competition-math

[prisma python]

oh, i missed something then

isn't the f(x) used convex?

oh wait no it's concave, i messed up the derivatives

[vernal axle]

yep, its concave for t<0 and convex for t>0.

[prisma python]

right, x in [0,1] is t in [-inf,0]

[novel karma]

Pls ping when reply

[ornate blade]

Well firstly a + c = 2b

[novel karma]

Yep

I did that after

[ornate blade]

Yeah so the insides of the logs must be equal

[novel karma]

Yeah

Idk how to get c in terms of b

[ornate blade]

Hmm I don't have pen and paper with me

[novel karma]

Oh

Just ping back whenever you can 🙂

[sly pawn]

Just solved this problem

👍

W idea

Probably should've figured it out faster though

[vernal axle]

put a=b-n and c=b+n. Then you should have (b-n)(b+n+6)=(b-1)^2. Don't forget that b and n are integers and b-n>=1 and b+n+6>=1.

[sly pawn]

does that question mean log base 10?

or log any base?

[vernal axle]

it doesn't matter. It can be any base.

[sly pawn]

ok

[sly pawn]

Ok

Well

E is wrong 😂

I found an answer through trial and error

[fickle torrent]

yo i am a informatics student but want to get into comatition math for fun

[sly pawn]

What is informatics?

[fickle torrent]

coding olympiad basicly

it is basicly imo thinking + coding

+no proof

you just need to vibeproof

like think it looks correct

[sly pawn]

Okay

[fickle torrent]

look at codeforces.com if you want to understand more

its basicly a collection of problems with a judge

so you submit code and it gives you a verdict

they also host contests and stuff

[sly pawn]

sounds fun

[near hill]

In some countries that is the local name for computer science in general.

[fickle torrent]

rly?

[sly pawn]

So if you already have the thinking and problem solving ability it should be pretty easy to get into comp math

It's useful to learn some theorems and methods for each subject still though

[fickle torrent]

ye i solved some problems given by my freinds i just dont know how to prove for shit tho

[sly pawn]

me neither 😂

[fickle torrent]

given a hard(ish) combinatorics problem i can get the answer but cant prove it

[sly pawn]

Planning on learning proof writing later because I heard it's pretty easy to prove if you can solve the problem

like format your solution into a proof I mean

[fickle torrent]

nah proof in general

[sly pawn]

Oh I see

[near hill]

Yes, such as Informatik in German or informatique in French.

[sly pawn]

Can you give an example of a problem you would struggle proving?

[fickle torrent]

mostly any problem

i solved the jbmo 2025 p4 but didnt prove it

like algebra step by step i can prove

in informatics you learn to have a good intuition but not prove

[sly pawn]

This one?

[fickle torrent]

yes

i got the sol but didnt prove

i have a intuative proof

but not concrete

[sly pawn]

hm

This is an interesting problem

[fickle torrent]

the sol: ||a cyclic shift for 1-n so it is 1-n, 2-n 1, 3-n 1 2||

[sly pawn]

Do you know the types of proofs?

[fickle torrent]

kinda yes

[sly pawn]

I think this is a direct proof or induction kind of problem

just by looking at it that's what makes sense to me

[fickle torrent]

contradiction induction exaughtion and normal

[fickle torrent]

||that isnt correct from what i heared||

[sly pawn]

But how can you put your intuition into a proof is what you struggle with?

[sly pawn]

its not direct or induction?

Ok

[fickle torrent]

the intuition is just a educated guess

[sly pawn]

What isn't correct?

[fickle torrent]

it is called ||double counting||

[sly pawn]

The type of proof or your solution?

[fickle torrent]

type of proof

[sly pawn]

I don't think I've ever heard of that as a proof method

[fickle torrent]

same

the guy that gave me the problem told me abt it

[sly pawn]

Wait

isnt this just the same as proof by bijection?

or whatever

[fickle torrent]

idk tbh

[sly pawn]

Isn't that basically just the same as a direct proof

showing that two things are the same

[fickle torrent]

ye

double counting from what i understand is counting somthing from two different ways

[sly pawn]

oh I guess it's a more abstract way

of saying they're the same

[fickle torrent]

yes kinda

[sly pawn]

Hmm

[fickle torrent]

i did smthn similar to somthing called the exchange argument

as a intuative proof

[sly pawn]

Well if you struggle to prove things, maybe it's best to start with easier proof problems and work up to the harder ones

[fickle torrent]

ye definately

[sly pawn]

Or also maybe watching videos where you do the problem along with someone else is also a helpful way of getting better

[fickle torrent]

ill prolly just solve problems that works mostly for me when doing informatics

also you should try informatics it may improve your intuition

i am doing compatition math to improve my proofs

[sly pawn]

I have good intuition but I struggle with proving mostly too

Like all of the logic or puzzle problems they put on competitions I usually solve those pretty quickly

I usually struggle more with problems that require good drawing/graphing skills or problems that are super wordy

[fickle torrent]

Informatics will help you with the wordy problems part

The problem can be pages

I once solved like a problem that was 7 pages or smthn

[sly pawn]

That sounds good then

[twilit peak]

hello

[sly pawn]

hi

[sturdy robin]

I’ve been stuck on this problem for a bit, does anyone know how to solve?

[shadow niche]

just a sketch but use that fact that f(f(y)) = y to show that this function satisfies cauchy equation(?) i could be wrong

[prisma python]

i can do the second one but i have to assume ln(x!) is convex. is there another way to find that out except for finding the second derivative?

i think maybe it's possible to make ln(x!)=ln(1)+ln(2)+...+ln(x) so as x increases, ln(x) gets bigger so it probably has a positive second derivative but i'm not sure if that's enough of a proof

[pallid tundra]

oh huh USA’s top scorer at IMO

I’ve met him at the contest I run at my uni

this is his second gold medal

goated

[maiden panther]

he is goated

[pallid tundra]

swept all the problems we threw his way without breaking a sweat

[maiden panther]

the whole team is goated tbh

[pallid tundra]

true and real

on last year's team there was someone i had directly competed against in the past when i was still in HS

in the local leagues where he was the ONLY person from outside my school who could actually put up a fight

[maiden panther]

wow

[pallid tundra]

then there's me who only barely made AIME my senior year of HS when i had all but given up on doing so

💀

[maiden panther]

congrats on finishing off your last year strong tho!

all achievements are worth celebrating

[pallid tundra]

ig lol

[maiden panther]

some people just spend more time on the contest stuff

[pallid tundra]

it feels so long ago im going into my final year of undergrad now

and my contest career consists of a whopping two (2) points on last fall's putnam 😭

[maiden panther]

putnam is a really weird contest tbh

it's easy to get overwhelmed with a lot of the problems in a small time interval

[pallid tundra]

from what i understand the top contestants are mostly just coasting off their HS olympiad training?

[maiden panther]

and the problems generally aren't phrased in a very unapproachable way

[maiden panther]

yep

[pallid tundra]

yea i've never heard of anyone seriously studying for it LMAO

[maiden panther]

i think the main thing is

most hs olympiad people don't know how to think about some college math concepts in an olympiad context

so it's useful to practice the problems that are out of the scope of high school contests

[pallid tundra]

i look at anything analysis related my eyes glaze over in soul crushing dread

[maiden panther]

lol yeah

i can't see myself solving a calculus or linear algebra problem on the putnam

at least one that requires deep knowledge about the subject

becasue i dont have a deep understanding of the material yet

[pallid tundra]

doesnt help that my intro linalg class SUCKED

covered nothing of substance, was all just rref spam

[maiden panther]

aw that sucks

[pallid tundra]

and left out a lot of critical topics

[maiden panther]

mine was like that too

[pallid tundra]

😭

[maiden panther]

at least we defined a vector space at the end

[pallid tundra]

and did the extremely boring "verify the 10 axioms"?

[maiden panther]

something like that

[pallid tundra]

ew

that's not rigor that's mindless symbol pushing

[maiden panther]

yeah exactly

[pallid tundra]

"proofs" but i can't be arsed to actually teach smth that isn't just symbol push

[maiden panther]

linear algebra has so many rly cool ideas it's kind of a shame it's taught poorly in so many unversities

[maiden panther]

exactly

that's why i hate high school proof classes like geometry

as well

it's all symbol pushing

[pallid tundra]

i need to see if i can fit in both second semester analysis and algebra my final semester 😭

otherwise i am giga cooked for grad school apps

[maiden panther]

are you planning to continue studying math?

[maiden panther]

algebra is a great subject

[pallid tundra]

looking at doing grad school

[maiden panther]

nice!

[pallid tundra]

but thats moot point if i can't even get in 😭

if i'm missing prereqs

[maiden panther]

good luck 🙏

[pallid tundra]

thank 😭

[sly pawn]

Ln(x!) is convex.

[prisma python]

yeah, i'm just wondering if my explanation is good enough

[sly pawn]

idk how to prove it but I know it is though

[sly pawn]

probably is good

[sly pawn]

why do you need ln(x!) to be convex though?

[prisma python]

it's for the Jensen's inequality

if it was concave, the sign of the inequality gets flipped

[sly pawn]

oh I see why

[sly pawn]

so that you can add them instead of them being multiplied?

[prisma python]

the ln was to turn addition to multiplication for the LHS

[sly pawn]

I thought its to turn multiplication to addition

for the LHS

[prisma python]

nah, since the ln is outside of the x!

like it's ln(x!) and not (ln(x))!

[sly pawn]

.

the LHS is multiplication initially though so wouldnt you use the ln to turn it to addition to use jensens?

[vernal axle]

@prisma python For the second one I used f(x)=ln(Г(x+1)) which is equal to ln(x!) for integer x. I proved that f(x) is convex by using the property of the digamma ψ-function, which is the derivative of ln(Г(x)) and which has a nice expansion into series. So, ψ'(x) is positive for x>0.
https://en.wikipedia.org/wiki/Digamma_function

[sly pawn]

hmm ok

[prisma python]

would that explanation of making ln(x!) into a sum of ln(x) be good enough for a proof though? since that would apply for all integers

[vernal axle]

first I proved this inequality for all real x>0 using ln(Г(x+1)) and since this function is increasing we can say that ln(Г(x+1))>=ln(Г([x]+1)) if I got your question right.

[sly pawn]

It makes sense because the sum is increasing at an increasing rate for all x

[prisma python]

I was asking about like how I kind of explained why ln(x!) is convex for x>0, because I didn't use the gamma function

and also since from the question, a_1,a_2,...,a_n are just positive integers

basically ln((x+1)!)-ln(x!)=ln(x+1), and so the difference between f(x+1) and f(x) is increasing since ln(x) is increasing

so since it's increasing at an increasing rate, I think it would be enough to kind of prove that ln(x!) is convex right?

and also, because I haven't learned about the digamma function, so this was what I thought of

[vernal axle]

If you mean this #competition-math message explanation I don't think it is valid. I don't how to solve the problem with Jensen staying in integers only. At least you cannot apply Jensen directly, because you will have some factorial of the non-integer number. So you do have to use ln of Gamma function and it being convex is proved by Г'(x)/Г(x) properties.

[prisma python]

oh ok, thanks

[undone ivy]

wsp, so i have an incoming olympiad this october and i feel like learning from just my notes is a horrible method of preparing for it. any websites, apps, or methods i can use?

[sly pawn]

previous olympiad problems

[thick sierra]

anyone here doing the MAT this application cycle

[pure rover]

the oxford one?

[thick sierra]

yeh

[pure rover]

yeah i’m taking that and tmua

[thick sierra]

nice what u applying for

[sleek ivy]

I’m doing tmua and the Cambridge one

[pure rover]

maths & phil at oxford and then maths at imperial

[pure rover]

step?

[thick sierra]

LMAO same bro

i've never met anyone else wanting to do maths & phil

[pure rover]

tbf i don’t know why i want to do it

cause i prefer applied maths but oh well

i found some pre covid data and why are the MAT score higher for phil then they are for normal maths

[thick sierra]

dang

i really hate applied with all my soul & love philosophy so i figured it'd be the best choice

straight maths at ox has a lot of stats so

[thick sierra]

maths & phil is more competitive i think

[pure rover]

it’s got a marginally higher acceptance rate, but i’m pretty sure how it works is that the maths department hold you to the same standard and they choose whether or not to accept you and then the philosophy department choose

[thick sierra]

what basis would the phil department even choose on though as we don't have to do the TSA

just an interview

[pure rover]

yeah i think it’s just interviews

[thick sierra]

ok fair

and the interview for both is decided by MAT score?

have you started MAT prep yet

[pure rover]

yeah so the way it works is they set upper and lower bounds after everyone’s taken the MAT, everyone above that bound is automatically interviewed, everyone below the lower bound is automatically rejected and then 2/3 of people in between are interviewed, so usually it’s only if you fall between the bounds they read your personal statement

but they don’t publish what the bounds were

[pure rover]

i started in march 😭, i only properly started in june though

[thick sierra]

i heard it's around 60

upper bound

[pure rover]

that doesn’t seem that high tbh

[thick sierra]

damn, how many hours have you done so far

[half arrow]

Same

[thick sierra]

tbh i haven't started yet

[pure rover]

honestly no idea, probably 80-100. but my natural ability is horrendous so i have to work to get to a good enough level

[thick sierra]

dude thats crazy

nice

i need to lock in

i doubt your natural ability is that bad

you're predicted 2 A*s in maths fm right

[pure rover]

like in UKMT with no revision i scrapped a silver, when i did revision i got a gold

[pure rover]

yeah

[thick sierra]

yeah that's not horrendous natural ability 🙏

[pure rover]

my procrastination is so bad i have to revise early otherwise ill never start as well

[thick sierra]

i was meant to start mat prep 10 days ago and i just procrastinated and now i've done nothing

i don't think i'll get in but i'm just shooting my shot yk

have you done a full paper yet @pure rover? if so what score are you getting

[sleek ivy]

Yh

[pure rover]

like if it’s hard maybe 65-70, am easier paper like 70-75

[pure rover]

good luck

[thick sierra]

that's insane

how much were you getting when you first started

[sleek ivy]

Hahaha thanks

Is it that bad

[thick sierra]

yes esp since cambridge want top grades in it

[pure rover]

like 40-45

but i was quite good at the long questions and they’ve got rid of most of them now so idk how ill do in the new format

[thick sierra]

thats crazy

im cooked then lol

[pure rover]

it was right after i had been doing UKMT so problem solving was still in my head

[sleek ivy]

Eh its fine ill start prepping in like february

[stark swallow]

How

[pure rover]

how did i revise?

[stark swallow]

Yes with topic wise practise

[pure rover]

basically i did like 6 padt papers, then i made a spreadsheet of which topics came up and then made some notes on like cheats, like fast expansions and last digit calculations

[sacred edge]

Can someone check my solution to this problem please?

[near hill]

I don't see how you conclude |S| > |N| at the end; even if you meant |S| < |N| that would suggest you're mistaken about cardinal arithmetic.

[near hill]

Instead I would say: Once you know there are infinitely many large sets, ||let K be large enough that at least 2014 of the sets A1, A2, ..., AK are large. That means those sets together contain at least K+2014 different naturals, so one of those naturals must be >= K+2014. But such a number cannot be in any of the A1, A2, ..., AK, since it would immediately make the sum of that set larger than it's allowed to be.||

[high goblet]

lmao they don't read personal statements

if ur somewhere in the middle region it'll depend on ur privilege

(i.e. if you go to a private school or not, how wealthy is ur area etc.)

or your a level results

literally most other factors will be more important than ur p.s.

[high goblet]

i mean you just need a 1,1

Cambridge is different to Oxford, ur offer will be tied to STEP

if you're interview went okay and everything else is normal, you'll probably just get an 1,1 offer

if there are extraneous circumstances then you might get offered something like S,1

[pure rover]

ah ok, so wealthy area + private is not a good combo for being in those bounds?

[high goblet]

yeah generally not

anyway gl on MAT, you still have like a while to prepare!

[pure rover]

just gotta be above the bound ig. so once you’ve got your interview, is that all that matters in giving an offer or do they combine your interview score and mat score?

[high goblet]

iirc i'm pretty sure that after you've got ur interview the interview is the main thing that matters

although if you do slightly below avg on the interview but you had a really strong MAT score i think they would factor that in

i mean it has been a while since i was applying to unis etc. and needed to care abt this stuff

[pure rover]

oh ok, cause i remember looking at score distributions and smth like 5 people got above 96 and only 4 of them were made offers. how badly do you have to mess up your interview for that to happen

[high goblet]

oh it's more likely that they cheated

[pure rover]

would their score not just be discounted?

[high goblet]

(it's hard to prove for sure that they cheated but like if they are just completely shit at the interview then it's pretty obvious they must've cheated)

(like my sister interviewed an international who scored 97 on the MAT, but like couldn't answer any basic maths questions in the interview)

[thick sierra]

yeah you don’t get 96 on MAT then flop the interview that badly

[high goblet]

(with candidates like that, it's easier to just reject them based on their interview score than to prove that they definitively cheated)

[pure rover]

oh right i see

that’s pretty cool that your sister runs interviews

[high goblet]

lol yh i'm fairly familiar with oxbridge

[ornate blade]

what kind of fuckass family do you have

your fricking sister!?

god are any of your parents or relatives academics

if so, at Oxbridge?

[high goblet]

lol i swear u knew abt that

[high goblet]

nah not my parents

[ornate blade]

no????

I don't know about your siblings at all

you never talk about them

[ornate blade]

also I'm an only chil

I know that I haven't brought this up before

[ornate blade]

you have some uncle then lmao I bet at Cambridge

[high goblet]

ok i swear u knew but maybe not

[high goblet]

nah my parents are chinese so like none of my extended family are oxbridge

[ornate blade]

wait your parents are Chinese? hot damn

ok so you are part of the helpfuls Chinese club

welcome again!

[toxic sorrel]

is this the proper place to ask how to start comp maths? i just did my a-levels so maybe it’s too late. i was always invited to math competitions by my school but turned them down everytime which i regret. So is it too late and where to start if it’s not?

or should i just study on university when i study maths in university

[pure rover]

is it your order sister? cause that must have been insanely helpful

[high goblet]

lol ty

[high goblet]

i mean up to you, like they're a fun thing you can do but like if u're stressed about like prepping for i.e. uni entrance exams then you could focus on that instead and only do like a few past papers for competition maths

[high goblet]

yeah older

[warm rock]

wait theres a surprising amount of people from britain here

didnt expect it honestly

[sour cobalt]

Why is the proof given in this particular way? To prove this, I noted that 3, 2 are prime and 3-2=1 therefore any natural n can be decomposed into its prime factorization and then multiply 3, 2 by n to obtain n. And 3n and 2n will have the same number of prime divisors. But the given proof is more complicated - why? And for (2n)-n, doesn’t 2n have one more prime factor than n?

[near hill]

It looks like they want the same number of distinct prime factors, whereas your procedure will produce the same number of prime factors with multiplicity. Those are different tasks.

[sour cobalt]

How does this relate to 2n-n? I expect that, breaking both down, you would find they share all prime factors but one (2) and yet still wouldn’t have the same number of them

[near hill]

That's for the case where n is even, so 2 is already a factor of n, so the set of primes that divide 2n is the same as the set of primes that divide n.

[sour cobalt]

Ok, I see - we are writing them (p_1)^a_1, (p_2)^a_2, …, (p_m)^a_m

and each p_i is considered distinct, and we don’t care if they repeat

[near hill]

Right.

[sour cobalt]

i don’t think the distinction is actually clear

What do you mean ‘by multiplicity’?

[near hill]

It only became clear to me after I read the first part of the solution.

[sour cobalt]

As in, if we do not care about exponents, then aren’t 2n and n considered to have the same prime factors?

[near hill]

The multiplicities in your factorization is a_1, a_2, ..., a_m.
Your solution expresses n as a difference of numbers with the same a_1+a_2+...+a_m.

[sour cobalt]

I see, So we really want the same p_i, but different multiplicity

Since ‘number of prime factors’ refers to p_i which we preserve

[near hill]

Not necessarily the same actual primes.

[sour cobalt]

Right

[near hill]

Your solution for n=25 would be 25 = 75-50, with the factorizations 3¹5² and 2¹5², but still 1+2 = 1+2.

[sour cobalt]

We could have them distinct, but in the case of an even they remain the same. For odds, the given solution adds an additional 2 to the second piece, while changing the exponentiation

[near hill]

The book's solution would also be 75-50, but would justify it as #{3,5} = #{2,5}.

[sour cobalt]

I see

thank you

And is this language common?

As in, referring to prime factors without regaars to exponents?

[near hill]

I think "number of prime factors" is ambiguous -- it could mean either one or the other.
I initially guessed it would be your interpretation too, until I noticed it was not what would make "2n-n" work as a solution.

[sour cobalt]

Ok, I got it

[spring zenith]

Hello

not sure if this is the right channel but struggling to show 1/a+1/b = 1/2

analytically

ive tried rationalizing the radicals but to no avail

[wicked hemlock]

idk but maybe try simplying the double surds?

as in expressing surds in surds as a sum or difference of two other surds?

[sacred edge]

I see, this makes perfect sense. Thank you so much

[ornate blade]

do you know how to denest square roots? search it up if you don't know

hmm that doesn't work here though according to WA

[ornate blade]

if you brute-force expand and simplify $ab$, you should get $\frac{2}{5} \left(25 + 15\sqrt5 - 6 \sqrt{75 - 30 \sqrt5} \right)$

[gilded halo]

south

[ornate blade]

https://www.desmos.com/calculator/y97vzpu4ae

then hopefully you can follow this wording for a + b

[vernal axle]

Since you know the answer, which is 1/2, verifying it is much easier than actually finding it. Given the expressions for a and b, you can easily write quadratic equations (with irrational coefficients) for 1/a and 1/b. Then substitute 1/2−x into the equation for 1/a, expand, and check that you get exactly the same equation as for 1/b. It's not difficult, as it only involves multiplying expressions of the form a+b*sqrt(5)​ a few times.

[stark swallow]

can someone tell me why gap and block method in relevant in combi

how does it help

[stark swallow]

now in this question should not there be 6c4 x 1 x 3 x 3 x 3 x3, 4 times 3 because each question has other 3 options which leads to incorrect answer if he chooses single option correctly

[kind pulsar]

there are more than 1 combination for each 6C4

for example the correct answers are A,B,C,D,A,B

one possible combination is the first 4 questions are correct

so A,B,C,D,?,?

the first question mark cannot be A and the second question mark cannot be B

so for the first question mark there are only 4-1=3 choices which are either B,C,D

similar thing for the second question mark

so for each 6C4 there are 3×3=9 possible combinations and not just 1

A,B,C,D,B,D
A,B,C,D,C,A
...
and so on

[oblique abyss]

I am entering a vedic math competition, how does one start to study for it? and what topics of mathematics usually comes up in these types of competitions? Any kind of suggestions, tips, or anything you think will help me is appreciated!

I should also add that I literally cannot do mental math and I still have not memorized the multiplication table! My understanding and memory of basic math is also bad 😀

[hushed charm]

How old are you and what grade are you in

[oblique abyss]

This might sound bad... I'm 16 years old and 11th grade! (please don't make fun of me, I'm trying to change my ways 💔)

[hushed charm]

no, it doesn't sound bad. when studying for math olympiads, the most important is to focus a lot on solving problems, hard problems. this can be helpful: https://blog.evanchen.cc/2014/07/27/what-leads-to-success-at-math-contests/

Try to start with regional and national olympiads, those are usually easier than international olympiads. Even though, national olympiads from some countries like usa are still really hard, so try to do what evan chen is saying in this blog, try to solve problems that are hard, but not impossible to be done by u. There are thousands of problems in this website, you are certainly going to find what you need: https://artofproblemsolving.com/community/c13_contest_collections

Eventually you might need some olympiad books, but for now, if you need help with a math topic, youtube and internet in general are probably enough, this server is really good either for getting help. Algebra, geometry, combinatorics and number theory are the most common topics in these olympiads.

I don't think there are any specific methods to study or anything like that, it just takes a lot of time and effort, but of course you have to study correctly.

[oblique abyss]

Alright, I'll be sure to check out what you recommended! Thank you for the insight and advice !! 😊💓

[pure rover]

@high goblet sorry for the ping but you wouldn’t happen to know how maths and philosophy admissions works at oxford? i was looking at jesus college and it said they give up to 8 offers for all maths courses per year, this leads to some years having no maths & phil, so if i was the only applicant lets say, how they differentiate between my value to maths and phil and someone else’s value to pure maths?

[torpid cairn]

I'm not sure this is helpful for a Vedic competition

[hushed charm]

Maybe, in my opinion those are pretty helpful for olympiads like aime, usamo and imo, but I am not really familiarized with the vedic competition.

In that case, what do you think would be helpful for a vedic competition

Oh, I just checked it better, indeed what I said isn't going to be the best way to do it.

[torpid cairn]

I literally had to Google the word lmao it's no worries

[torpid cairn]

I'm not sure what extensive prep you'd do but memorizing the multiplication chart would definitely help

[dusty mauve]

Oo, okay so for vedic maths, I think you can search for Vedic books pdf and you get numerous of free books which have techniques and practice questions, also look on YouTube for, "MathOGenius", he has a whole course on mental maths of around 25 videos, they can help too

But you would have to do a bit of research and buy some legit books, PDFs would help but I prefer to depend on a more liable source

[dusty mauve]

Yes, tables up to 30 and then squares, cubes and square root and cube root, these should be remembered atleastt

Vedic maths I think would basically be more about the time taking part, as it helps to deduct us time, time would play a big factor in it

Just get Vedic books or look for PDFs online, but I think practice is the most important, hope this helped, All the bestt!!

[uneven inlet]

what is vedic math

,w vedic math

[gilded halo]

Results provided by WolframAlpha

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[uneven inlet]

,w vedic

[gilded halo]

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Upgrade to WolframAlpha Pro!

Requested by usrcard

[oblique abyss]

oh 😭

[oblique abyss]

I'm not entirely sure about it but my brother did participate in a vedic math competition before and he said it was fast computation 😊

[oblique abyss]

alright thank you for the advice and all! I really appreciate it !! ^^🌸

[dusty mauve]

Vedic maths is like ancient tricks written in the Vedas, which helps calculate problems real easilyy

Like it mostly consists of multiplication and division and basic algebra

How to multiply big numbers and stuff like that

[oblique abyss]

they're still not telling me when the competition is and I'm so nervous 😭. They said they'd train me but it's been 4 days and I've heard nothing from them!!

[dusty mauve]

That's crazyy, I thought you had like a date in mind

What do you think the rough estimate time of when the exam would be ?

[oblique abyss]

maybe October?

I'm not sure though... for all I know, it could be in Aug or Sept

[dusty mauve]

Damn, then ig you would have to plan out

I took a vedic maths course during the lockdown, it wasn't that complicated but we gotta do so much practice istgg

[oblique abyss]

really??? ong 😭

I'm really REALLY bad at math, I genuinely don't know why they even signed me up for it

do you have any tips in general? hell, I'm bad at memorizing

[subtle sundial]

they must have signed you up for a reason

trust the process brother

[dusty mauve]

Just write it until you have it memorised and then use the blurt method or teach imaginary people to test yourself

But for maths question practice is what the most recommended

[dusty mauve]

Yess they must have signed you up for a reason, do your best and don't worry about the resulttt

[weary vortex]

found this problem on a Chinese olympiad:

Let {a1, a2, a3, ..., an} be a set of integers. What is the probability that the sum of their squares is divisble by 5? Give your answer in an g(n) function format, where n is the amount of integers in a set

tried to crack this but failed miserably. any potential solution?

[near hill]

We'll probably have to assume that the remainders modulo 5 of each of the n integers are independent and uniformly distributed.

[weary vortex]

neither the assumptions nor the exceptions were provided in the description

if the author of the translation to English did not miss anything

[near hill]

Without some assumption about probability distributions, it is downright impossible to answer anything about probabilities,

If you do make the uniform-modulo-5 assumption, then you can (by brute force) work out the probability distribution of the sum of two squares modulo 5; you can notice something peculiar about it that should let you extend it to the sum of any even number of squares.

[weary vortex]

1: 1
2: 3
3: 7
4: 19
5: 53
6: 153
7: 449
8: 1331
9: 3967
10: 11859
11: 35509
12: 106417
13: 319073
14: 956931
15: 2870327

wrote a programm for this task. for values higher than 15, python takes 15000+ ms

1: 1, # Default
2: 3, # * 3
3: 7, # * 3 - 2
4: 19, # * 3 - 2
5: 53, # * 3 - 6
6: 153, # * 3 - 10
7: 449, # * 3 - 14
8: 1331, # * 3 - 16
9: 3967, # * 3 - 26
10: 11859, # * 3 - 42
11: 35509, # * 3 - 68
12: 106417, # * 3 - 110
13: 319073, # * 3 - 178
14: 956931, # * 3 - 288
15: 2870327 # * 3 - 466

There is some Fibonacci type sequence after 10 but we cant yet be sure if it continues or not

[weary vortex]

took 20 minutes to compute the next 5 results. the exponential time complexity is no joke

[oblique abyss]

Thank you again!! I never thought people in this server are nice, I came here prepared to get bullied 😭. I appreciate the advice and all 🙇🏻‍♀️🙇🏻‍♀️ !!

I'm definitely gonna work hard for this comp (and in understanding math in general!)

[dusty mauve]

Hehe no worries at alll

[dusty mauve]

Good luck, rooting for you!!💖

[soft vigil]

can you possibly induct

on n

because everytime you add a new a_n you add 0, 1, or 4 to the total sum of squares

[weary vortex]

it cant possibly be a one liner function, it has to have certain conditions for which the function will change

[soft vigil]

yeah

but you can do like

a_n, b_n, c_n, d_n, e_n be the probabilities you have 0, 1, 2, 3, or 4 mod 5 as the sum of squares

then you can make a big recursion and solve it

not the most fun but it works

[near hill]

It actually turns out to be pretty nice, if one assumes independently uniform probabilities for the a_i.
For odd n it turns out the probability of 0 (mod 5) is always 1/5 exactly.

[soft vigil]

oh nice

[near hill]

For even n slightly more than that, but converging to 1/5.

[zinc tundra]

yo yall think if i completed alg 2, geo and alg 1 i could js practice papers to prep for amc 10

i lowkey forget most of geo

[orchid linden]

Could someone send a sol to this 😭

Correct answer ||81||

[ivory ember]

Note that if $b$ has $d$ digits, then the integer formed is $a \cdot 10^d+b=a(10^d+1)+1$

[gilded halo]

Civil Service Pigeon

[ivory ember]

The rest is trivial but annoying

Because ||10^9 + 1 = 0 mod 19||

[orchid linden]

yeah like do you just memorize that

or smth

mabe in the contest its from calculator is allowed

[ivory ember]

Divisibility rule for 19 ig

Or

9 = (19-1)/2

So it’s reasonable to suspect 10^9 is -1 mod 19

That’s why I said “but annoying”

[high goblet]

sorry i've been busy! i'll reply to this later

if you have any qs feel free to just DM me directly

i'll reply when i can

[maiden panther]

the answer is $\frac{5^n + 2(2w + 2w^4 + 1)^n + 2(2w^2 + 2w^3 + 1)^n}{5^{n+1}}$ where $w = e^{2i\pi/5}$

[gilded halo]

flying_fly

[weary vortex]

are you sure it's 5ⁿ and not 3ⁿ? x² ≡ 0; 1; 4 (mod 5). so there is only 3 possible outcomes for each independent variable

perhaps, would you like to share a solution?

[maiden panther]

there are still 5 choices for x, in particlar this makes x^2 = 1 and x^2 = 4 twice as likely as x^2 = 0

so you will expect your denominator to be 5

[weary vortex]

makes sense

[maiden panther]

in particular these values are wrong

or, you are solving a different problem

the probability should always be 1/5 for odd n

[weary vortex]

the programm calculated them by force

[maiden panther]

your setup was wrong

you treated 0, 1, 4 as equivalent

remember that 1 and 4 are each twice as likely as 0

[maiden panther]

btw this simplifies to $\frac{1}{5}$ if $n$ is odd and $\frac{1}{5} + \frac{4}{5^{n/2+1}}$ if $n$ is even

[gilded halo]

flying_fly

[maiden panther]

let me know if you want me to walk you through a solution and/or how to set up a correct brute force program

[weary vortex]

# {a1, a2, a3, ..., an} - set of integers
# What is the probability that the sum of the squares of the elements of the set is divisible by 5?

def main(r): 
    exec(
    f"""
def inner_task(r):
    num = 0
    for {', '.join([f"a{i}" for i in range(1, r+1)])} in __import__("itertools").product([0, 1, 1, 4, 4], repeat={r}):
        if ({' + '.join([f"a{i}" for i in range(1, r+1)])}) % 5 == 0:
            num += 1
    return num

print(str(r) + ": " + str(inner_task(r)))
    """)  
    
if __name__ == "__main__":
    for i in range(2, 11):
        main(i)

what seems to be wrong here?

[maiden panther]

what's wrong is you wrote your entire python program as a string and then used exec() wtf 😭

[weary vortex]

i dont feel like writing for-recursion function

ik exec() is a bad practice but its not a full prod project here, its excusable

[maiden panther]

give me a second to parse your code

um your code is right

[maiden panther]

it just doesn't print these values

it prints the correct values

💀

[weary vortex]

i know, the code i did with that included only [0, 1, 4]

not [0, 1, 1, 4, 4]

i did not realize that key part

[maiden panther]

ok. well obviously i wasnt saying that your new code was wrong because i had only seen the output of your old code

but anyway glad we cleared that up

[maiden panther]

now that your brute force program is correct you can see that this is the answer

[weary vortex]

interesting how all of the omegas cancel out leaving simple fractions

[maiden panther]

yeah it's nice

do you want me to explain the solution?

[weary vortex]

would appreciate that definitely

[maiden panther]

have you heard of a roots of unity filter?

[weary vortex]

you mean zⁿ = 1 cases?

[maiden panther]

the main idea of the solutoin is as follows:

if you take the polynomial (1 + 2x + 2x^4)^n

and expand it out

you want to find the sum of all coefficients where the exponent of x is a multiple of 5

[weary vortex]

hold on, isnt this the idea of generating function?

[maiden panther]

yes it is

the beauty arises when you relate generating functions to roots of unity

it is one of my favorite ideas in mathematics

[weary vortex]

this reminded me of one 3b1b video about an olympiad discrete math problem

[maiden panther]

ooh which video

[weary vortex]

the title is something along the lines "olympiad counting"

https://youtu.be/bOXCLR3Wric?si=lpVfcT-Uuu1O8VNe

got it

[maiden panther]

ah just skimmed through the video

it's the same idea we use here

[weary vortex]

right

what an amazing technique

[maiden panther]

yeah

[weary vortex]

thanks for explaining

[maiden panther]

no problem, let me know if you want me to explain more of the roots of unity part

[weary vortex]

i remember working on this problem for different inputs and making a programming task out of it

what i soon realized, the divisor has to be prime, i think in this Chinese problem, the requirement is also that the mod has to be prime

[maiden panther]

i think the technique can still work if the mod/divisor isn't prime

it will just be a little more complicated

the pth roots of unity have this nice symmetry

where everything except 1 "behaves the same" in a sense

[weary vortex]

i dont have those papers in my hand but i remember while working with complex plane, certain points were just missing, so i had to do some tedious factoring to figure out how to compute those. it only occured in the non-prime divisor cases

[weary vortex]

by "missing", i mean when i calculated f(ζⁿ), some points just didnt come up

[maiden panther]

as in f(ζⁿ) is harder to calculate?

[weary vortex]

found it (the answer at the end is incorrect btw lol)

and its just one of the examples of my pain

[maiden panther]

,rotate

[maiden panther]

are you trying to find how many subsets of {1,2,...,3000} have sum divisible by 6?

[weary vortex]

yes

[maiden panther]

ah i understand

so you need [(1+1)(1+z)(1+z^2)(1+z^3)(1+z^4)(1+z^5)]^500

but you also need [(1+1)(1+z^2)(1+z^4)]^1000

and [(1+1)(1+z^3)]^1500

[weary vortex]

okay, prime factorization. and after i compute them, what should i do with them?

[maiden panther]

does it make sense where these are coming from

[maiden panther]

this one is equal to f(z) and f(z^5)

[maiden panther]

this one is equal to f(z^2) and f(z^4)

[maiden panther]

this one is f(z^3)

[weary vortex]

they are already calculated in the papers no? or i need whole different functions for them?

[maiden panther]

yeah i was just summarizing

your answer should be right, if you put an extra 2^3000 in the denominator

*for the probability

your answer is correct for the number of subsets

[weary vortex]

are you sure?

i checked smaller values by brute force using computer, the cases with non-prime divisors turned out to be incorrect

[maiden panther]

oh let me see

which values were incorrect?

[weary vortex]

f(z³) = (1+1)¹⁵⁰⁰(1+z³)⁵⁰⁰

is this a problem? even tho the answer is 0 anyways

[weary vortex]

let me see the discourse on github one moment

[maiden panther]

no it's not a problem you correctly found that f(z^3) = 0

[weary vortex]

btw all of the answers are mod 1000009

otherwise the outputs would be in thousands of digits

[maiden panther]

hm do you have an example with mod 6

[weary vortex]

not reported

[weary vortex]

i still find it hard to believe this is the right answer, maybe the requirements for the divisor is something different then

[maiden panther]

i think it's right

your math looks right

and i wrote a program to check

what makes you think it's wrong?

[maiden panther]

this is a good thing to consider, but the roots of unity filter is still valid

[weary vortex]

its just that smaller cases with composite divisors always were wrong, now i doubt the higher values because it will take forever for python to compute it (the only language i know proficiently)

[maiden panther]

composite divisors other than 6?

[weary vortex]

there were some reports of 12 but they were deleted for some reason

[maiden panther]

im still confused as to where the issue is

i thought you only computed an answer for 6

and now you're saying it's wrong for other composite divisors

[weary vortex]

not me, other people calculating them through other fast languages

[maiden panther]

but your formula is for 6

how can your formula be wrong for other composites if it's for 6?

[weary vortex]

thats were i got stuck

maybe there are actually whole other requirements for the divisor

[maiden panther]

what do you mean it's where you got stuck?

you found a formula for the number of subsets with sum divisible by 6

you said it was wrong for small cases

but you didn't actually check any small cases for divisibility by 6?

[weary vortex]

wait

i feel so dumb, i forgot to check it on the pc, at that time i didnt have an access to it

[maiden panther]

do you want me to check cases for you? i have a program

[weary vortex]

yes

[maiden panther]

or i can give you my python program

[weary vortex]

my head is rolling around this problem right now, i dont feel like writing python rn

[weary vortex]

your summarization made me think. if the test cases show, that for composite divisors, there are less actual subsets than expected, maybe these are the error terms?

[maiden panther]

def subsets(l):
    if len(l) == 1: return [l, [0]]
    prev = subsets(l[:-1])
    return prev + [[l[-1]] + s for s in prev]

count = 0
n = 12
d = 6
for s in subsets(list(range(n))):
    count += (sum(s) % d == 0)
print(count)

n is the length of the set, d is the divisor

[maiden panther]

it's not less than expected im just saying

your formula for d=6 is correct

.

[weary vortex]

wow

[maiden panther]

it's not the most efficient but it's fine for small cases

[weary vortex]

i was so naive setting these requirements, not realizing the task would backfire on me 😵‍💫

[maiden panther]

lol

btw, you probably want divisor to divide n

otherwise you get weird boundary issues

[weary vortex]

In order to make the outputs exact, n % divisor == 0 condition holds for every test.

included that in the rules, dont worry

[weary vortex]

looking back at this, errors occur in the 2**n cases huh

[maiden panther]

error against what though?

what formula?

[weary vortex]

i dont really know the notation y'all use to tell the bot to display the formula nicely to you, so sorry

(2^n + (d-1) * 2^(n/d))/d

this works for every prime divisor, that's indeed true

[maiden panther]

oh i see

yeah this formula is wrong lol

for d=6 you calculated your own formula and it was right

there was never an issue with your calculations, only this formula

[weary vortex]

lol really?

[maiden panther]

do you want to work on other composite cases, like 4?

[weary vortex]

i did it here

for {1,2,3,...,2000} and the divisor=4, my answer is 2^1998

[maiden panther]

that should be right

when d=4 the answer is just 2^(n-2)

[weary vortex]

i even tried d=8

[weary vortex]

so, divisor does not need to be prime in order to be solved this way?

[maiden panther]

it doesn't

the central idea of the roots of unity filter is the following:

if $z$ is a primitive $d^{\text{th}}$ root of unity, then $1 + z^k + z^{2k} + \dots + z^{(d-1)k}$ is $d$ if $d\mid k$, and $0$ otherwise

this holds whether or not $d$ is prime

[gilded halo]

flying_fly

[weary vortex]

so i gave up on that problem just by believing the stupid formula, while the answer was in front of me?

lmfao that sounds so absurd

[maiden panther]

the formula was an overgeneralization from the case where d is prime

your methods for composite d were correct

HOWEVER there is a way to simplify your calculations a lot

[weary vortex]

i mean yeah, there has to be some

[maiden panther]

here is the general answer:

$\frac{1}{d}\sum_{k\mid d, k\text{ odd}} \varphi (k)2^{n/k}$

[gilded halo]

flying_fly

[weary vortex]

im assuming φ(k) is Euler's phi?

[maiden panther]

yep

[weary vortex]

is there any article or video about this answer? would love to see how its done

[maiden panther]

maybe i should make a video on it 🙃

it's kind of long

im too tired to explain it now, sorry

[weary vortex]

yeah no problem

[weary vortex]

in case you do, let me know

[maiden panther]

will do

[weary vortex]

thanks a lot for this conversation, i really do appreciate it

[maiden panther]

no problem :) always happy to help

[languid shale]

,help

[gilded halo]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[languid shale]

,list

[gilded halo]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

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Page 1/4

[gray pollen]

$$ \int \left( \frac{\cos^2 x}{1 + x^2} + 1 \right) dx $$

[gilded halo]

Glacivee Studios

[sour cobalt]

Hey, can someone confirm my proof to the following problem? I’ve attached the one given in the book as well, which is different

||Assume WLOG that m is odd, n is even, and that it is possible to have the conditions fulfilled when m, n have different parity. Note that the number of terms in a row and in a column will be odd. Therefore take the sum of all entries in the following way: subtract the odd number of -1’s from the even # entries in a row to obtain an odd number. Multiply this by odd m (m many rows) to obtain an odd number. Thus the sum is even. Then apply the following alternate method: subtract the odd # -1’s from the odd number of entries in a column, and multiply by even n to obtain an even number. By contradiction, m and n cannot have different parity.||
I think the stickiest part may be that I am not showing that the alternative - that they have the same parity - always works. However, the book does not prove this either so I am doubtful how necessary it is

[radiant jasper]

100a+10b+c=a^b-c
a≠b≠c
a,b,c>0
a,b,c<9

[torpid cairn]

no solutions

assuming theyre integers

[radiant jasper]

,, a,b,c \in \mathbb{N}

[gilded halo]

! ! ¿ N Ø R T H ‽ 🌟 ^w^

[radiant jasper]

as well as the other restrictions i put

[torpid cairn]

100a+10b+2c=a^b

LHS is even -> RHS is even -> a is even

2c is some even integer from 2 to 16 so we need to find 100a+10b - a^b to be greater than -18

casework reveals no such a= 2, 4, 6, or 8 can satisfy this

I mean I assume they can be 0 or 9 as well because this looks like a digit problem but I don't think that helps either

[haughty bolt]

idk if this is the right channel but I couldn’t really approach this problem, any ideas?

[gray pollen]

!help @haughty bolt

[scarlet root]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[torpid cairn]

it's the competition math channel what else are you supposed to post here lmao

[pallid tundra]

inclusion exclusion?

[civic gorge]

Anyone know a good video or article that explains vietas formula?

[pallid tundra]

https://gprivate.com/6hu4m

but seriously just google it

[vernal pasture]

help pls

[weary vortex]

this isnt really a competition problem but sure

[vernal pasture]

oops sorry

[weary vortex]

before explaining, quick question, do you know how matrixes work in linear algebra?

[vernal pasture]

hmm thats not un our portiin

its fine

ill ask cahtgpt

[weary vortex]

its not fine. youll still be clueless even after the explanation

[vernal pasture]

u make it sound scary

https://tenor.com/view/you-fa-real-you-for-reak-side-eye-are-you-serious-not-looking-gif-15654014

[rough fern]

you're cheating yourself out of understanding!

[vernal pasture]

ylw_pwease

but

um

chatgpt is good explainer

right

?

DepressedMexican

[weary vortex]

fuck no

[vernal pasture]

[vernal pasture]

umm gemini?

[weary vortex]

havent used it since the last update, cant really tell much

but before, it was not a complete garbage but had several issues

[vernal pasture]

its liek chatgpt but a genaration behind

[vernal pasture]

you have done ur ug or highschool?

[weary vortex]

graduated hs this year

[vernal pasture]

oh nicee

[weary vortex]

next month starting uni

[weary vortex]

thanks man

[vernal pasture]

cool.....

[vernal pasture]

unii in Kazakh?

[weary vortex]

Yeah

ill finish bachelor's here

[vernal pasture]

ahh nice i have been to almaty this year

[weary vortex]

then move abroad to study math deeper

[vernal pasture]

that soo cool

[weary vortex]

great

[rough fern]

LLM's and AI imo shouldn't be used to teach you this stuff

[vernal pasture]

im gonna do bachelors in india preparing for jee

droolstare

[vernal pasture]

why not...i mean they have mostly all the info in the world

[weary vortex]

well, you should learn from them but not rely on them

thats how i see it

even the errors they make, over time, if you comprehend the given subject enough, you will find the mistake neuron network did

[rough fern]

idk. im just sick and tired of them and the way they are being used rn

[vernal pasture]

that soudns lieka youn probem Drink

[weary vortex]

yeah, they are definitely being misused by.. the whole god damn planet

[rough fern]

yea

its painful and frustrating

[torpid cairn]

what's ur problem with ai

[vernal pasture]

exactly.

[rough fern]

google and bing don't even give you an option to shut them off

[weary vortex]

i dont have a problem with AI itself

[vernal pasture]

ai is cool

[weary vortex]

i have a problem with people who rely on it

[rough fern]

they are making content confusing and unreliable

they require massive amounts of energy

[torpid cairn]

fine but why does that mean you can't use it to teach you math? (even though I wouldn't trust it with anything too complex)

[torpid cairn]

it's less than people make it out to be

[rough fern]

because i wouldnt trust it with anything complex in mathematics

even remotely

its just bad rn

[weary vortex]

few months ago, it couldnt solve quadratic equations

[torpid cairn]

I don't think the pre university people in #competition math are using it for complex math

[weary vortex]

from that point forward, ive lost faith in it

[torpid cairn]

maybe like a year ago

before reasoning models

[weary vortex]

nah, i checked it by the end of this winter

[torpid cairn]

u probably weren't using a reasoning model then

[rough fern]

i just don't like how unregulated it is, ig

[torpid cairn]

that's fair I'm not a huge fan

but it has its usecases

[weary vortex]

you see, they learn constantly from the content on the internet

[rough fern]

they start learning from themseleves lol

[weary vortex]

but we have SO MANY dumbasses on the internet and AI doesnt validate their content

they just take it as granted

[rough fern]

yea, that can cause issues

[weary vortex]

sure, im not saying to stop using it entirely

im a CS student, I use it to debug the code sometimes

but that doesnt mean, I rely on it, cause I learn from the mistakes it points out

[rough fern]

im frugal when i use it. recently, i switched search engines (and browsers on mobile) to ecosia

because like i said, i dont want google and bing giving me ai answers on every single search that i make

or taking the data for training or something

[torpid cairn]

oh yeah that's a big one too

I stopped using twitter when the ai bot started calling itself hitler like I don't wanna contribute to this thing

[rough fern]

yea

i also try not to use amazon anymore

im fine going directly to the vendor and waiting a few days

[weary vortex]

"X" 🤓 ☝️

[torpid cairn]

hell no

[rough fern]

simpler times when it was twitter

[weary vortex]

95% of the people in my country dont just not use it. they never even heard of amazon

[rough fern]

good lol

[torpid cairn]

what'd

[rough fern]

i think 100% of people have heard about it in the united states

[torpid cairn]

what'd Amazon do

I mean I know they're shitty but

[rough fern]

well, there ya go

lol

[torpid cairn]

did they do something ai related

[rough fern]

no (not that i know of)

[torpid cairn]

oh

[rough fern]

why support them if you know they're shitty

[weary vortex]

i dont think we are discussing ai in the right channel

or mods dont give a shit?

[rough fern]

idk, its late

i have to go soon anyways

[weary vortex]

its literally morning for me lol

[torpid cairn]

I don't think I've ever seen a mod in this channel

[rough fern]

haha, same, just that its 2 AM

[weary vortex]

Icl, Im impressed

gpt just solved quartic equation lol

[weary vortex]

@maiden panther just in case you hop on the server, I'd like to ask you smth about this formula you provided

[tender spade]

The problem is not very good. I am curious which Chinese Olympiad it is from, though. It is just a lot of computations. Assume that each of $a_1, a_2, \dots, a_n$ is in $\mathbb{F}_5$. (This is for two reasons. It doesn't effect the answer and picking random integers in general is not a well defined process). Suppose $n_0$ of them are $0$, $n_1$ of them are $1$, etc. Then root of unity filter across [p(x) \coloneqq 2^{n_0}\cdot \left(1+x\right)^{n_1+n_4} \cdot \left(1+x^4\right)^{n_2+n_3}.] Now average these values out over possible choices of $n_k$.

[gilded halo]

coolguy

[tender spade]

oh wait u just care about the entire set, not the subsets

[weary vortex]

the subsets are about another entirely different problem from discrete math

[tender spade]

Then filter across [p(x) = \frac{1}{5^n}\left(1+ 2x + 2x^4)^n ] and you will get something like [\frac{1}{5} + O(1/5^n)]

in which case the answer is very clear

[gilded halo]

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

[near hill]

The trick is to do the summing two numbers at a time. When a and b are independently uniformly distributed in F5, the distribution of a²+b² turns out to be

0 with probability 9/25
1 with probability 4/25
2 with probability 4/25
3 with probability 4/25
4 with probability 4/25
And you can get that very same distribution by a different procedure:
Roll a red d5 (with values in F5)
If it comes up 0, then the result is 0.
Otherwise, roll a green d5 and use its value.
So if you have a whole list of 2n random numbers so sum, start with rolling all the red dice. If even one of them comes up nonzero, then the end result is uniformly distributed. If they're all zero, then the result is zero.
So the probability of zero sum with 2n squares is exactly (1-(1/5)^n)·1/5 + (1/5)^n = 1/5 + (1/5)^n·4/5.

[tender spade]

Yeah that is the same thing as above once you combine like terms in the polynomial.

[weary vortex]

the answer has been already provided

[tender spade]

Also the 3b1b video is not very good. Read Zachary Abel's article about it. For whatever reason, 3b1b way overcomplicates the algebra.

[near hill]

Ah, I see. I took Coolguy's "it's just a lot of computations" to mean the simple solution had not been presented yet.

[tender spade]

I think any way you do the problem, it will require lots of computations. Once you get the right polynomial it is just lots of algebra. This is the clear estimate but since the problem requires it be exact there is lots of work to be done.

[weary vortex]

i've read a few articles about generating functions and their connection to complex analysis. basically the same thing Grant Anderson explained with a bit more examples

[near hill]

I don't think what I did qualifies as "lots of computations".

[weary vortex]

i dont know if i have read Zachary Abel's one, I dont really memorize the authors

[tender spade]

Oh yours is nice. I didn't notice that works so well.

[tender spade]

The zachary abel one makes it more clear why the problem does not depend on the divisor being a prime.

[weary vortex]

i have read an article that states that whether the divisor is prime or composite, the overall technique works but the generalized formulas differ

[near hill]

For 2n+1 squares, start by summing the first 2n squares. In the 1-(1/5)^n of times where the first 2n are already uniformly distributed, the last square doesn't matter. For the remaining (1/5)^n times, the result so far will be 0, and the final sum is 0 too iff the last F5 number was 0, which also has probability 1/5, so all in all the probability of success will be exactly 1/5 in the odd case.

[tender spade]

I don't believe they differ at all?

[weary vortex]

they do

well, not for all cases

there is some context need to be mentioned

[tender spade]

What formula do you mean exactly?

I mean the [S = \frac{1}{n} \sum_{k=0}^{n-1}f(\omega^k)] one.

[gilded halo]

coolguy

[tender spade]

Because this should be true regardless of n being composite or not. I'm sure the actual output is different, but the expression is the same.

[gray pollen]

Can someone teach me how this is even read

S is the inverse of n multiplied by the summation of f into w to the power of k where k has 0 intervals and every number is n-1 valued

[mystic shale]

No

S is the reciprocal of n multiplied by the summation of f(w^k), which is a function dependent on w^k, from k=0 to k=n-1

[ornate blade]

inverse of n is already wrong, as inverse refers to functions

or to matrices, same thing (linear maps)

[ornate blade]

S is 1 over n times the summation of f of omega to the power of k, from k equals 0 to n minus 1

[soft vigil]

is ts just rou filter

[astral jacinth]

||The product of all values is (-1)^m=(-1)^n. Then (-1)^(n-m) =1 -> n-m is even -> n has same parity as m.||

[tender spade]

Yes. I was clarifying that it does not depend on n being prime.

[versed bobcat]

guys any tips for mathcounts

?

i competed in california state competition but got absolutely crushed by some other people

[soft vigil]

Cali is cooked

Just do tests I guess

[worldly swift]

Practice a lot probs

[subtle sundial]

true

[versed bobcat]

Yeah that’s what i did

A bunch of practice tests

[soft vigil]

Including past nats?

[versed bobcat]

Well, not nats

[versed bobcat]

Just state and counties

[sour cobalt]

Yes, that’s the proof given by the book

[pallid tundra]

back when i did MATHCOUNTS (and barely missed nats qual from CT)

my preparation was

drill every state and nats test you can find

[torpid cairn]

wow I vastly overestimated gpt's ability to do very easy problems lmfao

these are from the same message it loves confusing itself

[pallid tundra]

it doesn't actually know what it's doing lmao

[torpid cairn]

apparently u can add 0 a bunch of times and your sum will increase by 500

[torpid cairn]

I knew it wasn't reliable but this is sad lmao

[prisma python]

i'm pretty sure gpt is like "yeah sure whatever" most of the time

[subtle sundial]

mhm try giving it a standard problem, and a very wrong answer

it will just agree with you most of the time

[high goblet]

sorry about the late reply!

generally joint degrees are harder because you need to be good at both the subjects u apply for

the ppl who assess ur maths abilities just assess maths

so basically like you have to be good enough to apply for just maths and just philosophy

so basically you'll be assessed on ur maths abilities just like any other maths applicant

if say, your maths was really good, good enough to get an offer but like the philosophy ppl thought it could be a little better, like you might get an offer to study just maths instead

hope that clears up ur q

[hushed charm]

Hello! I wrote this solution for a Brazilian math olympiad, and I’d really appreciate any feedback you can give me. Let me know what I can improve and if you notice any mistakes.

It was badly translated, sorry, Pot = Pow

[glacial berry]

hi, how should i prepare for things like amc 12?

just grind problems?

[zinc tundra]

this better not be amc 10 problems

[zinc tundra]

yes vro

[mental rose]

Maybe do like 2 practice tests and like grind alcumus if you do aops

and then repeat until all practice tests go bye bye

[ember veldt]

ohh, this i the good channel

every other one sucks

[timid flicker]

Black

[ember veldt]

nahhh, do mathdash

[timid flicker]

Bro mispelt it

Oh nvm

He clutches up

[ember veldt]

can't react two t's

[torpid cairn]

the problem itself was probably about high AMC 10 level but it was easy casework and I wanted to learn the general rule

[zinc tundra]

no way I’m gonna have to learn this by November

bro I’m not even done my internship yet

shibal this🫩💔💔

[torpid cairn]

you'd be fine without this lol

I can try to find the problem again it wasn't AMC 10 but it was similar difficulty

[open viper]

oof

[zinc tundra]

bro amc 10 had like super basic single variable algebra

why do I see a sigma

💔💔🥀

[zinc tundra]

this finna be a problem I leave blank for that easy 1.5 points

[open viper]

amc10 algebra usually trivial

[zinc tundra]

bro I js finished alg 2

[torpid cairn]

I can't find the problem but iirc it was pretty much "john can either walk 1 step or 2 steps up a 14 step staircase, how many ways can he walk up?"

[zinc tundra]

am I cooked

[granite stump]

I love MATHCOUNTS guys

[open viper]

intermediate?

[granite stump]

Recurrence relation

[open viper]

in what state

[zinc tundra]

wdym

[granite stump]

California

[open viper]

algebra 2 as in intermediete algebra?

[torpid cairn]

do they not call it algebra 2 in other places

[zinc tundra]

what I’m saying

[open viper]

californias really competitive

[zinc tundra]

least obvious permutation problem

[errant latch]

HI

[open viper]

hi

[granite stump]

Ik but I made it to states

[open viper]

nice

[errant latch]

how r you all?😙😙👴🏻😟😜😟

[mental rose]

hai

[granite stump]

Yes but also recurrence would work just fine

[mental rose]

bro this is the only place I can chat without getting a 30 minute slowmode

[mental rose]

good?

[errant latch]

Thats good bob

[zinc tundra]

js use Fibonacci😓

[mental rose]

tanks april

[errant latch]

you welcome

[mental rose]

https://tenor.com/view/you-are-too-polite-do-you-know-that-percy-diggstown-102-you-realize-youre-too-kind-right-gif-23833127

[open viper]

is there not a welcome forum

[zinc tundra]

yall think I can do AIME qual with just math up to alg 2 and practice problems I think it’s highkey doable

[mental rose]

yuh?

[zinc tundra]

unless I’m underestimating this test💔🫩🫩

bro im tryna prep usabo

[open viper]

sure but how r ur geo skills

[mental rose]

oh dang

[zinc tundra]

uh I can do like all the triangle geo stuff and proofs and like trig

[open viper]

thats good

[zinc tundra]

am i cooked

with number theory

and the other weird sub topics

or should I js learn them on the way

[open viper]

dont be a number theory main

number theory is non existent

[zinc tundra]

also is there a cut off on amc past test where doing them won’t be effective

[open viper]

😭🙏

[zinc tundra]

it is light work

[meager pawn]

Why can't we speak in discussion

What happened lloa

lmao

[torpid cairn]

I thought I liked number theory but when I try the high AIME problems I honestly struggle for like 30 minutes and get nowhere it's humiliating

combinatorics>>>

[deep roost]

anyone got any tips for amc 10

[open viper]

mathdash i guess?

idk

[deep roost]

u responded quickly.

[torpid cairn]

why does math dash want my parent's email what is this 🫩

[open viper]

yea

[deep roost]

.

[open viper]

u heard of it?

[deep roost]

no

[zinc tundra]

yall actually like math comp

or nah

[open viper]

sorta

[deep roost]

i like

any comp

[open viper]

its kinda fun to outscore ppl

but i think olympiad is actually fun like the questions

[zinc tundra]

🫩

[open viper]

wait whats the blue diamond next to my name

[torpid cairn]

pronouns

[open viper]

oh ol

*ok

[deep roost]

im js

tryna get into math comps

[torpid cairn]

wait math dash seems cool is it like live comps

[zinc tundra]

ur doing amc 10v

?

[deep roost]

ye

[zinc tundra]

it’s doable

[open viper]

no

actually sorta

theres past and live comps

but u can also train the minis

theres a discord server for mathdash i think

[torpid cairn]

this is cool ty

I'll look thru all of it later

[open viper]

ur welcome

they also have a book

but half of it is locked unless u get premium

[torpid cairn]

yeah I'm not paying for this ✌️

[zinc tundra]

vro js grind AoPS

[torpid cairn]

I'm gonna run out of aops problems in the somewhat distant future it's nice to have other options

[timber hamlet]

There’s no way you run out of aops problems

[high goblet]

lmao what

that's not happening

[glacial berry]

oh god alcumus ptsd

[mental rose]

lol

[torpid cairn]

are there really that many aime difficulty questions

[glacial berry]

so hypothetically, if i did amc 8 before and did so so, and now i'm getting back into competition math to try and get aime, how should i practice and is this realistic?

[mental rose]

yuh totally realistic

and turns otu that doing just practice tests are not enough so try doing mathdash

its just mathdash.com

then sign in

[glacial berry]

alr tysm

[mental rose]

then it has a ton of more problems

[mental rose]

nop

ahhh

np

[glacial berry]

lollll

[mental rose]

Also quick question, will knowing a ton of geometric formulas help me on the amc10?