagle]

how old are you (if you dont mind)

[untold urchin]

?

14

[severe eagle]

[untold urchin]

okay

im 14 but the time im doing amc 10 again ill be 15

[severe eagle]

ah

i didnt take amc 10

[untold urchin]

estimation

[untold urchin]

?

how did u do aime then

[severe eagle]

amc 12

[untold urchin]

SWEATTT

you study atleast 2 hours a day

i study 15

[severe eagle]

duh

[untold urchin]

15 minutes

mb

[severe eagle]

i got 111 only tho

[untold urchin]

i dont study

[severe eagle]

friend got 120 or smth

[untold urchin]

bigger sweat

[severe eagle]

(for amc 12)

hes orz

[severe eagle]

best in my country in my year

[untold urchin]

thats good

[severe eagle]

wait he got 132 for amc 10

ya

[untold urchin]

im top 30 for amc 10

in my country

BUT

[severe eagle]

what is that

[untold urchin]

thailand

its okay

but but but

[severe eagle]

oh im 500 km south

[untold urchin]

listen to this

[untold urchin]

oh my gosh bro

ik

cus yk chinese

char

[untold urchin]

also

[severe eagle]

?

[untold urchin]

how u so good

teach me

although i dont study

:/

[severe eagle]

then study

[untold urchin]

and my practice test for amc 10 always around 110-130

[untold urchin]

yeah what do i study

[severe eagle]

math

[untold urchin]

i get nervous for my tests yo

[severe eagle]

any geom mains

ts is good question

[untold urchin]

ill see

[severe eagle]

its utterly trivial for a China TST Q3

[untold urchin]

tst?

china tst

[severe eagle]

but in absolute terms its meh difficulty

[untold urchin]

we srs

[severe eagle]

its a good source of problems

[untold urchin]

i know a 16-17 yo

and hes doing better than you

but not much

[severe eagle]

duh

[untold urchin]

hes my math friend and half tutor

[severe eagle]

im not good

[untold urchin]

he tried to smash tst question

at me

i cant solve any

[severe eagle]

theres many people better than me

[untold urchin]

i know

[severe eagle]

im not locking in at all for tstst in 2 months

[untold urchin]

i know a person who was my age

got caught to japan to study for math

in a uni

and was tst contender

[severe eagle]

are japan unis good?

[untold urchin]

he was japan tst contender

[untold urchin]

kinda

[severe eagle]

i dont see them in top 20

[untold urchin]

i balance my life not all about math

[severe eagle]

uni ratings

of course

[untold urchin]

no international students

also

[severe eagle]

if youre only good at math youre cooked

[untold urchin]

tsinghua

[severe eagle]

unless youre really good

[untold urchin]

is top 20 but

should be top 5

[severe eagle]

清华 orz

[untold urchin]

cus 5% is international

[severe eagle]

so?

[untold urchin]

95% all chinese hongkong taiwan

bad international reputation

[severe eagle]

also ratings are subejctive

[untold urchin]

in terms of student

exactly

princeton top 20

but math top 2 or 1

[severe eagle]

i will disregard your argument...?

okay

[untold urchin]

ok go on yo

[severe eagle]

what

im not gonna make any counterpoints im too tired for a debate

[untold urchin]

also

why u so smart

?

[severe eagle]

im asian

[heady summit]

Makes sense

[severe eagle]

[thin stratus]

Whats 1st/2nd if not tsinghua?

[inland totem]

hmmmm

[sturdy knoll]

Purple comet tjis year gard asf

[ivory ember]

lol old aime problem

[spark scaffold]

Yep

Can a be 1

[spark scaffold]

And b as 2

[novel karma]

I love logarithms

Does anyone have a list of logarithm problems

[spark scaffold]

That’s also 2^7

[spark scaffold]

Me too

[alpine depot]

Don't you guys have logarithm problems and exercises on your school books?

[novel karma]

Yeah but I like creative problems like the one Jake showed

The school books are just introductory stuff

[neon summit]

I'm confused what the question means by 2 equilateral triangles

[spark scaffold]

I’ll find it

[neon summit]

wait I might have one lowk

[spark scaffold]

Here

[neon summit]

these are all pretty much 15-30-60s problems so time urself if u want

I randomly have 196 oral logarithm problems idk why I have that much

¯_(ツ)_/¯

[neon summit]

aside from 2 from here, I hate the rest for being tedious asf

[zinc ridge]

what abt b=250, 500

[neon summit]

oh wait

nm

nvm

I'm stupid

[tiny eagle]

It says all possible values right? So like b = 250, 500, and 1000 would also work

Oh someone already said that whoops

||125+3+250+2+500+1+1000+0=1,881||

[novel karma]

Totally forgot about all possible values thanks!

[novel karma]

Ty can we dm?

[neon summit]

no

[zinc ridge]

wait nooo

a>0

so 1000 1 doesnt work

[tiny eagle]

Oh u right

Well I put b as 0 but yea I get what u mean

Cuz 2^b is still 1

[zinc ridge]

o

[tiny eagle]

And it’s fried

[rare quarry]

Guys how do I prep for olympiad math?

[near hill]

The typical advice is to train with old problem sets from the contest you're aiming for. Use them to identify topics you need to learn and seek out material about those.

[pallid dragon]

the odd advice is to not prep so you can feel superior to people who get ahead cuz they prepped

[neon summit]

u don't the knowledge just spawns in ur head

trust

[rare quarry]

I am trying qualify for the IMO, if possible somehow, in the future and am attempting a qualifier later this year. I am not very familiar with the olympiad specific math or much formal math, but I've delved alot exploring a few math fields from first principles, and attempting alot of problems unrelated from olympiads

[rare quarry]

I hope that is true

[neon summit]

what country r u from

[rare quarry]

India

?

[neon summit]

I dunno what to say ur basically cooked if ur starting late
I'm not familiar with the rules of team selection at ur place so /shrug on that but uh
I take it you have roughly 2 or 3 chances left to get into the team for that, if you aren't too well versed uhhhhhhhhhh ur basically cooked but um try to be a generalist for everything, read a lot of olympiad books, they're boring asf but they help a lot, especially read MONT and EGMO and do AoPS whatever helps

js read a lot of book s

books

and practice

practice and building up the 'sense' is important

[rare quarry]

Thanks

[neon summit]

what else

[rare quarry]

Is 15 late?

[neon summit]

as people I know have stated
"these people eat and breath math, if you haven't been doing math since you were a baby and every moment you are asleep, you have no chance"
most of everyone, especially on countries that tend to fair better in IMO would probably have tougher competition and this is for my country which does horribly at IMO and I've done my fair share of olympiads and this is what my teacher stated I had to do in order to do good against them super smart people
if u haven't started ur at a lower starting point than I was when I tried for my country and u might be cooked
but hey, starting too late is a horrible mentality, you can probably learn a lot in a year

if I'm remembering this, selections start now for most places

or releases

[neon summit]

focus forward, essentially

[rare quarry]

I'm prepping for next yrs, whose first qualifier takes place in September in my country

[neon summit]

there is an insane amount of theories
many MANY random ahh theories that appear out of nowhere especially in number theory, I guess geometry sometimes, inequalities are fairly easy as long as you can do problems and apply them altho at a higher level I doubt they'd actually add inequalities because they're "too easy" what else
there's like an insane lot of stuff u just randomly come across and u also have the thing the funny function thing, idk what the name is but it's basically having to find functions given a condition, it's pretty hard to "assume" particularly with nested functions but eh those r in prospect prolly easier than most other questions u might see

[rare quarry]

That is demoralizing, but I don't intend to actually get on the team and win, I just wanna have my best go at it when I have the chance

[neon summit]

start preparing

[neon summit]

gl with that

[neon summit]

combinatorics I dislike, it bad

[rare quarry]

Ok, but what are some good resources for them, and am I supposed to just read em and do a bunch of problems, or is there anything more to it?

[rare quarry]

I like it tho, geometry is what I dislike

[neon summit]

EGMO, MONT, AoPS, Books u also discover along with those, watch videos, check the server literally in the channel desc. rn that'd be more specific. Look through previous questions in your country's math olympiad

and practice those

[neon summit]

geometry is so made up 😔

[rare quarry]

Thanks for everything man

[neon summit]

ye

how many triangles r in this

[rare quarry]

Asking me?

[neon summit]

any1 in here

[pallid dragon]

2 very large triangles, 6 triangles and 6 small triangles

[neon summit]

NO USING COMBINATORICS U HAVE TO MANUALLY COUNT

[neon summit]

there's actually 3
hexagons r made up of 3 triangles trust

and triangles are also made up of 3 triangles

hence nonagons are made of 3 hexagons trust

[echo ocean]

19 points so 19 choose 3, there are collinear points and also sets of 3 points that's not fully connected

So maybe subtract that

13 points inside the hexagon doesn't seem to form any triangle so subtract 13 choose 3

okay that's a lot of invalid triangles, maybe there is a better way

[heady summit]

Combinatorics is kinda my favorite topic-

Any channel recommendation for number theory?

[neon summit]

we should eradicate combinatorics and probability

[heady summit]

I'll admit I haven't studied probability that well
But combi is way more fun to do than algebra (my least favorite)

[severe eagle]

?

Combi is uncivilised

[heady summit]

I mean
It is possible it gets harder or more boring
Because right now I'm only studying stars and bars
I have a looooong way to go

[olive haven]

The hate on combinatorics is unjustified 😞 I like combinatorics, it’s quite fun and has interesting situations and problems to study

[azure wadi]

Is olympiad math ded now

[viral sierra]

Yeah, combinatorics is fun. Cant say the same about geo

[viral sierra]

True

[olive haven]

W h a t. I love geometry (my #1 in math). What do you not like about it? Geometry is like an intuition-based puzzle

[viral sierra]

As you said, it is an intuition based puzzle. My intuition doesn’t work

But specifically, the 3d and 4d cordbash 😭

[olive haven]

I’d say math (at a higher level) and physics are all about intuition

I love math when it’s like a puzzle that requires you to think outside the box. I hate when it’s just a ‘plug-in a formula’ or ‘speedrun it’ thing

It’s beautiful how it can be so abstract + intuitive but at the same time rigorous and formal

[inland totem]

wait jmo scores are out?

[olive haven]

Pretty sure Zelix is not here anymore

[inland totem]

imma ask him in gc

[inland totem]

4d coordbash?

[viral sierra]

I remember there was this one problem that i had to do. It was so bad 😭

[inland totem]

i sometimes use triple integrals for probability problems ig

idk if that can be considered "4d", since it's basically just calculating volume

[viral sierra]

I js remember there was this one 3d masspoint thingy which you had to convert to 4d coordinates and then bash

[inland totem]

hmmm

[thorny tusk]

wth

[inland totem]

<@&268886789983436800>

[thorny tusk]

wha happened

[inland totem]

mrbeast's top secret crypto

no one shall know

[thorny tusk]

...

[radiant shoal]

what if i break in

[zinc ridge]

was it one of those messages

[inland totem]

ye

[neon summit]

combinatorics is a weird midground of similar standing to inequalities
if ur new and u don't know, it feels hard to do
if you know abt it's like just more of a mild annoyance

[neon summit]

geometry sucks when they provide no diagrams cuz I will NOT be drawing accurate triangles and the ' things on the lines and angles

they almost always never provide diagrams 🥀

[viral sierra]

At first, combi is like prison, but then its just intuitive and mildly bashy

[viral sierra]

real.

[neon summit]

combinatorics is 3 formulas trying to desperately claw their way into competitive relevance

inequalities r at least a bit more diverse and goofy asf combinatorics is just having experienced the problem before

[viral sierra]

inequalities are fun, but abstract...

is torture

[neon summit]

erm

those r the goofy ones '<'

[viral sierra]

no abstract alg. i hate it to its core

[neon summit]

no idea what that is

seems like some math I'm never touching

goofy

it's relatively easy tbh

mm was it titu's lemma

[viral sierra]

isnt this just qm-am

[neon summit]

it's pretty easy ye

[inland totem]

optimization

[neon summit]

I

I'm not knowledgeable enough on

uh

whatever that is but ain't that a calculus thing get that outta here

[inland totem]

hmmmmmmm

u could use calculus ig

[neon summit]

we hate calculus here(I used to solve for minimum values with calculus)

[viral sierra]

i was lowk gonna bash with derivatives 😭

[neon summit]

reminds me of that one question

[neon summit]

can't find it

[inland totem]

what was it abt

[inland totem]

calc bash

just this once

[neon summit]

https://klipy.com/gifs/ryu-ishigori

it doesn't even work w calculus cuz calculus sucks

I think

I might've tried it but that is a 60s question

[inland totem]

ook

then

[neon summit]

and 0 obviously isn't an answer

[inland totem]

ye i'm not finding roots of a quintic

[viral sierra]

wait can this smh be done with roots og unity

*of

[neon summit]

erm

no?

I'm pretty sure that is more than 60 seconds and this is a 60s question

[inland totem]

wait huh

how

[neon summit]

dead ahh hated this
we can disregard the 1 to get a(a^3-1)(a^2+1)=a(a^2+1)(a-1)(a^2+a+1) and now how do we minimize
easy, a^2+1 is always positive, so is a^2+a+1 since=(a+1/2)^2+3/4 so we basically only have to consider the value that minimizes a(a-1) which is 1/2 which turns it into -1/4 so the thing is negative
trust guys my logic is flawless frfr

trust trust frfr

[viral sierra]

i feel like the only way to do so is to js calc bash

im pretty sure like 4 of the roots of the quintic are complex or smth

[neon summit]

,w (0.7098)^6+(0.7098)^4-(0.7098)^3-0.7098+1

[gilded halo]

Results provided by WolframAlpha

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

Requested by nautilus2143

[neon summit]

,w 0.5^6+0.5^4-0.5^3-0.5+1

[gilded halo]

Results provided by WolframAlpha

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

Requested by nautilus2143

[neon summit]

bruh 🥀

,w x(x^3-1)(x^2+1)+1

[gilded halo]

Results provided by WolframAlpha

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

Requested by nautilus2143

[neon summit]

so the minimum is 0.31431

strange that isn't the solution on the paper

asnswer

answer

[inland totem]

hmmm

[neon summit]

boo algebra or something

(yes, solve the problems I haven't done-) wait a minute this isn't that hard

oh it is

yeah it is hard

[inland totem]

nooooo

i actually hate problems like these

[viral sierra]

Noooooooooooo

Whyyyyyyyyyy

I rlly hate such problems

Also isn’t this similar to some of the problems in the awesomemath algebra textbooks

[tiny eagle]

I would love a solution to ts problem im so bad at these

[thorn heron]

I found one

The zero function f(x) = 0

The identity function f(x) = x works too

[radiant shoal]

how do you even solve these

[near hill]

Generally by repeatedly plugging "special" pairs of x and y into the equation and simplifying, hoping that you'll get something that can improve your knowledge of the function.

[radiant shoal]

does not sound fun

[near hill]

For example, if we plug in x=y=0, the equation simplifies to f(f(0)) = 0 -- which isn't much, but it's something.

[loud narwhal]

i was told that hints about competition maths question will be entertained here ?
or concepts related to it ?

are there any constraints on this channel ? like what i can not ask , related to mathematics

*i am new

[near hill]

Then plug in (x,y) = (0,1) and we get f(f(1)) = f(1).

[loud narwhal]

for starters can anyone drop me hint on this

[ivory ember]

Answered here: #prealg-and-algebra message

[near hill]

Plugging in (0,f(0)) gives (unless I've made a mistake) 2f(0) = f(0)², which means f(0) = 0 or f(0) = 2.

[ivory ember]

welp I'm bored enough to try an FE for the first time in forever

[near hill]

Supposing f(0)=2, and thus f(2)=0, then plugging in (1,1) and then (1,-1) gives f(1)=1 and f(-1)=3.
Hmm, those values lie on a nice line, I wonder if ...
Ha! f(x)=2-x is a third solution!

On the other hand, if f(0)=0 then plugging in (0,a) gives f(f(a))=f(a), which will restrict at least continuous solutions considerably.

[ivory ember]

Welp here's my solution/scratch. The writing could be a lot better, but I wrote it as I was fucking around so leave me alone.

,texsp ||I claim the only solutions are $f(x)=0$, $f(x)=x$, and $f(x)=2-x$. \

Setting $y=0$ yields
$$f(0)+f\left(f(x) \right)=xf(0)+f(x) \implies f\left(f(x) \right)=xf(0)+f(x)-f(0).$$

Setting $x=y=0$ yields $f\left(f(0)\right)=0$. \

Setting $x=y$ yields
\begin{align*}
f\left(2x^2\right)+f\left(f(2x)\right) &= 2xf(x)+f(2x) \
f\left(2x^2\right)+2xf(0)+f(2x)-f(0) &= 2xf(x)+f(2x) \
f\left(2x^2\right) &= 2xf(x)-2xf(0)+f(0)
\end{align*}

Replacing $x$ with $-x$,
$$f\left(2x^2\right)=-2xf(-x)+2xf(0)+f(0).$$

Thus, for $x \neq 0$, we have
$$2xf(x)-2xf(0)+f(0)=-2xf(-x)+2xf(0)+f(0) \implies f(x)+f(-x)=2f(0).$$
Of course, this also trivially holds for $x=0$. This means that $g(x) \coloneqq f(x)-f(0)$ is an odd function. \

Rewriting $f\left(f(x)\right)=xf(0)+f(x)-f(0)$ yields
$$f\left(f(x)\right)=xf(0)+g(x) \implies f\left(f(x+y)\right)=(x+y)f(0)+g(x+y)$$

We can also rewrite the original equation as
\begin{align*}
g(2xy)+f(0)+f\left(f(x+y)\right) &= xg(y)+yg(x)+(x+y)f(0)+g(x+y)+f(0) \
g(2xy)+f\left(f(x+y)\right) &= xg(y)+yg(x)+(x+y)f(0)+g(x+y) \
g(2xy)+(x+y)f(0)+g(x+y) &= xg(y)+yg(x)+(x+y)f(0)+g(x+y) \
g(2xy) &= xg(y)+yg(x) \
\frac{g(2xy)}{xy} &= \frac{g(y)}{y}+\frac{g(x)}{x}
\end{align*}

This motivates us to define $h(x) \coloneqq \frac{g(x)}{x}$ for $x$, $y \neq 0$. Since $g$ is odd, $h$ is even. We clearly have that
$$2h(2xy)=h(x)+h(y).$$

Setting $y=1$, we obtain $2h(2x)=h(x)+h(1)$. Sending $x \mapsto xy$, we have $2h(2xy)=h(xy)+h(1)$. So,
$$h(x)+h(y)=h(xy)+h(1) \implies \left(h(x)-h(1) \right)+\left(h(y)-h(1) \right)=h(xy)-h(1).$$

Define $u(x) \coloneqq h(x)-h(1)$. Clearly, $u(1)=0$, $u(xy)=u(x)+u(y)$, and $2u(2x)=u(x)$ for all $x,y \neq 0$. Thus,
$$2u(2)+2u(x)=u(x) \implies u(x)=-2u(2).$$

We see that $u$ is constant on $\mathbb{R} \setminus {0}$. Since $u(1)=0$, $u \equiv 0$. So, $g(x)=xh(1)$ for all $x \neq 0$. And since $g(0)=0$, we have $g(x)=xh(1)$ for all $x \in \mathbb{R}$. Thus, $f$ is affine and the rest is straightforward.||

[gilded halo]

Civil Service Pigeon

[near hill]

Neat. If your conclusion in the last line holds, then it means there are exactly the three solutions Kookie and I found.

[heady summit]

I hate inequalities they're not fun they're a prison

[copper hare]

Can someone help with this? I'm confused about the question and what it means, I got the answer but I wanna check if I interpreted the question wrong

[neon summit]

ultra fun 🔥

[neon summit]

vieta's theorem and discriminant frfr
also that's supposed to be a^4 and B^4 not a^4+B^4 looking at the original paper

original paper for this

[neon summit]

I'm getting weird deja vu

[copper hare]

Oh thanks, I took literally and assume second eqn was perfect square

[neon summit]

suppose x=0 then
you get f(0)+f(f(0))=f(0) => f(f(0))=0 which can obviously give us f(x)=0
suppose y=1/2
f(x)+f(f(x+1/2))=xf(1/2)+f(x)/2+f(x+1/2),,, wait! this looks familiar
going back to the original condition, substitute x=x+1/2 and y=0
=> f(0)+f(f(x+1/2))=xf(0)+f(x+1/2), you can subtract this from the original giving
f(x)-f(0)=x(f(1/2)-f(0))+f(x)/2 => f(x)=2x(f(1/2)-f(0))+f(0)
since f(1/2) and f(0) would have constant values, then this looks like a linear equation hence f(x)=mx+b and then solve for the other kinds of the thing

[neon summit]

as a general way to do these types of things, peops do x=0, 1, y and y=0, 1, x and so on, there's also probably many other ways to do this for example with x=2x and y=0, and x=x w/ y=2

peops just substitute and see, u try to generally find a 'trend' or something that tells u that the function is either usually linear quadratic or rarely sometimes cubic or if it even is a polynomial function

[neon summit]

honestly thought p and a^4+B^4 were the roots lol

[neon summit]

solve it. NOW!!!

[neon summit]

dawg really did x=y

[copper hare]

Answer changes, if I take root as a^4+b^4 as "roots" of the eqn that means only 1 root so d =0 and after doing everything at the end answer comes to be imaginary roots for the third eqn but if take a^4 and b^4 as seperate roots answer change to equal and non zero

[neon summit]

whatever u say(too much yap[I'm asleep])

[neon summit]

prove it only has linear solutions NOW

cuz just guessing ain't enough

[neon summit]

note that for f(x)=0, it's rather distinct? I'd say it's not linear so just a thing there or smth just something to like rather be determined outside of the thing determined in that 2nd part where y=1/2

[neon summit]

u can find functions via trial and error but u kinda still do have to prove that they follow this typical trend(f(x) is linear)

[neon summit]

mm another function but this time actually better than before

if I'm remembering this, I gave up on trying to do this

wait a minute

squares

also a pretty question

[neon summit]

there is a quite elegant solution to this I quite like this one

[ivory ember]

,texsp ||Fix integers $x_2<x_1$ and $y_2<y_1$. Summing the identity
$$f(x+1,y+1)-f(x+1,y)-f(x,y+1)+f(x,y)=1$$
over all $x=x_2, \dots, x_1-1$ and $y=y_2, \dots, y_1-1$ yields
$$f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1)+f(x_2,y_2)=(x_1-x_2)(y_1-y_2).$$
If we had $\left \lvert f(x,y) \right \rvert \leq 2024$ for all $x$, $y \in \mathbb{Z}$, then $\left \lvert (x_1-x_2)(y_1-y_2) \right \rvert \leq 8096$. But choosing $x_1=8097$, $x_2=0$, $y_1=1$, $y_2=0$ gives a contradiction, and so the bound does not hold for all $x$, $y \in \mathbb{Z}$.||

[gilded halo]

Civil Service Pigeon

[neon summit]

ooh you can maximize the value of f(x1,y1)-f(x1,y2)-f(x2,y1)+f(x2,y2) which isn't restricted to (x1-x2)(y1-y2)

I knew it was wrong but I was unsure how to prove it

thanks

[ivory ember]

[stuck tartan]

wait isn't the question asking whether the situation can happen, and not if there exists a scenario where the requirement ISN"T held?

[ivory ember]

Note this is a proof by contradiction. The question asks if $f(x,y)$ can be bounded everywhere while also satisfying the given functional equation everywhere. The proof assumes the situation can happen, then shows that this assumption forces $8097 \leq 8096$ for specific choices of $x_1$, $x_2$, $y_1$, $y_2$. This is false, so the assumption must be false.

[gilded halo]

Civil Service Pigeon

[neon summit]

I have an ugly question

the answer would be the no, of subsets of {1, 2, ..., n} that only contain 1 as a square no?

[stuck tartan]

oh mb i didn't see the last part of the original problem

[ivory ember]

[ivory ember]

What if $n=3$? Then you propose ${1,2}$ works. But $f$ just cycles this between ${1,2}$ and ${1,3}$.

[gilded halo]

Civil Service Pigeon

[pallid dragon]

you would have to start with all primes, or no primes

otherwise it ocillates

[neon summit]

I'm confused, don't primes have 2 factors

[pallid dragon]

they don't all count

they have to be in the set

{1,2} becomes {1,3} because 3 only sees 1 divisor

[neon summit]

at most n? so at most 2 no?

I'm still confused

[pallid dragon]

n doesn't change it's 3

[neon summit]

ok that makes sense but how would it get to 3 from {1, 2}

[pallid dragon]

because there's 1 divisor of 3 in {1,2}

so 3 appears

[neon summit]

am I misreading the question

[neon summit]

oh

at most n that have an odd number of factors in S not at most n that

[pallid dragon]

if you have some primes, but not all, you never get to {1}

that's all i got, i can't really solve it

[neon summit]

if we remove that "in S" part and say in general it would be the no. of subsets of {1, 2, ..., n} that only contain 1 as a square ye?

wow yeah that is hard asf

[tiny eagle]

Ty

[neon summit]

what's a penguin

[tiny eagle]

Bird, but not bird, fish but not fish

[neon summit]

what's a bird

[tiny eagle]

Good question, government agents perhaps?

[neon summit]

what's a perhaps

[tiny eagle]

Yeah

Thank you

[near hill]

Wow, demanding much?

[neon summit]

as if I didn't like alr show the way to find out it only has linear oslutions like 5 messages before lmao

[near hill]

And here too. What charming manners.

[sacred silo]

does anyone have the intro to algebra by jermey king pdf possibly

[opaque heart]

i need that too

[inland totem]

hmmm

[fallen apex]

no one has pirated material in this server

however,

google might

[radiant shoal]

what

[near hill]

The server rules forbid using the server to exchange copyright-infringing material, and the moderators actively enforce this.

[radiant shoal]

ik but

nvm

[near hill]

So the situation is observationally equivalent to nobody having it.

[tired raptor]

yoo guys i am having a doubt

what cyclic intergrals

you know the with circle in the middle

[near hill]

Contour integrals?

[tired raptor]

yes

but in my thermodynamic book it's written as cyclic intergrals

and could tell me how to calculate it

[fallen apex]

well depends

on context

[little wraith]

how would this be done? it’s the only one i can’t figure out i tried just calculating the sum of the sequence and multiplying by 201 and then adding one but it wasn’t even close to an answer i think im missing something

is it that when the sequence repeats it’s then 1, 4, -9, -16 because of the signs

[ivory ember]

Take the terms four at a time

[tiny eagle]

All solutions, all real solutions, or what

[spark scaffold]

What’s the value of x

it can be anything whether it’s integer, complex

Or not

[near hill]

Allow me to doubt the correctness of that heading ...

(brought to you from the people behind only people with a [redacted] of at least 180 can solve this!)

[wet hare]

,w (x/2)^6 = 3^6

[gilded halo]

Results provided by WolframAlpha

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Requested by altanis_

[wet hare]

🤯

[spark scaffold]

My idea was

Sub x/2 = u

[ashen hatch]

the simplest solution would be undoubtebly to say x/2=plus or minus 3

[spark scaffold]

Then u get x=+-6

[ashen hatch]

i mean like u sub is usually reserved for repetition or more complex equatiosn

[spark scaffold]

Y sub

[ivory ember]

fwiw this is just $x^6=6^6$, so take the sixth roots of unity and scale them all by a factor of $6$

[gilded halo]

Civil Service Pigeon

[ashen hatch]

what are roots of unity? i haven't heard of those before

[stuck tartan]

the complecks roots of x^k = 1 are the kth roots of unity

[spark scaffold]

I wondered too

[neon summit]

funnily enough you can just tell from the units digit since it's multiple choice and they're all different

[ashen hatch]

oh

i... kind of get it

[ivory ember]

insert circle

[neon summit]

cistheta magic

[inland totem]

hmm

[neon summit]

(x/2-3)(x/2+3)((x/2)^4+9(x/2)^2+81)
and then u=(x/2)^2 then quadratic cuz I'm too lazy
be lazy kids and NEVER learn cis

[gilded halo]

Slomenist

Slomenist

[neon summit]

yeah cis theta magic x^3-1 has 3 solutions in R, C or whatever

[near hill]

Many more solutions in H, by the way.

[blazing pilot]

no solutions in many other fields
infinitely many in other fields

[near hill]

It can't have more than 3 solutions in a field.

[neon summit]

✌️ I ain't doin that for a competition

[neon summit]

does harvard even have an entrance exam

[inland totem]

uhh i don't think so

[blazing pilot]

it can
in non commutative fields

[near hill]

I mentioned H myself, but those are not usually called "fields" in English.

[blazing pilot]

some mathematicians consider them fields
like bourbaki

[neon summit]

calm down this is comp math prolly hs comp math not field theory(if I'm using the term right)

i've no idea abt other genres of math

[blazing pilot]

there are a lot of field theory problems in comp math
you probably didn't check the romanian high school comp problems

[inner hazel]

arent these harvard entrance exams usually like clickbait

[spark scaffold]

What should I sub

k=x+3?

[ivory ember]

yeah sure that works

[spark scaffold]

Then it’s k(k-1)(k+1)=120
k(k^2-1)=120

k^3-k=120
k^3-k-120=0

[stuck tartan]

?

i think u multiplied it out wrong

[spark scaffold]

I fixed it

[spark scaffold]

After did synthetic it’s
(k-5)(k^2+5k-24)=0

I take the real solution k=5

x+3=5 x=2

[stuck tartan]

i mean there are other solutions u don't only hjave to tak the real one

[spark scaffold]

Check

[stuck tartan]

uh

how did u get -11/2 as the real part of the solution'

ohj wait nvm

yheah that looks right

[little wraith]

is it c

[ivory ember]

,w sum of (-1)^\frac{n(n+1)}{2} (n+1)^2 from n=0 to 2010

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[little wraith]

oh lovely 😄

[spark scaffold]

Find x

[smoky urchin]

x = 8

[spark scaffold]

How would u use roots of unity to solve this x^4 power

[ivory ember]

I'd use the equation directly

$x^5=8^5 \implies \left(\frac{x}{8} \right)^5=1 \implies \frac{x}{8}=e^{\frac{2\pi i k}{5}}$ for $k=0,1,2,3,4$

[gilded halo]

Civil Service Pigeon

[ivory ember]

irl people don't really do the middle and jump to $8e^{\frac{2\pi i k}{5}}$

[gilded halo]

Civil Service Pigeon

[ivory ember]

but the division probably helps to make the roots of unity connection clearer idk I just had a random idea to do that in the moment

[past quest]

yo how do you type that fast

do you do coding

[ivory ember]

I really don't lol

[ivory ember]

I mean I occasionally write code snippets and programs and all

but tbh (or at least for me) I type relatively slow when I code compared to when I write a longer form of text for example

because with coding, you need to think about what you type

and/or be really careful for typos

dropping one semicolon and the whole thing breaks

[past quest]

no cause like the way youre perceiving the eq into simple language that fast, with special symbols too

amazed me

[ivory ember]

actually I do remember I sent this in the past

[ivory ember]

eh if you use latex enough you talk in it too so

[past quest]

ohh i didnt know that

ill look into latex

[ivory ember]

tbf if you get exposed to anything enough to the point where it becomes second nature, it'll influence your speech lol

[past quest]

true

[ivory ember]

but yeah all the triggering of the texit bot is latex

[past quest]

i see isee

cool!

[ivory ember]

stem fields use latex a lot because math and all

[past quest]

wow

thats cool

ill look into that too

thanks

[ivory ember]

[humble schooner]

Is it normal to struggle in number theory? (Elementary)

[neon summit]

I'm not sure I got ||E, since the unit digit just cycles every 20 so 1-1-4+4-9+9-6+6+5-5...=0 so 2011mod20=11 and so consider the units digit of the first 11 which would end up being -2? I think||

[neon summit]

ye

E.

[neon summit]

yes but practice and exposure makes perfect

[neon summit]

not my problem

it's in the math olympiad server

I'm kinda bored and it looks possible frfr

[neon summit]

update: I tried m=f(n), I have no idea what I'm doing

[severe eagle]

wait this question seems familiar

[severe eagle]

its d5 so should be that bad

gl

reminds me of BMO 2002

[neon summit]

😔 I can't do it I'm too bad

[static torrent]

@livid gust

yooo

[mental cove]

ill try this

[inland totem]

hmmm

might give this a shot

[edgy orchid]

wait it look like IMO 2007

[oak tartan]

its from the shortlist

[neon summit]

my best answer is 2007 and 4013, iffy on the 4013 I kinda pulled that outta nowhere

<@&268886789983436800>

[mental cove]

i know the answer but i cant prove it yet using mathematical logic

[neon summit]

since it's f N->N I can only assume it's mostly polynomial(and a few other kinds of functions) for the 'easy' answer and basing from how it looks it has a linear solution(f(x)=x, which is how I got 2007) but there's more than just polynomial functions so /shrug I threw around basic assumptions for 4013 '

that was js about it on my end

[mental cove]

try with fx = c and fx = ax + b

[turbid venture]

the server have a chat in other languages?

[ashen hatch]

!english

[scarlet root]

Hwæt! We Gardena in geardagum

[ashen hatch]

ack maybe the wrong command.. supposed to be english only, if you dont speak english try translation

[pliant wagon]

[
\frac{a^2+1}{a} = 5
]

[
\frac{a^2}{a} + \frac{1}{a} = a + \frac{1}{a} = 5
]

[
\frac{a^4 + a^2 + 1}{a^2}
= \frac{a^4}{a^2} + \frac{a^2}{a^2} + \frac{1}{a^2}
= a^2 + 1 + \frac{1}{a^2}
]

[
a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2
]

[
= 5^2 - 2 = 25 - 2 = 23
]

[
a^2 + 1 + \frac{1}{a^2} = 23 + 1 = 24
]

[gilded halo]

xev

[pliant wagon]

or if you look at it properly

the equation you need to find is basically $\frac{a^2+1)}{a}$ squared minus $\frac{a^2}{a^2}$

[gilded halo]

xev

[pliant wagon]

which makes it 5²-1

[mental cove]

quadratics

[radiant shoal]

how do you even come up with an answer to this

[spark scaffold]

What will happen if I use exponent property like negative exponents

[frank hornet]

no I don't think that's needed, it'll make it more time-consuming than doing it in a simple way

[spark scaffold]

Ok

[neon summit]

((a^4+a^2+1)+a^2-a^2)/a^2=((a^4+2a^2+1)-a^2)/a^2=((a^2+1)^2-a^2)/a^2=(a^2+1)^2/a^2-a^2/a^2=((a^2+1)/a)^2-1=5^2-1

[neon summit]

from the get-go since it's natural numbers, always consider lines, f(x)=x makes sense since it satisfied m+n__>m+n-1 for all m, n epsilon N
I'm realizing now that "wait, constants exist" and (m+n)c>mc+nc^2-1 => nc>__nc^2-1 => c=1 since f must lead to a natural number so f(x)=1 must also be a solution
I'm prolly wrong tho idk

[neon summit]

another POTD that also seems possible frfr

[neon summit]

(a+b+c)=ab/2
a^2+b^2=c^2=> (a+b)^2-2ab=c^2
(a+b)=ab/2-c
Note: either a or b must be even so a+b and c are an integer
ab^2/4-abc+c^2-2ab=c^2 => (ab^2/4)-(2+c)ab=0
ab/4-c=2, ab does not equal 0 because either a or b>0 => ab=8+4c => a+b+c=4+2c => a+b-c=4, this now obviously gives 6 8 10, 5 12 13

idk aside from that tbh

I've seen a similar problem before

[vernal axle]

If a,b are legs of a right triangle, then your equation is a²+b²=(ab/2-a-b)² which is equivalent to (a-4)(b-4)=8. So (a,b)=(6,8) and (5,12) only

[fair bronze]

Try this next:

Find all right triangles with integer side lengths.

[knotty yoke]

Pythagorean triples?

With side lengths u^2 - v^2, 2uv, u^2 + v^2?

We show it via examining it in $Z[i].$
Let $x^2 + y^2 = z^2.$ Since we are looking for primitive triples, we let $\gcd(x, y, z) = 1.$
Now, $(x+iy)(x-iy) = z^2.$
Now, convince yourself that $x+iy$ and $x-iy$ are relatively prime in $Z[i].$ (Consider the fact that $Z[i]$ is an UFD. Consider some common divisor $\delta$ of these factors, and then use the fact that $\delta$ divides their sum and difference. It follows quite straightforwardly.)
Thus, both $x+iy$ and $x-iy$ are perfect squares. Let $x+iy = (u + iv)^2 = u^2 - v^2 + (2uv)i.$
Comparing: $x=u^2 - v^2,$ and $y = 2uv.$
Just put it in the original equation to get: $z = u^2 + v^2.$

[gilded halo]

LemmaLover

[knotty yoke]

The approach is quite similar to showing Fermat's Last Theorem for $n=3$ in $Z[\omega]$

[gilded halo]

LemmaLover

[knotty yoke]

You can then kinda go along the direction of Cyclotomic Fields by adjoining higher roots of unity.

However, not all of them are UFDs.

[knotty yoke]

Here, $u, v \in \mathbb{Z}.$
And so are $x, y, z$

[gilded halo]

LemmaLover

[inland totem]

hmmm

[neon summit]

let a and be the legs of a triangle and c be the hypotenuse of a triangle, and let a be an odd number __>__3 and b>a, then the legs of the right triangle would be b=(a^2-1)/2, c=(a^2+1)/2 so na, n(a^2-1)/2, n(a^2+1)/2 where n epsilon N

note: this doesn't work to give you stuff like (8, 15, 17), (20, 21, 29), (20, 99, 101) ... cuz I cannot be bothered to prove that like let c=b+k, then, (b+k)^2-b^2=a => 2bk+k^2=a => k(2b+k)=a wherein k epsilon N and k> nah I suck too much for that

oh, k is an even number yadda yadda and then a must be divisible by 4 or smth like that

[radiant shoal]

fun

[inland totem]

the potds are always pretty fun

[radiant shoal]

uh

[inland totem]

unless it's like d6+

[radiant shoal]

idk about that one

[neon summit]

reminds me this only works for b>a and a is odd but like when a is even is goofys

[radiant shoal]

especially the ones where you have to find all functions that apply for a certain requirement or w/e

[neon summit]

I suck too much to enjoy them 😔

[inland totem]

noo

too orz

[neon summit]

those are usually the easy ones ngl unless it's 15 bajillion nested functions or > type shi 😭

well no those are possible I guess but like

f:N->N is so restrictive

[fair bronze]

You did something, congrats 👏

do the rest later

[inland totem]

hmm

[daring sparrow]

[novel karma]

Does anyone got good ideas for this officer position for my math club

I would genuinely appreciate it

[neon summit]

how does one solve this in 30 seconds even

[raven oyster]

whats mini mu? If you could tell i would be happy to give suggestions

[spark scaffold]

What should I substitute

[fleet drift]

where is the competition

[lime steppe]

(x-1)(x-6) then (x-2)(x-5) then (x-3)(x-4)

[lime steppe]

idk if this can be the proof but this is how my teacher used to solve it

[spark scaffold]

0 works bc negative*negative = positive

7 works bc it will look like 6!

[lime steppe]

yep

[novel karma]

mini mu is an event hosted by the high school to bring middle schoolers and get them to do math competitions and sort of teach them about math. tbh i couldn't be there tho.

[twilit bane]

Who's gonna help me out??

[neon summit]

well for one I can only assume for x=m that ax^2+bx+c is the same sign as ax^2+bx+c so we can assume that ax^2+bx+c is always positive or 0, so b^2-4ac>0 from the discriminant formula but aside from that idk
m must not be a root
SCRAP THIS^
you can also assume from the choices lmao, that's even easier
we know that ax^2+bx+c has to be positive so a probably concaves up
so if a>0, then m>B and m<a because that would lead to a positive answer either way because it's like that(let ax^2+bx+c=a(x-a)(x-B) and substitute it in)