competition-math

[radiant jasper]

IN my school

[prisma loom]

¯_(ツ)_/¯

Shit happens

Someone is always gonna be better than you

[radiant jasper]

And my parents are well...

[prisma loom]

Don't let that demotivate you

[radiant jasper]

not pleased

[prisma loom]

The more you let externalities impact you the more toxic your mindset will get and the less you will improve

I know it's hard but just focus on the math

[radiant jasper]

Thanks

Like really

Thank you so much

[prisma loom]

No problem

[glass pelican]

Very wise words indeed

[warped wedge]

I was 1.5 points away from AIME qual this year in 12B

What can I do to qualify next year

[pallid tundra]

wdym "what can i do"

shouldn't you know what you can improve on?

[tame burrow]

Lock in next year

[warped wedge]

I meant what should I study

Bc I exhausted most of the previous tests already

Btw are the current cutoffs on aops official

[pallid tundra]

if they're posted by the official AMCDirector account

then treat them as official

[ornate coyote]

i was surprised at 10a scores

[radiant jasper]

the difference between 10a and 10b was huge

had I only got 4.5 marks more....

😭

[ornate coyote]

i made dhr barely

by 1.5 points each

[radiant jasper]

orz

i can't even make AIME

[ornate coyote]

thing is i have no life

[radiant jasper]

better no life than no skill

[pallid tundra]

i didn't make aime until my senior year of HS

and even then only barely made it

[radiant jasper]

My parents are really mad which only makes it worse

[pallid tundra]

if you took the 10 that means you have a few more chances

[radiant jasper]

They were just like "This is completely your fault."

[pallid tundra]

bruh 😭

[radiant jasper]

This one kid in my garde who I beat last year got Honor Roll on 10

And my parents keep comparing me to him

[pallid tundra]

the specific problems on the test are out of your control, you can be the most prepared you could possibly be

and get blindsided by some really nasty tricky problem

it's happened to me a few times too many 😔

[radiant jasper]

I somehow got 3 of the last 5 right but missed the 5th question

[pallid tundra]

:o

[radiant jasper]

If I'd gottn all 18 questions I attempted right I would have definitely made AIME

But no

Ofc I get 3 wrong

[pallid tundra]

did you leave a lot of blanks in the middle questions?

[radiant jasper]

Yep

[pallid tundra]

i get the sense that those questions especially

[radiant jasper]

I only got like 3 of them right

[pallid tundra]

have been creeping up in difficulty the last few years

and sap your time so that the end becomes even harder bc time pressure

[radiant jasper]

If I'd gotten more of the middle questions

I would have made AIME with flying colors this year

[pallid tundra]

im like 3-4 years retired from doing contests atp

[radiant jasper]

What makes it worse is watching other people celebrate because they made AIME even though I beat them last year

[radiant jasper]

Cool

Where are you now?

[pallid tundra]

3rd year of college

[radiant jasper]

Nice

and even worse is the fact that some of these kids make fun of me and my friends in nsfw ways

which is why I ditched everyone in my school on discord

[pallid tundra]

bruhhhhhhhhhhh 😭

man that kinda toxicity is just

uncalled for

[radiant jasper]

ikr

[pallid tundra]

but alas the stereotype of the antisocial math nerd is not without its basis in reality

i knew a few such assholes myself while in middle/high school

[radiant jasper]

You'd think they were just quiet and calm

But some of these kids start verbally attacking you

[pallid tundra]

i used to be a bit of a self-absorbed asshat myself lmao

[radiant jasper]

My parents are like "If they really say things like this online, how do you explain their results in competitions"

[pallid tundra]

but got humbled pretty quickly once i got destroyed on my first few contests

[pallid tundra]

people can be book smart but absolute assholes (not social smart) at the same time

not rocket science 😭

[radiant jasper]

Fact is the guy in my grae who made honor roll for amc 10 this year scared my friend into keeping her mouth shut after saying creepy things to her in both dms and gcs

[pallid tundra]

aw hell nah what 😭😭😭😭

[radiant jasper]

I saw the things he told her and she won't even let me tell a teacher

She's that scared

[pallid tundra]

and with school admins being completely toothless wrt bullying

😭

[radiant jasper]

Yet if I try to stand up to some jerk after finally cracking

Then their "protectiveness" comes flying

ANd hits you in the face

[pallid tundra]

is there a teacher that you or your friend trust to talk to about this or nah

[radiant jasper]

Nope

[pallid tundra]

man

[radiant jasper]

Too scared

[pallid tundra]

😔

[radiant jasper]

Trying to avoid trouble

With other kids

[pallid tundra]

i get it yea

[radiant jasper]

Especially kids who can do pretty much anything

And are unpredictable

[pallid tundra]

i hear it's gotten way worse post covid

[radiant jasper]

yep

[pallid tundra]

i was in late HS already when it happened so i dont think i really saw the full extent of what it did

for the younger kids

[radiant jasper]

I was in like elementary school post covid

[pallid tundra]

damn

[radiant jasper]

And some boys were trying to trick me into calling one of them "daddy"

This was 5th grade

[pallid tundra]

bruhhhhhh

the brainrot is too strong goddamn

[radiant jasper]

It gets worse

In our school Nationla Science Bowl server

These idiots added an e-girl bot that says weird nsfw stuff

I left that server a while ago just to get away from these idiots

[pallid tundra]

😭 bruh

[radiant jasper]

And yet

These same kids end up doing well in competitive exams

[pallid tundra]

y'know how older people are always like "omg we're so worried about the younger generation?"

[radiant jasper]

They have a right to

[pallid tundra]

now im starting to see why 😭

[pallid tundra]

when they focus their energy on academic stuff

it can come at the cost of their social development

[radiant jasper]

It's ok to be like socially awkward

But like

This is not socially awkward

[pallid tundra]

actively antisocial

[radiant jasper]

This is purposely trying to disturb people just to get a rise out of then

[pallid tundra]

ye

[radiant jasper]

exactly

Yk about gartic phone right?

Like a game here?

[pallid tundra]

yeah i've played it a couple times

[radiant jasper]

These kids

[pallid tundra]

drawing lewd shit?

[radiant jasper]

A month or two ago

[pallid tundra]

😭

[radiant jasper]

Exactly

About my friends

And describing this stuff

My friends who were being described were traumatized

[pallid tundra]

damn

[radiant jasper]

They literally compete for who has the most trophies on brawlstars

And they each have like 50,000 trophies

[pallid tundra]

do they ever

actually go and touch grass

or talk to people

[radiant jasper]

they talk to each other

about weird stuff

[pallid tundra]

how about people outside their immediate circles

where talking that weird shit is

not accepted

[radiant jasper]

evidently not

[pallid tundra]

😭

[radiant jasper]

Because if they did

Everyone would figure out that soemthing is up

With their studious nerd facade

[pallid tundra]

does your school try to restrict cellphone usage during class

[radiant jasper]

They try

But tbh the tecahers do not care

[pallid tundra]

then what's the point of such a policy 😭

[radiant jasper]

Kids can literally curse and say b-word

[pallid tundra]

if it isn't even enforced

[radiant jasper]

And they do nothing

[radiant jasper]

Ikr

They just take out thier phones in class and teachers might say

"put your phone away" and gove up

The only teacher who actually tries is the art teacher

and he already is a major target for the "popular athletic boys"

[pallid tundra]

oh god that kinda student 💀

[radiant jasper]

Why do they even bother taking certain classes I wonder

[pallid tundra]

yea i tended to just avoid them while i was in HS

i didn't really see them much anyway bc i was APmaxxing

[radiant jasper]

Sadly my art class is full of them

[radiant jasper]

Yeah in hs

These idiots tend to pop up less in more academic settings

Since they are grade level their whole life

[pallid tundra]

ye

[radiant jasper]

Yeah

Idk

I'm in a ranting mood today

Really sorry if I bothered you

thanks though

[pallid tundra]

npnp

[high goblet]

practise olympiads instead

the common trap ppl fall into is that they keep practising for their current stage

but this has diminishing returns

when u've done all the past papers for ur current stage, u should just start doing the harder stage

i'm not american, i've never done any of the american competitions but if AIME is next then probs start preparing for that even though u haven't qualified yet

https://bmos.ukmt.org.uk/home/bmo.shtml

i'll link the british maths olympiad here, BMO1 isn't too hard and should be doable (i'm not sure how this compares to american competitions tho)

[shadow spruce]

practise

[shadow spruce]

how

[ornate blade]

$\frac{(a/2R)^2 (b/2R)^2}{(c/2R)^2} = \frac{a^2 b^2}{c^2 (4R^2)}$

[gilded halo]

south

[ebon dock]

I'm kinda interested on this one

[ebon dock]

it's from my country's math olympiad of the national stage

[tulip orbit]

Inequality spam

[gusty verge]

Write it as a product, if it's a positive integer then there will be n+1 factors of 2 on the top.

[dapper magnet]

what does cyc mean lol

combinations of A, B, C?

[prisma loom]

It cycles through a,b,c in that order

(a,b,c) -> (b,c,a) -> (c,a,b)

E.g. $\sum_{\mathrm{cyc}} a^2b = a^2b + b^2c + c^2a$

[gilded halo]

daniel

[dapper magnet]

what about 4 tuples

[prisma loom]

then it's (a,b,c,d) -> (b,c,d,a) -> (c,d,a,b) -> (d,a,b,c)

[dapper magnet]

ohh

im thinking about using that notation for symmetric polynomial

is it applicable

[prisma loom]

for that you can use $\sum_{\mathrm{sym}}$

[gilded halo]

daniel

[prisma loom]

but it's not exactly the same as symmetric polynomials

sym means go through all permutations

so e.g. $\sum_{\mathrm{sym}}a^2b = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2=\sum_{\mathrm{cyc}}a^2b + \sum_{\mathrm{cyc}}ab^2$

[gilded halo]

daniel

[dapper magnet]

woah

well i was wondering if theres a way to shorten this

[prisma loom]

I don't think there's any established notation for it

I just think of it as the sum of every product involving $k$ distinct elements

[gilded halo]

daniel

[dapper magnet]

yeah but i wish theres an easier notation

maybe set??

[tulip orbit]

Ofc bruh

The thing abiut symmetric notation is that

It gets complicated after 4 terms

[vernal axle]

It's easy to prove by induction that (2^n-1)!!=(2^n+1) mod (2^(n+1)). Just partition it into pairs (2^(n-1)-a)*(2^(n-1)+a) for n>2.

[digital hedge]

Source?

[versed bobcat]

if y'all need some competition math

[tame burrow]

What level MATHCOUNTS is this?

[burnt trail]

For how many natural numbers less than 100 is the product of the number's distinct prime factors equal to 6?

[prisma loom]

what is necessary and sufficient for the product of the distinct prime factors to be equal to 6

[burnt trail]

All of its prime factors to either be 2 or 3

Or what do you mean, idk

[prisma loom]

yes

so how many numbers less than 100 are only made up of 3's and 2's

just count that

[burnt trail]

How

[prisma loom]

3^4 > 100 so just take 3^3, 3^2, 3^1 separately

there aren't that many

[gusty verge]

What 3^4=81?

[burnt trail]

, calc 3^4

[gilded halo]

Result:

81

[prisma loom]

Sorry that's true

So just add 1 to the cases 3^3, 3^2, 3^1, since 2*3^4 > 100

[burnt trail]

can you help me compute the combinations

[burnt trail]

[prisma loom]

You want to count how many twos you can add before it exceeds 100

3^2=9, if we multiply this by something greater than 11, it's gonna exceed 100

So how many powers of two are there less than or equal to 11? 2, 4 and 8

So there are 3 valid numbers with 3^2

Repeat this for the others

[burnt trail]

,calc 3^3

[gilded halo]

Result:

27

[burnt trail]

,calc 2^4

[gilded halo]

Result:

16

[burnt trail]

,calc 3^3 * 2^2

1. 3^3 . 2^1

,calc 3^2 * 2^3

[gilded halo]

Result:

72

Result:

108

[burnt trail]

1. 3^3 up to 2^1

because 3^3 . 2^2 > 100

2. 3^2 up to 2^3

,calc 3^2 * 2^3

[gilded halo]

Result:

72

[burnt trail]

,calc 3 * 2^5

[gilded halo]

Result:

96

[burnt trail]

3. 3^1 up to 2^5

i) 3^3 up to 2^1

ii) 3^2 up to 2^3

iii) 3^1 up to 2^5

@prisma loom

is i) ii) and iii) correct?

how do we know we are not double counting?

[inner ermine]

theres only one way to write a number as its prime factors

[burnt trail]

okay so let me see if I did it right

i) 1 x 5
ii) 3 x 1
iii) 1 x 5

or am I tripping?

@inner ermine

[inner ermine]

huh

1 is not a prime

[burnt trail]

my badd

wait what?

how is that related

1 is the number of 3s

or the number of 2s

basically 1 x 5 + 3 x 1 + 1x5 numbers that have a product of distinnct prime factors that are equal to 6

which are less than 100

i) 3³ up to 2¹ ==> 3³ x 2¹ = 27 x 2 = 54
ii) 3² up to 2³ ==> (3² x 2³) + (3² x 2²) + (3² x 2¹) = (72) + (36) + (18)
iii) 3¹ up to 2⁵ ==> (3 x 2⁵) + (3 x 2⁴) + (3 x 2³) + (3 x 2²) + (3 x 2¹) = (96) + (48) + (24) + (12) + (6)

[burnt trail]

so in total is 9 numbers, right?

[prisma loom]

Yeah sounds true

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[burnt trail]

can I get some hints?

[mortal cedar]

You need to find the greatest power of 10 that divides this number

[woven cape]

Well I know it is 20! but how do we get the answer without a massacre of numbers

[mortal cedar]

Think about which numbers in the product 20! contribute towards the power of ten dividing it

[ornate coyote]

10=2*5

keep that in mind

[ornate blade]

there will always be more 2s than 5s

so the trick is to count the number of 5s, then there will always be a corresponding 2 to make 10

also you don't have to worry about 5^2 = 25, whereas you do need to consider powers of 5 for other questions

[versed bobcat]

random sample

[radiant jasper]

can some one sovle this

[pallid tundra]

i dont get the notation on the first line

which argument is the base of those logs?

the x or the x_i?

[deft wraith]

Well what have you tried and we’ll give you a hint

No point solving it for you

[deft wraith]

I believe it’s base 10 I.e just lg

It’s log(x * x1)

[radiant jasper]

i tried graphing |lnx-lnx_1| +(|lnx +lnx_1|

thats it

and

yea

idk what to do next

i am soo confused @deft wraith

[pallid tundra]

try bounding the LHS using triangle inequality?

[jade widget]

Hii

[radiant jasper]

alr

[lilac niche]

Can anyone please solve this

<@&286206848099549185>

[clever skiff]

use taylor expansion of e

[dry quartz]

WTH

[clever skiff]

break your term into three parts and compare coefficients of each part with above

[silver lagoon]

simplify the expression

you can either use taylor series or only partial fractions

[pallid ginkgo]

for prime $p$, find all positive integers $(a,b,c)$ such that
$$a^p+b^p=p^c$$

[gilded halo]

Skissue ping4response

[pallid ginkgo]

by fermats little theorem, a+b mod p=0

for p=2, we see that (1,1,1) works, WLOG a>=b, since a+b mod 2=0 thwn either a,b is even or a,b is odd, if its even thwn we can take out a 2 to make it such that arleast one of them is even, if both of them arent even then a+b mod 2=/=0 so we can factor out 2's so that they are odd, (from the image) mod 4 we can get 1=3 mod 4 (for c>1) which isnt true, for c=1 then it would hold, which only happens at a=b=1, note that since we factored 2's from the start, we can multiply them by say 2^n and it would still hold, thus one of the solutions are (2^n,2^n,2n+1) for p=2, and those are the only solutions for p=2

not sure how to continue this, by lte lemma $c=1+v_p(a+b)$

[gilded halo]

Skissue ping4response

[pallid ginkgo]

#help-28 message
theres some ideas thrown around here

[prisma loom]

Bound using this

Also that only holds when p doesn't divide both a and b

So you need to consider that case as well

[mortal cedar]

How do we get this?

[prisma loom]

The lifting the exponent lemma

[jolly solstice]

if p divides both then you can divide it out, and subtract it from c

[prisma loom]

Yes but then you don't necessarily have that p divides a + b

[jolly solstice]

once you have a solution (a,b,c) then you can make more with (pa, pb, c+p) so you can boil it down to these types of solutions

[jolly solstice]

we do

[prisma loom]

Why

[jolly solstice]

reduce a^p+b^p=p^c mod p

[prisma loom]

oh yea I'm a moron

[jolly solstice]

all good lol

[jolly solstice]

I suppose jumping off and expanding from here with other stuff we can make these solutions: (ap^n, bp^n, pn+1+v_p(a+b))

but not super exciting without a "primitive" solution to start from

I think technically some of the LTE is part of factoring for odd p

since a+b=p^i and a^{p-1} - a^{p-2}b + ... - b^{p-1} = p^j with i+j=c

kind of vaguely relates to the proof of LTE if I remember

[normal raven]

can some help me with this

[ornate blade]

15^n is never evn

*even

[normal raven]

but when you prime factorize it

you also get 5

which can get you 0 right

[gusty verge]

To end in a zero you need a factor of 10

[radiant jasper]

Prime factorising it, we get 5 but 2 isn't there which is needed, so it isn't going to have da zero

[hexed oak]

can anyone give me a neat problem?

[prisma loom]

Prove that if you have a sphere tangent to a plane then the projection through the north pole of any circle on the surface of the sphere not passing through the north pole to the plane is a circle

[hexed oak]

any neat non bashy solution?
(i want to find the minimum top length, the top is parralel to the bottom, but the method i thought of is too bashy)

[hexed oak]

so you have a cone

through the northpole

no

not a cone

a slanted cone?

let us take any random point as the north pole

and the bottom plane as the plane

a circle on a sphere is the intersection of a plane and the sphere

let us look at it in the other way

we have a circle

and we have a plane

[prisma loom]

true

[hexed oak]

and we want the set of points that will take the circle to a circle on the plane

[silver lagoon]

i saw this question in aops

want the solution?

[pallid ginkgo]

sure

[gentle rose]

meth>maths

[radiant jasper]

ngl this channel is the most focused channel fr fr

Except like #calculus

[jolly solstice]

wish they would focus on this problem 🥹

[high goblet]

doesn't zsigmondy work lol

if p | a+b, then a^p + b^p has a prime factor not equal to p (apart from some specific counterexample) - qed

otherwise if p | a & p | b, then i sps you can factor out the p until one of them is coprime to reduce to the p | a+b case

[prisma loom]

True

Here is a solution without zsig: We have $\nu_p(a+b)= c-1$ as discussed above, hence we are assuming $p\nmid a,b$. Let $a+b = p^{c-1}k$. For $p\ge 4$, we have $a^p + b^p > (a+b)^2$, hence we get $p^c > (a+b)^2 = p^{2c-2}k^2\implies 1 > p^{c-2}k^2$, contradiction. Hence $p<4$. The case $p=2$ was already solved above. Hence let $p=3$. Then $a^3 + b^3 = (a+b)(a^2-ab + b^2) = 3^{c-1}\cdot k \cdot(a^2- ab + b^2) = 3^c\implies k(a^2 - ab + b^2)= 3$. Hence either $a^2-ab + b^2 = 1$ or $a^2-ab + b^2 = 3$. It is easily seen that the only solutions to $a^2-ab + b^2 = 1$ is $(a,b)=(1,1)$. But that gives $2=3^{c-1}$, contradiction. Hence $a^2-ab+b^2=3$. Again it is easily checked that the only solutions to this are $(a,b)=(1,2)$ or $(a,b)=(2,1)$. This gives $a+b = 3\implies c=2$, so the only primitive solutions for odd primes are $(a,b,c,p)=(1,2,2,3)$ and $(a,b,c,p)=(2,1,2,3)$. (which are precisely the exception to zsigmondy)

[gilded halo]

daniel

[silver lagoon]

https://artofproblemsolving.com/community/c6h3041815_diophantine

[pallid ginkgo]

thank you

[high goblet]

not quite the same lol

that's p!, not p^c

[prisma loom]

It's completely different in fact

[pallid ginkgo]

oh wow thanks, that makes alot of sense

i had a feeling finding an upper bound was key but couldnt think of how to make it

[silver lagoon]

ohh

in that case, zsigmondy gets the job done

[digital hedge]

Find all sequence a_1, a_2,... of integers such that for all k≥2024, the following expression is an integer.

[marble grove]

has anyone here ever solved all imo questions in a comp setting

[pallid ginkgo]

imo perfect scorers?

[prisma loom]

https://artofproblemsolving.com/community/c6h3463368_infinite_integer_sequence

[radiant jasper]

i did nto get the last line of this

i mean how a1 is divisible by every prime greater than N hwo didi they deduce that

[prisma loom]

Because k doesn't divide (k-1)!, if k is prime, hence it must divide a1

Because we can take k to be arbitrarily large primes, a1 must be divisible by arbitrarily large primes, hence it's zero

[radiant jasper]

ohhh

[radiant jasper]

i did nto get it what did they do with p taking common out

sorry i am not familiar with number theory and proofs

[prisma loom]

I don't get what you mean

[radiant jasper]

what did they do after p>N,m line

[prisma loom]

It's just factoring and creating common factors

[digital hedge]

Why can't I analogously prove that ai=0 for other i=2,3,... like I prove a1=0

[prisma loom]

You can

That's what the solution does

[digital hedge]

Oh ok I just realized

[radiant jasper]

Umm

So since p is prime

Unless it's 2

It can be factored

Yeah irdk

[sonic gale]

wtf am i looking at

im genuinely so cooked

the shit im seeing in this server looks like something from another planet

i can barely do the simplest things

[tame burrow]

lwk it's not that deep

[ornate blade]

Not everyone is good at NT

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[burnt trail]

Can I get some hints

[jolly solstice]

I guess one way is extended euclidean algo to solve 3b+2a=1, get all the solutions to that, then multiply through by 8 and then check the condition

might be quicker ways than that but I think that would get it done at least

[burnt trail]

why 3b+2a=1

and then multiply by 8, also idk extended euclidian algo, is it possible to learn or do I need math maturity

[jolly solstice]

I think you just have to learn the algorithm, I guess to elaborate what I'm really talking about is Bezout's Identity if you want to look that up

basically the idea is 3b+2a=gcd(3,2) is solved by the extended euclidean algorithm, and it turns out there are infinitely many pairs (a,b) that you can get once you have one solution

then multiplying through by 8 means you get a solution to the original problem you asked

[barren grail]

hi guys how do i learn competition maths

[radiant jasper]

Look for competitive exam first? Olympiad perhaps?
Googling itself will help a lot, or if you have decided to prepare for a programme like olympiad as I said, specify your interest first... there sure will be a lot of people who could help you

[pulsar terrace]

give hint please

[gusty verge]

You can evaluate the sum, write it in sigma notation

If you haven't done it before, look at how to evaluate: $$\sum_{k=0}^{n} k^m a^k$$

[gilded halo]

Max

[gusty verge]

||Integrate wrt a, then multiply by a, then integrate wrt a, then multiply by a... do this k times||

[radiant jasper]

where have you been

before whenever I looked at this channel at any given point of time I saw at least one message from you

[digital hedge]

Let m,n be integers greater than 2. In a m×n grid, floor(mn/3) -1 of the cells are marked. Prove there exists a 3×2 subgrid which contains at most 1 marked cell.

[radiant jasper]

what grade are you in at school

[barren grail]

grade 10

[radiant jasper]

Blud did not react to my essay though

[barren grail]

didn't see this mb

hmmmm

[radiant jasper]

Read now atleast

[radiant jasper]

i would reccomend the aops books https://artofproblemsolving.com/store/list/aops-curriculum?srsltid=AfmBOor6REpa6LO5kiB8jWTbKnXZhNg6zYMvS6BJBg_hp_sitNnIp89_

[barren grail]

i mean yeah but how do i know learn how to do the questions (responding to bakuza)

[radiant jasper]

What are you trying to crack??

[barren grail]

how do i actually learn how to do the problems, thats the question i intended

[radiant jasper]

Kay... for your school exams?...

You guys call it graduation right?

After 10th or 12th...

[radiant jasper]

I don't think 10th grader in any country need calculus ?

@barren grail This is how I would start off, once you have completed these books then you'll be able to understand most other advanced materials. All books I am referencing can be found here (https://artofproblemsolving.com/store/list/aops-curriculum?srsltid=AfmBOor6REpa6LO5kiB8jWTbKnXZhNg6zYMvS6BJBg_hp_sitNnIp89_)
Algebra

1. Intermediate Algebra. (start with Introduction to Algebra if this is too hard).
2. Precalculus

Combinatorics

1. Introduction to Counting & Probability
2. Intermediate Counting & Probability

Number Theory

1. Introduction to Number Theory

Geometry

1. Introduction to Geometry

My bad, when I revisited the website, I noticed scroll too hard at the first time

[radiant jasper]

what books do u reccomend

[radiant jasper]

Since I'm not from any western country, I have no idea about the level questions asked in there

So I don't

[radiant jasper]

the level of questions asked in what

[radiant jasper]

In schools of your country

[radiant jasper]

I'm talking about IMO and competetive math preperation

I just know that the syllabus is same for all

[radiant jasper]

I've never prepared, infact I'll start after some time when my teachers say that we have our basics cleared

Idk much about olympiad and all also

[deft wraith]

When will bmo1 results come out

[radiant jasper]

Ig I should prepare for olympiad as well....

Can somebody help me?

Lol

[barren grail]

thanks man

[radiant jasper]

Somebody?

[plush abyss]

Guys

If tan Is - and they give you cos then sin is -

[radiant jasper]

Please write the question a little more readable way

[radiant jasper]

Somebody?

[ornate coyote]

Im failing the aime

[radiant jasper]

what?

[ornate coyote]

[radiant jasper]

if tan is -ve it means it is either in 2nd or fourth quadrant

so cos can be -ve or sin can be -ve

if cos is given to be +ve and tan is -ve it means angle is in fourth quadrant

[radiant jasper]

At least you didn't miss it by 4.5 points

[novel karma]

if anyone has the solutions book for everaise academy math 1, can you dm me

thanks

[burnt trail]

can i get some help

like I tried to learn extended eulcidian but maybe I lack the mathematical maturity

Suppose that $a$ and $b$ are integers such that$$3b = 8 - 2a.$$How many of the first six positive integers must be divisors of $2b + 12$?

[gilded halo]

938c2cc0dcc05f2b68c4287040cfcf71

[burnt trail]

maybe this problem is too much for my level of understanding of number theory?

how do I know if I am not good enough?

this was suggested in alcumus as the next exercise to solve , should be appropriate but idk, why is so hard

[ivory ember]

This is too much effort. Best bet (this is a general problem solving strat) is to try to narrow down your possibilities off the bat with a simple pair like when a=1.

Then you can consider those possibilities instead

Hint: ||parity (aka whether something is even or odd)||

[radiant jasper]

try modulus??

That normally works for divisibility problems

LIke find a(mod 3) and then fond b(mod 3)

because we like know that a and b are integers

[marble grove]

yes

[pallid ginkgo]

theres a list of them

[burnt trail]

how

[burnt trail]

when a = 1

3b = 8-2a
3b = 8-2
3b = 6
b = 2

so b is even?

[ivory ember]

Concluding that in general based off of one case isn't a sufficient proof, but it's easy to show by ||taking both sides mod 2||.

[radiant jasper]

Wait

lemme try it

[burnt trail]

yeah I am trying to take both sides mod 2 but I am having trouble with the modular arithmetic

3b mod 2 = 1b, for left hand of the equation?

like if we divide 3b by 2 we get remainder 1b

or at the very least dividing 3 by 2 leaves remainder of 1, idk what to do with the b tho

like for the right hand side

8 mod 2 = 0 because 8 is even right?

and for 2a:

2a mod 2 = 0

aswell because 2a is always even right?

like substituting back the results after taking mod 2 I get

1b = 0 - 0

idk what to do with that though

is someone following? I think I might be tripping here

@ivory ember

[burnt trail]

did you give it a try hypercube?

[ivory ember]

$3b \equiv b \pmod{2}, 8-2a \equiv 0 \pmod{2} \implies b \equiv 0 \pmod{2}$.

[gilded halo]

Civil Service Pigeon

[ivory ember]

This is too much effort. Best bet (this is a general problem solving strat) is to try to narrow down your possibilities off the bat with a simple pair like when a=1.

using this, you should get that the only possibilities after this are 1, 2, and 4

1 is obvious

2 is also obvious

[burnt trail]

for a = 1 , b is even

[ivory ember]

For 4, b being even -> 2b is a multiple of 4, and 12 is also a multiple of 4

[ivory ember]

and b is also even for all values of a

[burnt trail]

when a = 2

3b = 8 - 2a
3b = 8 -4
3b = 4
b = 4/3
not even

[ivory ember]

[burnt trail]

wait but 4/3 what is it

[ivory ember]

what is 3/2 supposed to mean here

[burnt trail]

typo, I was trying out a =2

[ivory ember]

.

[burnt trail]

b is integer

[ivory ember]

yeah

[ivory ember]

so how is this relevant?

[burnt trail]

a = 2 doesnt work ig

because b is integer

I might be tripping

when a = 3
3b = 8 -6
3b = 2
b = 2/3
b is integer a = 3 is useless

[ivory ember]

3b = 8 -6
2b = 2

what

[burnt trail]

my bad

when a = 4

3b = 8 -8
3b = 0
b = 0
mmm what

this is nor even nor odd

[ivory ember]

0 is even

Recall that a number n is even iff n=2k for some integer k

since 0=2(0), 0 is even

[burnt trail]

ok thanks

[burnt trail]

how did you got that b is divisible by 2

am I reading correctly?

[ivory ember]

left hand side is b mod 2, right hand side is 0 mod 2

aka take both sides mod 2

[burnt trail]

1. 3b mod 2 = b

2. (8 -2a) mod 2 = 0
   2.1) 8 mod 2 = 0
   2.2) (-2a) mod 2 = ((-2) mod 2) * ((a) mod 2 )
   2.3) (-2) mod 2 = 0
   2.4) 0 * (a mod 2) = 0

3. ==> b mod 2 = 0?

[ivory ember]

yeah

[burnt trail]

first six positive integers is 1 2 3 4 5 6

why do we only check 1 to 4

[ivory ember]

This is too much effort. Best bet (this is a general problem solving strat) is to try to narrow down your possibilities off the bat with a simple pair like when a=1.

[burnt trail]

when a = 5

3b = 8 - 10
3b = -2
b = -2/3
b is integer a = 5 is useless

when a = 6

3b = 8 -10
b = -2/3

b is integer a = 6 is useless

if b is even

then 2b + 12 is even

and thus 2b+12 is only divisible by the values of a such that b is even since 8-2a is even

for a = 1 b is even
for a = 2 b is not even
for a = 3 b is not even
for a = 4 b is even
for a = 5 b is not even
for a = 6 b is not even

[burnt trail]

we know 8-2a is even because we showed that in 2 to 2,4

idk really tbh

[ivory ember]

Let b=2k for some integer k

[burnt trail]

here you are assuming b is even

[ivory ember]

We legit proved it earlier

[burnt trail]

yeah, seems like a legit proof

but idk, I lack math maturity

I cant tell if its legit

also I never understood why we took cases a from 1 to 6

so we can see if b is even or odd

but then we proved it

I mean it feels surreal, I am new to modular arithmetic sorry if I sound like a clownass

[ivory ember]

???

This is too much effort. Best bet (this is a general problem solving strat) is to try to narrow down your possibilities off the bat with a simple pair like when a=1.

We have the pair (a,b)=(1,2)

So 2b+12=16 for this case

My point doing this was that we can immediately exclude the possibilities of 3,5,6 being divisors since they don’t divide 16

And can just consider 1,2,4

1 is obvious

For 2, note that 2b+12=2(b+6), hence 2b+12 is always even

For 4, note that taking both sides of the equation modulo 2 yields b=0 mod 2 (we did this earlier).

So, b=2k for some integer k -> 2b+12=4k+12=4(k+3), hence 2b+12 is a multiple of 4

[burnt trail]

ok

thank you for the help, I think I got it, I appreciate it

[mighty swallow]

If the qth term of an AP is p and the pth term of an AP is q then prove that the nth term is p+q-n

Challenge!!

[ornate coyote]

erm what

the words are not wording

[mighty swallow]

Its perfect

Solve it

[ornate blade]

hint: let q = p + n

[ocean narwhal]

you cant just do that

q,p and n are all given to you

[ornate blade]

oh right

[ocean narwhal]

just write the pth and qth term of the AP

you got 2 eqns and 2 variables

the first term and the common difference

solve for them then it just plugging in the values in a_n

[opaque citrus]

hi

i have a problem

can i share it here?

[ornate blade]

I guess then, in q - p terms the difference is p - q, so the common difference is -1

[ocean narwhal]

yea

that gives first term = p+q-1

[prisma loom]

Mod 3 and mod 4

[radiant jasper]

Not yet

I was grinding physics

I'm trash anyway

[high goblet]

this isn't really modular arithmetic, the problem itself isn't really that challenging, it's more testing ur mathematical maturity to understand what you have to do

the problem asks which of the first 6 positive integers must always be divisors of 2b+12

so we need to see if 1,2,3,4,5,6 always divide 2b+12

a=1 gives b=2 so that tells us that 3,5 and 6 cannot always be factors of 2b+12

obviously 1 and 2 are factors of 2b+12

then just taking mod 2 => b is even, so 4 is always a factor of 2b+12

so the answer is 1,2,4

[mighty swallow]

Perfect

[mighty swallow]

Perfect

[mighty swallow]

👏🫡

Lemme give you another one

[random thicket]

any tips how to prepare on math olympics?

[radiant jasper]

study

[random thicket]

but there is so much

[radiant jasper]

[prisma loom]

There is not that much

[radiant jasper]

study yor weak spots

[random thicket]

hmm

[radiant jasper]

do a practice test

[random thicket]

then geometry ig

[radiant jasper]

see what you SUCK at

[random thicket]

I did, I am waiting for results

[radiant jasper]

ok

[random thicket]

anything for first class of gymnasium?

[radiant jasper]

umm.

first

im american

second

im high school

[high goblet]

just do problems

start with easier olympiads then make ur way to harder ones

it'll be slow at the start but it'll be worth it

[gusty verge]

If you're looking for a start point:
Set y=0 and x=-f(0)

You'll get f(-f(0))=0

Then set x=y=-f(0), you'll find that
f(0)=0 or 1, consider each case separately

Not sure how to prove the f(0)=0 has no solutions as the function does not necessarily need to be continuous, if it was continuous you could show the only possible function is the 0 function which ofc doesn't work

||the only solution is f(x)=x+1 ?||

Nvm I got it, lmk if u want a full sol

[radiant jasper]

yes

[high goblet]

where's the problem from may i ask?

[radiant jasper]

8th European Mathematical Cup

Senior Division

[high goblet]

huh ic ty, not aware of that competition

[radiant jasper]

np

[gusty verge]

First set $y=0, x=-f(0)$, you get that $f(-f(0))=0$ as already discussed. We can make use of this by setting $x=y=-f(0)$, the equation reduces to $f(0)=f(0)^2$, so either $f(0)=0 or 1$.

Consider Case 1: $f(0)=0$.

Set $x=0$, you get $f(f(y))=f(2f(y))$, inductively you can hence show that $f(f(y))=f(2^nf(y))$. Set $c=f(y)\in\mathbb{R}$, you hence have $f(c)=f(2^n c)$. (Made a simple flaw in my proof, I think this is only valid for natural number $n$, if we could show that it was valid for all integers $n$, you could make a continuity argument arguing that we must have either $f(x)=0$ or $f(x)$ is piecewise constant, both wouldnt satisfy the original equation)

Consider case 2: $f(0)=1$. Set $y=0$, you get $f(x)+f(1)=f(x+2)$. In other words $f(x)$ must be linear. So set $f(x)=ax+b$, input this into the original equation and compare coefficients (as both sides must be equal for ALL x,y), solving you get $a=b=1$, so $f(x)=x+1$ is the only solution.

[gilded halo]

Max

[high goblet]

ur deduction for f being linear isn't quite correct

since f : R -> R rather than f:N -> N

[gusty verge]

Yep, but I have made the sturdy assumption that we have continuity otherwise the question wouldnt have a satisfying answer

[high goblet]

you can prove f(x) = x+1 without assuming continuity

[gusty verge]

I was thinking recurrence relation

but i havent done that stuff for ages

[high goblet]

i'm kinda tired rn, maybe even slightly ill so i can provide more detailed suggestions tmr if u want me to

but you have an exposed xy - you should be able to prove f is injective

once you have f is injective, plug stuff in to prove f(x) = x+1

[high goblet]

you can use the fact f(0) = 0 or 1 to get 2 cases

[high goblet]

also f(x) + f(1) = f(x+2) doesn't imply linear even for continuous functions

i.e. something like f(x) = cos x but you need to shift stuff and stretch stuff so everything works

[gusty verge]

Yeah now thinking about it I havent used the xy fact apart from comparing coefficients which isnt really significant enough

[prisma loom]

If that were true it would be written in the problem statement

[gusty verge]

ik im making a joke lmao

[prisma loom]

Ok sorry

[gusty verge]

nah its fine, but its true, the solution will likely be continuous or at the worst piecewise discontinuous

[prisma loom]

Yes of course

But continuity is extremely powerful

So it is faulty to assume it

[gusty verge]

yep

[high goblet]

my face when:

[gusty verge]

https://tenor.com/view/what-are-you-challenging-me-scooby-doo-shaggy-matthew-lillard-gif-17770189

[high goblet]

no it's just that this is like a functional equation on a shortlist

[gusty verge]

https://tenor.com/view/pasina-chhutna-swamod-swatipramod-shorts-break-darr-ke-mare-pasina-chhutna-gaur-se-dekhna-gif-27434016

me tryna resist the urge to have a go at this instead of sleep

[high goblet]

||that has uncountably many points of discontinuity lmao||

||in fact, the cardinality of the set of solns is 2^|R| lmao||

[prisma loom]

Is it ISL?

[high goblet]

yh

[prisma loom]

Bruh that's cooked

Who let them cook

[high goblet]

also like this is one of the hardest FEs that exist lol

[prisma loom]

What year is it?

[high goblet]

2004 A6 i believe

[gusty verge]

this one is probably really nice, but on the face of it cancelling things looks like a pain because of the squares

[prisma loom]

Did you know 2014 A6 is also an FE

There are rumors 2024 A6 will also be an FE

[gusty verge]

No, ones that exploit the Mobius transform trick are the hardest

[high goblet]

huh interesting

did u do the IMO?

[prisma loom]

No I am still in high school

I will probably next year and year after that since my country is quite easy

[high goblet]

ah fair, gl!

[prisma loom]

Thank you

[high goblet]

i almost did IMO but didn't quite make the UK team, my years were fairly competitive

the UK does fairly well in competitions so it's on the harder end to get in

[prisma loom]

Yes that is true

It is like how China TST is more difficult than IMO

[gusty verge]

So f must be even or odd

[high goblet]

that's not necessarily true

there exists a non-even non-odd function that satisfies the FE

[radiant jasper]

Whaaaat

Orz

[pallid ginkgo]

f(x)=0 :3

[radiant jasper]

Welp I'm stupid

[pallid ginkgo]

oh wait nvm

[radiant jasper]

So people who go to IMO use China TST to practice?

[gusty verge]

Unless it's the sleep derprivation:
$$f(x^2+y^2+2f(xy))=(f(x+y))^2$$

Set $x=y$, you get:
$$f(2x^2+2f(x^2))=f(2x)^2$$,

Then set $x:=-x$, you also get:
$f(-2x)^2=f(2x^2+2f(x^2))$$

So $$f(2x)^2-f(-2x)^2=0 \Rightarrow f(2x)=-f(-2x) \text{ or } f(2x)=f(2x)$
$

[gilded halo]

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

[radiant jasper]

AINT NO WAY

I'VE HEARD OF THIS GUY

MY FRIEND WON'T STOP TALKING ABOUT HIM

[gusty verge]

Seriously why is this wrong? Or am I losing my mind

[high goblet]

it's the piecewise trap

you've shown that for every value of x, either f(2x) = -f(-2x) or f(2x) = f(-2x)

[gusty verge]

Ah

[high goblet]

but that doesn't exclude i.e. f(1) = f(-1), f(2) = -f(-2)

[feral canyon]

Helloings chat

[gusty verge]

Hello

[feral canyon]

Damn

IMO

I just bought the AoPS textbook, i really wanna do the AMC10 next year

So yeah i suck basically

[pallid ginkgo]

no wonder bro is so damn cracked

[high goblet]

lol ty

i mean... a lot of my friends actually did IMO

cus like the entirety of the ppl who made IMO/almost made IMO for the UK study maths at trinity cambridge lol

[feral canyon]

Where do you find your friends😭

[high goblet]

i mean i met them at the selection camps so that's a bit cheating

but like we all study at cambridge together now

[feral canyon]

The only place where i've met people interested in maths is that one summer camp at UChicago

Awesome experience tbh, sadly i doubt that i will get accepted into the actual university

[gusty verge]

No shame in that, it's a very good uni, but just hit the grind and see if u can improve ur odds

[feral canyon]

Probably what i'll be doing for the next 3 1/2 years of my life

[ornate coyote]

hmmm

imo qualifiers are such sweats

[acoustic nova]

Without a calculator find the leading digit of 9875^24

[feral canyon]

Uhhh

[mighty swallow]

Real

[high osprey]

Anyone wanna give this a go?

[jolly solstice]

partial fractions I guess... 😬

[high osprey]

I think so lol

[radiant jasper]

why is it 38C2? Not 40C2?

oh im stupid they're picking the pair again..

[dense lagoon]

counting gotta be harder than calculus man

[tranquil pilot]

source

[high goblet]

2004 ISL A6

[severe eagle]

Damn

IMO has always been my dream

[thin patio]

Terence tao thought 27 is prime

[hexed oak]

Any non bashy ideas? (this is a russia problem, so there is a non bashy soln)

[feral canyon]

Seems like smth with trig

[feral canyon]

We could make an equation for finding alpha and beta based on time by using trig and the height of the plane

[deft wraith]

grothendieck thought 57 was prime

[feral canyon]

Ha i’m better than Terence

[feral canyon]

Say that $a_0$ js the initial angle from point A and $a_1$ is the angle from point A after one second.
Then, $|a_0-a_1|= a$
for later use in trig, we can drop the absolute values since most trig functions are even
$a_0 - a_1 = a$

That’s what i came up with

[prisma loom]

Epic latex fail

[feral canyon]

Oops

[prisma loom]

(use backwards slashes)

[gilded halo]

Bible Camp Victim

[feral canyon]

Bruh i forgor i’m on my phone

[hexed oak]

hm

[vernal axle]

hint: prove that the minimum base in that trapezoid is when the angle between its diagonals is 90-(alpha+beta)/2. The rest is easy.

[hexed oak]

thanks

how did you get that idea?

[vernal axle]

geogebra 🙂

[hexed oak]

oh

locus?

[vernal axle]

first i fixed smaller base. Then the problem is equivalent to maximizing the larger base. Then we have a common chord for two circles with inscribed angles alpha and beta. And then I just used coordinates to find that the radiuses of these circles to points A and B should be parallel. That is a fairly short solution. tho it still requires finding the derivative. So, it's not purely geometric

[hexed oak]

hm

maybe i can get an equivalent geometric proof

[vernal axle]

maybe

[vernal axle]

in my solution the key point is the fact that if f(x)=sqrt(x^2+2ax+b), then f'(x)=(x+a)/f(x). So if you manage to interpret this geometrically then you are done. Or probably there is some other more pleasant shortcut.

[hexed oak]

hm

i got it

no one sec

[tulip orbit]

I'd say combi algorithms are def harder

[dense lagoon]

Yea

[acoustic nova]

Find, without a calculator, the leading digit of 9875^24

[deft wraith]

You can mess around with logs but that seems really cumbersome

[pallid dragon]

it's ~5 operations to do ^24

so maybe you can just do it, multiply the first digits and ignore the rest somehow

could it work?

taking first two digits didn't work

well it did i guess

you get 7744

and the right asnwer is 7

it works if you start with 99

and doesn't if you start with 98

very coincidentally right

[round jackal]

Anyone here good at geometry

[shadow spruce]

https://www.quantamagazine.org/videos/2024-biggest-breakthroughs-in-math/

[shadow spruce]

Geometry is Pure Mathematics

[slender monolith]

amc scores for those who qualified came out

[digital hedge]

Is this solution for imo 2010 p1 valid?

Did I pass pointwise problem with this solution?

[thin crypt]

question:find 2-6+12-20......-9702+9900 and my answer is 7352 is that correct

[high goblet]

your soln is fine, but i do have some comments about your write-up

• you should probably almost always start your write-up with "we claim the functions which work are [etc]"

• clearly plugging them in we see they work (or whatever)

if you don't highlight that you've checked ur functions work, u'll lose 1 mark

(just look at 2010 P1 results and see the number of ppl on 6/7)

starting off a write-up with like "we claim these work, clearly these work" is like a really nice way to guarantee you don't forget

[high goblet]

i'd probably put slightly more detail on why floor(f(y)) = 0 means f(x) = 0 for all x, it's not immediate

and also probably be a bit more explicit with why C has to be 1 <= C < 2

[ornate blade]

Hi

[high goblet]

hi!

[ornate blade]

Does Cam do anything special for Xmas

[high goblet]

cambridge term finished like ages ago lol

our term lengths are 8 weeks 8 weeks 7 weeks which is like :zoom:

but we have bridgemas which is on nov 25th lmao

anyway yeah there's like bridgemas/xmas related stuff near the end of term, i don't think i really did any of those tho

[ornate blade]

Olay

[digital hedge]

Yeah I wrote it hastily I forgot several things😅

[digital hedge]

Thanks 👍

[ornate blade]

LY is not that old

[high goblet]

nw!

[high goblet]

wait what was the original message lol?

[ornate blade]

Did you do the 2010 IMO or something

Ask them

[high goblet]

oh ahaha lmao

[ivory ember]

,w \sum^{99}_{n=1} (n^2+n)(-1)^(n-1)

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[sterile cloak]

Would solving math Olympiad problems make me a better problem solver (in general, not just in math)?

[jolly solstice]

could I guess

[plain jacinth]

Ofc

[plain jacinth]

Mostly in stem fields tho prob

[sterile cloak]

Yeah I’m a CS student and a decent programmer

[modest marten]

where do you go

if you dont mind me asking

[radiant jasper]

does anyone want some free MO handouts?

i have 2 handouts that have AIME problems

i think they also explain the answer

[marble grove]

yes

plz

send

@radiant jasper

plz

[radiant jasper]

I didn't make AIME but according to my sources, my score is high enough that people who would get a score similar to mine almost always make AIME the next year

So yes please

Maybe not 2025 AIME

But if I prepare from now I can cook on the 2026 AMC and AIME

[inner ermine]

what have you tried so far

i feel like a nice factoring will be possible with a bit of manipulation

maybe subtracting the equations

[zealous violet]

hello!
my score on the amc 10A was an 85.5, and I was wondering how I could improve that in time for next year?

any ideas?

[deft marsh]

Prove that there are infinitely many pairs of positive integers $(x, y, z)$ such that:

$$x^2 + y^2 + z^2 + xy + yz + zx = 6xyz$$

[gilded halo]

saintyzy

[deft marsh]

ive seen a solution using pells equation but i think theres a way easier solution

[radiant jasper]

They say if you do all AMC practice tests, all AIME problems in last 40 years(4x from what i've heard), all Intermediate level algebra, comb, number theory, and geometry problems on AoPS, and all ARML/Purple Comet problems, you're guaranteed 130+ score next year and maybe even USA(J)MO

I got 100.5 on AMC 10B so I missed AIME by 4.5 points

I'm planning to try it it, and if it works, then we'll know for sure

It seems like a tall task, but you'll end up becoming so orz it will seem like the same problem at some point

Then you know you have become orz

[radiant jasper]

Of course the solutions must be read and understood in detail, and you need to know every single one

But uh I mean I am an 8th grader and I haven't made AIME myself as of now so I might not know what I'm talking about 🤷‍♀️

[radiant jasper]

square both sides????

[zealous violet]

alr I’ll try that ty

[untold thunder]

Anyone got hints?

[radiant jasper]

What grade are you in just curious?

It's ok if you're not comfortable sharing

[radiant jasper]

Write out a few terms?

Like

rationalize the denominator

[untold thunder]

Did so

Couldn't find a pattern :/

[radiant jasper]

Interesting

[untold thunder]

I think it was like 1,2,5/2,29/10,941/290 or something

[radiant jasper]

$x_{n+1} + 2 = (sqrt(x_n) + 1)^2$

[untold thunder]

THESE {}

[radiant jasper]

Ah ty

[untold thunder]

oh yeah

[radiant jasper]

no wait

[gilded halo]

Hypercube

[sleek ravine]

XD

[radiant jasper]

dangit

im trash at latex

[sleek ravine]

naw 😭

if you expand it doesnt work

:l

[radiant jasper]

Yep

im trash :/

[sleek ravine]

not like we have any better ideas

[radiant jasper]

sqrt(x_n) + 1/sqrt(x_n)

[untold thunder]

My parents are gonna deconstruct my toilet if I don't solve this problem

[sleek ravine]

wait yall literally a cube and hypercube

[radiant jasper]

what was i actually thinking ☠️

[sleek ravine]

das crazy

[radiant jasper]

It's a coincidence

[untold thunder]

VAMOS

AHORA

[sleek ravine]

there are no coincidences

[radiant jasper]

Bruh

"Vicious Viper"

Skibidi sigma

[untold thunder]

Hey

[sleek ravine]

[untold thunder]

It's supposed to be an AMC level question

[radiant jasper]

The 25th one apparently

[untold thunder]

Unfortunately there's no answer provided

yes

[radiant jasper]

Did y'all make AIME

I missed it by 4.5 points on AMC 10B 😭

[untold thunder]

I wasn't even close

[sleek ravine]

what is it with yall and cubes

😭

[untold thunder]

Get that one outta here

chatgpt came up with an absolutely beautiful solution

Actually crazy

[sleek ravine]

what is it?

[sleek ravine]

[untold thunder]

This problem was by RSM

so maybe it was just guessing but I hope not

[sleek ravine]

well no thats the obvious way and the way you're obviously not supposed to do it

😭

[untold thunder]

Chatgpt using calc

[sleek ravine]

just compute every term until you get it above 8 and then add the digits

😭

[sleek ravine]

calculus or calculator

☠️

[untold thunder]

Calculus

What if you do it backward

[sleek ravine]

this aint involve calculus doe :l

[untold thunder]

$x_n>8$, $x_{n-1}+1/x_{n-1}>8$, so $x_{n-1}>4+\sqrt{15}$

[gilded halo]

Cube

[untold thunder]

How nasty would the recursion get

LAS PERSONAS QUE SABEN

RESUELVAN

AHORA

AHORA

[sleek ravine]

is it better this way tho

💀

[untold thunder]

$x_n>a, x_{n-1}+1/x_{n-1}>a, x_{n-1}>\frac{a+\sqrt{a^2-4}}{2}$

[gilded halo]

Cube

[untold thunder]

How hard

$b_{n}=\frac{b_{n-1}+\sqrt{b_{n-1}^2-4}}{2}$

[gilded halo]

Cube

[untold thunder]

Yikes

$b_{1}=8$

[gilded halo]

Cube

[ivory ember]

Hint: Consider $x_{n+1}^2$.

[gilded halo]

Civil Service Pigeon

[untold thunder]

I'll try it but I resolved the answer came about through estimation or calculus

[deft marsh]

How would that help ?

[deft marsh]

Same bro

[severe eagle]

Hi

[deft marsh]

Its my first year in math comps but j rlly enjoy studying for them

[severe eagle]

Been doing them since 8 yr old

[modest marten]

same

[severe eagle]

Singapore stuff

[deft marsh]

Nicee

[modest marten]

do you guys reccomend RSM for AIME. I want to quali for USAMO in 2026

[severe eagle]

What rsm

[modest marten]

russian school of math

I almost made AIME this year

[severe eagle]

Idk sorry

[modest marten]

so I really need something that can boost me

[severe eagle]

I'm not familiar with Russia stuff

Sry

[modest marten]

do you reccomend anything

else

[modest marten]

no its in the U.S

the name of the school is just russian school of math

[deft marsh]

My goal is to get a medal in the national olympiad since its my first year and frmn next year maybe compete in the jbmo

[modest marten]

nicee

rooting for you

im just tryna get AMO

USAMO

because I dont think I have time for it

def not a medal

but maybe a qualification

I would be happy with that

[deft marsh]

Ure a junior too right?

[modest marten]

ofc a medal would be amazing

[modest marten]

sophmore

[severe eagle]

I live in Singapore