competition-math

[soft vigil]

High key

I have like every single past year’s solutions

[wooden ice]

kangaroo 2024

[wooden ice]

oh really

[pallid dragon]

it's what gamba said

it's easier than what you seem to imply

well ok, there's maybe a trick with 8

[karmic rune]

I know. I wanted to give a way of how to go about solving similar problems

that 8 makes it easy here, but its not reliable to depend on such stuff everytime

[pallid dragon]

what do you mean 8 makes it easy

i meant that you can't tell that 13 is what's under 8, maybe 8 is upside down in pic 2, but it doesn't cause problems because we can also tell that the "wrong number" is 34, so 13 is right

[karmic rune]

8 clearly shows 34 and 13 are opposite faces, and similarly shows 22 and 5 are opposite to eachother

coz the 8 is oriented like that

If there were some symmetrical figures, it would make this tougher

Imagine the faces were painted instead of printed

in that case you gotta rely on information of adjacency

[pallid dragon]

that's not an 8 thing, it's like you're correctly saying the numbers make it easy but you call it "8 makes it easy"

[karmic rune]

yea, mb. Its not an 8 thing

[pallid dragon]

the same applies to 22 and the rest

[karmic rune]

I got baited into thinking that by this /j

[pallid dragon]

yeah a trick in the opposite direction, 8 is almost a painted side

[hazy reef]

math kang is like a logic comp ngl

[sand river]

yo does anyone want to work on some problems together?

[hushed charm]

I'd like to try

Where are these problems from?

[pallid tundra]

!status

[scarlet root]

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

[pallid tundra]

also this is not even comp math

#prealg-and-algebra

[coarse sigil]

Ohh mb

[ornate blade]

<@&268886789983436800>

[wooden ice]

yeah true

[cedar heart]

!status

[scarlet root]

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

[summer roost]

???

[fallow elm]

well we know that it's 13 on the left dice because we can see from the rightmost dice, and for the middle one it's 22 on the table from the left dice. we know that for the right dice, it's a 8 on the table. hence it's 43 (i think?)

[radiant jasper]

Why mods being pinged everywhere what's going onnnnn even i had to report something recently

[pallid tundra]

we ping mods when there's some particularly egregious shit going on

like someone posting obvious scam links

in which case mods immediately swoop in and ban the offending user once we ping them

[radiant jasper]

I have been seeing way too many of those recently

Every channel

[pallid tundra]

well yea the scammers spam their links in every channel they can get to

it never fools anyone

lmao

[snow elm]

What

[chrome parrot]

Competition math!

Find a formula for the series $0, 1, 4, 8, 16, 32, 64, ...$

[gilded halo]

Mega Angelo13

[pearl holly]

Looks like a geometric sequence

0 - 1: 1
1 - 4: 3
4 - 8: 4
8 - 16: 8
...

wait

Is this

$2^{n-1}$

[gilded halo]

Shawn

[pearl holly]

where the first index starts from 0 and so forth to 0,1,2,3,4,...

[summer roost]

No 2

Though

[chrome parrot]

And that's the thing! Does a formula for the series 0, 1, 4, 8, 16, ... exist or not? With every natural number being possible?

[ivory ember]

$(n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}$

[gilded halo]

Civil Service Pigeon

[ivory ember]

,w (n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}, n=1

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

,w (n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}, n=2

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

,w (n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}, n=3

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

,w (n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}, n=4

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ivory ember]

,w (n-1)^2-(n-2)^2 \cdot \frac{1+(-1)^n}{8}, n=5

[gilded halo]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[chrome parrot]

Oh my god

That's incredible

Civil Service Pigeon solved my question! Not only did he found a formula, he also found a different then mine! Congratulations!

Um, @ivory ember can I use your formula in my personal math book?

[ivory ember]

All imma say is “check out oeis”

[chrome parrot]

Yeah I found it out 3mins ago from a different server...

[kind pulsar]

T(n) = 2^(n-1) - max(0, 2-ceil(n/2))

T(1) = 2^(1-1) - max(0, 2-ceil(1/2))
= 2⁰ - max(0, 2-ceil(0.5))
= 1 - max(0, 2-1)
= 1 - max(0, 1)
= 1 - 1
= 0

T(2) = 2^(2-1) - max(0, 2-ceil(2/2))
= 2¹ - max(0, 2-ceil(1))
= 2 - max(0, 2-1)
= 2 - max(0, 1)
= 2 - 1
= 1

T(3) = 2^(3-1) - max(0, 2-ceil(3/2))
= 2² - max(0, 2-ceil(1.5))
= 4 - max(0, 2-2)
= 4 - max(0, 0)
= 4 - 0
= 0

[chrome parrot]

No formulas in formulas pls 🫡

Wait

That doesn't have to do with anything I said

[kind pulsar]

what formula?

[chrome parrot]

No no I misunderstood

[timid saddle]

s(infinity)=a/1-r

@gilded halo

[kind pulsar]

oh okay

so correct right?

[timid saddle]

is it not the fuormula for infinte sum of exponetial series

or sum of gp

[wicked thorn]

6th degree polynomial can obviously fit this but that's essentially trivial

[high goblet]

i mean $2^n - {1}_{n \leq 1}$ lol

[gilded halo]

LY

[chrome parrot]

What? No?

Oh wait

No no no it's correct

[high goblet]

i mean u can always find a formula for whatever sequence you want lol by doing smth dumb

[chrome parrot]

So you can really find a formula for the series 3, 7, 1, 9, 6, 283, or what?

[pallid dragon]

yes

it's like, mechanical

[chrome parrot]

Ok how tf

[wicked thorn]

bro doesnt know the secrets of the sums

[chrome parrot]

I do but it's crazy to see it

[pallid dragon]

e.g. 8, 19, 18
the difference is 11 and −1
the difference of that is −12
divide −12 by 2, means it starts with −6x²
that would start as 0, −6 so −6x²+17x
makes 0, 11, 10

−6x²+17x+8 is the formula

4 numbers would have an extra step with cube term etc.

i used this calculator https://polynomialregression.drque.net/online.php for yours

[urban valley]

<@&286206848099549185>

[ivory ember]

Reposted: #prealg-and-algebra message

[sand river]

is it just 19 + 18 + ... + 1?

then -1

ah no im getting something wrong

[sand river]

oh it's 18+18+17+16+...+1

so yeah it is 19+18+...+1 -1

this is because it starts with 1-18, then 20-37, then 40-56, ...

which is 18 + 18 + 17 + ...

[sand river]

thats why its -1 because there is no 0

[violet wasp]

is there a mu alpha theta discussion?

[icy pollen]

hi guys,
I want to participate in purple comet contest and I want to make a team. Is there anyone interested?

Do you still need?

[raven void]

No, unfortunately, my classmate gathered a team

[icy pollen]

np, good luck

[sullen blaze]

How to prove?

[ornate blade]

fill up the bottom layer with height 1 unit and you'll quickly see that you need 4 blocks that are standing vertically, with heights 4 units

alright, so using this method, you can definitely pack 36 bricks into a 6x6x4 box

then repeat the same process, repeating the bottom layer twice, but now there will be a hole you cannot fill

if you're clever, you can realise that 53 bricks is 1 short and so trying to put a 1x1x4 hole in the brick is impossible without creating a 1x1x2 piece

[gaunt mural]

Make a checker board pattern where each color fills the 2x2x2 box. Now every piece will hit the same number of black and white cells. WLOG there are 8 more white cells. So after 52 pieces there are only 8 white cells left, therefore we can't place the 53th piece anywhere.

[brave dirge]

yo i need help

[hushed charm]

You are in a high school competition math channel, please send your question in #calculus or anything more appropriate than competition math

[brave dirge]

mk i did

[gusty verge]

It is competition math tho 🤔

[chrome parrot]

Yeah

[sleek ravine]

(this must be a very old problem that occured in the year 1311 cuz all competition math uses the year it occured when it has a 4 digit number)

[deft wraith]

It’s fine

Like Putnam problems are fine here too

[high goblet]

yk on my 2022 TST this question came up

and i was like "what's the significance of 1829"

"surely this isn't some super old contest problem from 1829 is it"

(fyi 1829 is the year abel died)

[coarse shuttle]

i was wondering if there are MAA official solutions for AMC 12/AIME after 2007

like the contest problem book

idk i really liked those solutions

[zealous violet]

anyone got a good tutoring service for amc 10

preferably 1on1 but if its a really good group class then thats fine too

[pallid tundra]

not sure you’re gonna find anyone here

[pallid dragon]

you're supposed to say "a 700 year old problem pog"

[sleek ivy]

whoever invented it had a vision of integrals

bro was ahead of his time

[acoustic nova]

which solutions?

[soft vigil]

aops.com/community

[radiant jasper]

Can anyone assist me in my calculus work?

I'm learning about l’hôpital’s rule

I have no idea what I'm doing

And I'm so shit at derivatives

[sleek ravine]

#calculus , also you don't have to ask to post a problem 💀

[pallid tundra]

lol hi viper

[sleek ravine]

lol hi how are ya

💀

dang it

you ca'nt cross out skull emoji :l

[pallid tundra]

😔

suffering through algebra

(abstract algebra)

[sleek ravine]

dang bro, how far in? (I say like I would even know anything about the answer you give)

[pallid tundra]

we’ve been covering rings and fields

most recently we’ve been speedrunning introductory linalg again lmao

[sleek ravine]

is it like when they review something right before they do something that uses it? or are they just covering it more "rigorously" or with a new perspective or smthn 💀

[pallid tundra]

just speedrunning the gist of it

vector spaces, linear maps, etc

[sullen blaze]

how can you say wlog there are 8 more white cells left, when each piece covers both equal number of white and black cells

[half basalt]

The 6x6x6 cube is broken into 27 2x2x2 cubes. Then, the checkerboard pattern is applied on these 2x2x2 cubes (and not the 1x1x1 cells), so one of the colors will occupy 14 2x2x2 cubes, and the other 13 2x2x2 cubes. Since each 2x2x2 cube contains 8 individual 1x1x1 cells, whichever color (WLOG white) has 14 2x2x2 cubes will contain 8 more cells.

[late elm]

Initially, only one elf is in a good mood. Every day, a cheerful elf has a chance to make a grumpy friend smile (and thus become cheerful) with probability p (not equal to 0). The elves form a complete graph, meaning every elf is friends with every other elf.

A grumpy elf becomes cheerful if at least one cheerful friend smiles at them. Importantly, an elf in a good mood can smile at multiple grumpy elves in the same day.

We define τ as the total number of days required until all n elves are cheerful. Compute E(τ).

I found some weird sum at the end by considering the variable giving the nb of days to go from k-1 cheerful elves to k cheerful elves, but Python simulations give me another result.
For p=1/2 and n=10, simulation gives 2.07 and formula gives 9 and decimals...

[scarlet root]

Show your work, and if possible, explain where you are stuck.

[wooden ice]

hey what do u think this year's kang will have?
can u tell me the things that repeat that i should review please?
Anyone who sees this message, and thank you so much i have the competition tomorrow

how did i fnd 54

hm yea

[wooden ice]

yall

need some help

[pallid tundra]

!status

[scarlet root]

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

[wooden ice]

i mean its obvious

like explain

[pallid tundra]

what’s obvious

by symmetry all four trapezoids have the same area

(100-16)/4=21

the area of two of them is 42, which is 42% of the square’s total area (100)

[surreal willow]

I can’t read Spanish so no it’s not obvious lol

Oh actually I sorta understand mb

Actually I don’t understand bc I can’t apply the 10 and 4 :/

[hazy reef]

yeah i took it like a couple weeks ago

i cant rlly say

well i dont remember but

you should try to do some mock tests to see the level

then plan how much you will spend on each

and try not to spend too much time on each questions

but imo math kang is rlly ez

just do ur best its not life changing

[brave dirge]

Ok the only thing that i actually know how to do in this problem is understanding Sn
just a compliance way of saying:
S0 = 1 S1 = x_1+x_2+... +x_n S2 = x_1x_2 + x_1x_3 + ... + x_(n-1)x_n ... Sn = x_1x_2... x_n

idk how to solve the problen doe

[brave dirge]

wait is this actually n E N

[ivory ember]

Should be, but excluding n=1

[wooden ice]

its not spanish😭

[wooden ice]

i see thx

[wooden ice]

whatt how is that possible??

it should be secret

ive

[late elm]

blud I did

[late elm]

<@&286206848099549185>

[boreal shadow]

💀

[brave dirge]

im fairly certain that ||n = 0, 1, 2 have finite but idk about n >= 3||

[static summit]

eeeeeeeeeeeewa

f

wesg

erg

er

hrt

jhry

j

ytj

tyj

ty

kj

rd

@everyone

lol

@dawn gyro

[boreal shadow]

wtf was this

[surreal willow]

Wait then what language was it?

[summer roost]

Feemch

French

[late elm]

bud its french

[hazy reef]

Math Kang isn’t very formal

And isn’t considered as important as like amc 10 12 aime Olympiad’s etc

Just like amc 8 it doesn’t lead to anything

So score doesn’t rlly matter

Also did u do good?

[surreal willow]

Oh

That was my next guess lol

[late elm]

in spanish there is plenty of j and the n have accents like this ñ. In french we have a lot of agcents on the es

[ornate blade]

if you see el or los it's definitely not French

[ornate blade]

how can I dumb this down cause I know a fair few Romance languages
cause "de" absolutely exists in both Spanish and French, but only French has "de le -> du", and similarly only Italian and Portuguese have "de la -> da"

probably just look at all the words that don't end in a vowel
grand, not grande
deux (-eux is very French), not dos
sur and not sobre

instead of French pourcentage, you would have porcentaje
instead of French est, you would have es or está (estar)

instead of centre you would have centro

I would have thought that the feel of French and Spanish is very very obvious

certainly Italian can get confused for Spanish if you don't know a whole lot

[half arrow]

It doesn’t look Spanish or Italian

[ornate blade]

exactly

it's trying to explain how Romance languages work at a beginner beginner level that's the challenge

[ornate blade]

@surreal willow

[edgy vapor]

What's the harder between USAJMO and ELMO ?

[wooden ice]

ohh

[wooden ice]

i did okay, (but i hope i did good)
Did not answer everything but that can be good

cuz wrong answer -1

no answer 0

[half arrow]

I think it’s ELMO

[hazy reef]

ELMO

usajmo leads to MOP

and ELMO is in MOP

[soft vigil]

ELMO definitely

[edgy vapor]

Thanks

[tough patrol]

help

[ornate blade]

n^3 + 9 = n(n^2 + 13) - (13n - 9) for starters

then you must have |13n - 9| >= n^2 + 13

oh yeah and of course Euclid's algorithm for the gcd of numerator and denominator

[tough patrol]

didnt get you

pls explain

[tough patrol]

yeah next

[ornate blade]

numerator must be greater than denominator in absolute value

otherwise it'd be a fraction not an integer

[tough patrol]

yes

[ornate blade]

(or equal to)

[ornate blade]

you don't know what Euclid's algorithm is?

[tough patrol]

yes ik

[ornate blade]

Euclidean algorithm to find gcd

oh ok

[tough patrol]

it is dividing remainder till you get 0

a=bq+r

ohhh

so u were saying division lemma?

[ornate blade]

yes

if you do it right you should get that 13n - 9 must divide into 2278

[tough patrol]

ohk ill have to attempt it once more

[ornate blade]

ok nw

[primal glen]

Can someone help me with this question🥲

[polar patio]

notice how in $\triangle OFB,$ $\overline{OD}=\overline{DB}$ and $\overline{FD} \perp \overline{OB} \iff \triangle {OFB}$ is isoceles

[gilded halo]

parabolicinsanity

[polar patio]

honestly i'd go on by assigning a radius $r$ and then noticing that $\ \overline{OC}, \overline{OD} = \frac{r}{2}, \overline{OE}, \overline{OF} = r, \overline{FD}=\sqrt{\overline{OE}^2-\overline{OC}^2}$

same for $\triangle DFA$

[gilded halo]

parabolicinsanity

parabolicinsanity

[outer igloo]

how do i exactly master no.theory

[hushed charm]

Guys, what books do you like for studying geometry? I think introduction to geometry by aops wouldn't fit perfectly for me

[summer roost]

Did you try the ioqm paper ???

[half arrow]

Euclidean Geometry in Math olympiads by Evan Chen, A beautiful journey through Olympiad geometry by stefan would be two great books

With the latter having beginner as well as advanced concepts

Does anyone have book recommendations for graph theory for math competitions?

[hushed charm]

Interesting, going to check these latter, thanks a lot

What about combinatorics? I was going to read the introduction to combinations by aops, what do you think about this one?

[half arrow]

It does a good job explaining, but it is too basic for olympiads

If you feel like the book is around your level, then do have a read at it; for combinatorics it’s mostly about solving problems. I feel like doing the problems in chapter 1 of “principles and techniques in combinatorics” by Kim should be enough to lay a strong foundation

After this move on to doing problems

[hushed charm]

Sure, thank you

[hushed charm]

Probably that is why so many people recommend the art and craft of problem solving by paul zeitz and problem solving strategies by arthur engel

[half arrow]

Problem solving strategies is good if you know the content already — it is like a buffet of classical problems about the topic. It skips through the examples fairly quickly and its main focus is on problems

[hushed charm]

then i should know the content first. are you studying for math olympiads either?

[half arrow]

Yeah, I am. The content for combinatorics isn’t too much, the first chapter of “principles and techniques in combinatorics” should be enough to get you started.

After this, learn the various applications of “pigeonhole principle” and you can start doing the problems. Doing problems is very helpful because it gives you hands on experience, it builds your intuition which is very important.

[icy pollen]

I want to participate in purple comet contest and I want to make a team. Is there anyone interested?

[sleek helm]

I'd use thales theorem ...

[west mural]

ill try

tho im not the best mathematician

[west mural]

thats honestly one of the best

[pallid tundra]

art and craft is a bit too surface level for my taste

it was written only a few years after the original art of problem solving texts by rusczyk and lehoczky

before targeted contest prep books that went into lots of depth on more specialized topics were a thing

[shadow magnet]

My question is how can you come out or observe that the sequence a is slightly larger than n/2 and come out with that solution 🤔

Is there any trick to do so or it’s just observation and imagination?
Thank you

[pallid tundra]

pattern recognition

same reason you’ll sometimes see “by inspection” in solutions

[shadow magnet]

Take years to do so?

I think there will always be some logic behind these sorts of situations but I always can’t figure it out 🥲

[hushed charm]

I see, do you like any specific combinatorics and number theory books?

[icy pollen]

let's go

[half arrow]

Combinatorics — “problem solving methods in combinatorics”. Number theory — “modern Olympiad number theory”

[hushed charm]

Sure, just asking to know different opnions. I am aware that there isn't the best book, but it is nice to know different books and different arguments

[hushed charm]

Do you know 104 number theory problems? People say it is a good book for number theory either

[sacred edge]

I have heard of it, but I haven’t seen it yet

[west mural]

Introduction to Theory of Number by G.H.Hardy 😎

[formal trout]

$lcm(x,y) - gcd(x,y)=243$ Solve in $\mathbb{N}^{*}^{2}$

[gilded halo]

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

[pallid tundra]

!status

[scarlet root]

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

[pallid tundra]

don't expect us to do the whole question for you

[outer mural]

i have an idea
let d = gcd(x,y), x=dr, y=ds
then use lcm(x,y) = xy/gcd(x,y) and substitute

you get d(rs-1) = 3^5

then you can solve this but i think you get extraneous solutions

i think you have to keep in mind r,s are coprime in order to avoid them

[formal trout]

Even if you solve for r and s coprime you’d still get 20 couples

I was wondering if there’s any way to make it easier

[ivory ember]

me when there’s actually 20 solutions

[reef condor]

Idts, I don’t think there is any fast way
Well, if d=3^k then rs=3^5-k + 1, so then rs=3^i+1 for the same range of i. Now just partition prime factors of 3^i+1 since r,s coprime

[formal trout]

Oh nice

[urban valley]

common sense maybe?

[vapid delta]

Is x and y coprime?

Oh yeah it is

Wait but if d(ab-1)=243

So then d and ab-1 are integers

Can't you do all combinations of integers that multiply to 243

Then consider when X and y are coprime

[outer mural]

x and y aren't necessarily coprime, for example (6,123) is a solution

[radiant jasper]

I was preparing for this exam and came upto this problem can anyone help me

[jovial cypress]

alr lemme try solve

[radiant jasper]

👍 if u solve please explain in easy language my english is little bit poor

[jovial cypress]

it will be solved using telescoping identity

[pallid tundra]

product to sum

[fickle sigil]

Divide and multiply by sin(pi/6)

Take the numerator sin(pi/6) inside the summation ( as it's constant ) and rewrite it as sin ( { π/4 + [(k-1)π]/6 } - { π/4 + kπ/6 } )

Call the bracket of π/4 + [(k-1)π]/6 as A and π/4 + kπ/6 as B

=> you have Sin(A-B) in numerator

Which can be expanded via the identity sin(A-B) = SinACosB - CosASinB

When you you will write it this way and split the denominator as :-
[ (SinACosB)/ denominator ] - [(CosASinB)/denominator ]

And now you just have to notice that the denominator is SinASinB

[fickle sigil]

So on writing it this way, you'll be left with CotB - CotA

And now you can expand the summation and all the guy above said it is sort of a telescopic series

Sort of because usually telescopic series go upto infinity or n

But yeah..things will cancel out and you'll get a nice answer

[dire portal]

Here's a nice riddle:

Is there a set of sets of natural numbers A, such that for any two sets, one must be contained in the other, while satisfying the property that |A|=|R|?

[boreal sky]

Why this is the maximum?

[vague temple]

Find how many prime squares are of the form 2p^2-1.

[reef condor]

Pell equation

[mental pivot]

How would i start preparing for international math modelling competition???
I am in grade 11 rn need beginner book or courses to make my foundation strong

[pallid tundra]

stop spamming this question across multiple channels.

[hybrid grove]

hey yall, what are some good slightly-beyond-beginner practice questions for calculus (specifically integration)? I find a lot of them are either way beyond me (the ones from my local math competitions) or use targeted techniques (I've been practicing on khan academy)

like places where I could find them

[wintry phoenix]

Same im in gr 11 too and looking into math contests but i am BEYOND cooked school did not prepare me for ts 😭

[fallow elm]

honestly it's kinda too late for u to start rn

[mental pivot]

in my country its normal

[soft vigil]

What is math modelling

[deft marsh]

$$xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$$

[gilded halo]

saintyzy

[deft marsh]

find all functions f: IR -> IR

[ornate blade]

maths fashion show

[hexed oak]

R being the set of reals?

and contain meaning subset?

[narrow isle]

Ts pmo rn ong fr fr

[hexed oak]

first let us construct such a set which is maximal (as in any subset of N, B, either belongs to A, or there exists an elemnt of A, C, such that B is not a subset of C, and C is not a subset of B) (we can do so by axiom of choice)
the answer to this question is yes If and only if this set can have cardinality |R|
this means if X,Y belong to A, then X union Y belongs to A, and X intersection Y belong to A

hmmm

[hexed oak]

let us define X-Y, as the set of all elements in X not in Y unioned with the set of all elements in Y not in X

the answer is no

i just have to formalise my proof to make sure it is correct

[dire portal]

For now I haven't seen reason why no

[vague temple]

For q^2 - 2p^2 = -1 it gave 7 and 41, the next few don’t work, what do I do after? Are there no more solutions after, I don’t know how to proceed

[high goblet]

||lol the answer's yes||

||so ur proof doesn't work||

[high goblet]

where did you get this problem from?

[deft marsh]

someone help

fe

[dire portal]

A math teacher in my school gave it to me

And I thought about it

And I assumed the answer to be yes

Although it's counter intuitive

And then I found out the solution a couple of hours later

[pure mauve]

I mean, consider a set of even numbers, a set of multiples of 4, one of multiples of 8, etc...

In general let A_n be the set of multiples of 2^n

Let A be the set of those A_n-sets

@dire portal Can I ask if that was the approach you had?

[high goblet]

ah fair

it's just that this was on one of my first year problem sheets so i was curious

[dire portal]

This doesn't work

It's not even a set of sets

[pure mauve]

I mistyped

[high goblet]

it's countable

i can list them: multiples of 2, multiples of 4, multiples of 8, ...

you need an uncountable collection

[pure mauve]

Oh it says compared to |R|

I thought |Z| fuuuuu

[outer mural]

this is ridiculous

they want you to come up with this identity

[quick frost]

Ah, this is an enlightening puzzle. ||Pick an bijection f: ℕ → ℚ and for each a ∈ ℝ, let X_a = {n : f(n) ≤ a}. Then A = {X_a : a ∈ ℝ}, essentially because a < b ⇒ X_a < X_b. Basically X_a shows that you can "represent" real numbers by sets of rational/natural numbers "faithfully" (without "losing information").||

[dire portal]

Gg's

[hybrid grove]

bump

[frail root]

do you have an example?

[acoustic nova]

$\int \frac{\sin(x)}{\sin(x) + \cos(x)} ,dx$

[gilded halo]

hockeydude85

[hexed oak]

yeh, it was unable to formalise my proof

[reef condor]

Not that ridiculous

This idea is not new

[hybrid grove]

interesting

will do

that seems to be around the difficulty I can probably manage lol

[storm dove]

try this

[hybrid grove]

skul

[stark ridge]

<@&268886789983436800>

[opaque meadow]

Find all functions f: Z-> Z such that if $a+b+c=0$ then $f(a)+f(b)+f(c)=a^2+b^2+c^2$

[gilded halo]

Eric

[opaque meadow]

All I could deduce is that f(0)=0 and f(x)=x^2 + kx, k integer, is a solution

[near hill]

You should be able to derive some more consequences, e.g. a relation between f(a) and f(-a).

Or, if your hypothesis is that f(x)=x²+kx are the only solutions then set f(1)=1+k for some k, and see if you have enough to derive the values of f(-1), f(2), f(3), ...

A pattern should show up.

[outer mural]

i think those are the only solutions

using f(a+1)+f(-a)+f(-1), f(a)+f(-a), and f(1)+f(-1) you can find the recurrence f(a+1)=f(a)+2a+f(1)
then you can show by induction that f(a)=a^2+a(f(1)-1) holds for a >= 0,
and you can use f(a)+f(-a) to show it holds for negative a
so f must be one of the functions you described

[gilded halo]

Imkoma

For a polynomial $P(q)$ with real coefficients, find $12a$ where $a$ is the coefficient of $x^7$ and $P(n) = \sum_{k=1}^{\n} k^8$ where $n \in \mathbb{N}$ ($0 \notin \mathbb{N}$)?
```Compilation error:```! Undefined control sequence.
l.49 ...icient of $x^7$ and $P(n) = \sum_{k=1}^{\n
                                                  } k^8$ where $n \in \mathb...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```

[stable plank]

For a polynomial $P(x)$ with real coefficients, find $12a$ where $a$ is the coefficient of $x^7$ and $P(n) = \sum_{k=1}^{n} k^8$ where $n \in \mathbb{N}$ ($0 \notin \mathbb{N}$).

How do I solve this?

[gilded halo]

Imkoma

[pallid tundra]

P(n) can be computed directly for arbitrary n

[stable plank]

Yes, but how would that be done?

[pallid tundra]

I’m going to sleep so can’t fully explain rn but have a look at this wiki page and see if it helps

http://en.wikipedia.org/wiki/Faulhaber's_formula

[stable plank]

Thank you

I see how you could compute the problem with this! Thank you. But how do you even know these formulas? I never even heard of them before. Is this something like you're just supposed to know?

[outer igloo]

Hi guys its my first high school year and I have just entered 9th grade and want to give prmo/ioqm which is the first stage of math's Olympiad in India I am starting and wanted to seek help of someone more experienced than me in preparation for math's Olympiads pls dm if u can help

[balmy hound]

People
Can I know the important stuff for ioqm

Pls

I have around 140 days left

[pure mauve]

https://www.google.com/search?q=ioqm+contents

[balmy hound]

Tysm

[pure mauve]

I mean it was literally a google search

[dawn agate]

[pallid tundra]

they’re not covered in standard curricula besides the easy cases (p=1,2,3)

there are a couple ways to derive those formulas, one involves another similarly obscure result known as “Newton’s little formula”

[low fiber]

when doing comp practice problems what should i be analyzing and taking note of in the solutions for future reference

[pallid tundra]

https://blog.evanchen.cc/2017/03/06/on-reading-solutions/

certainly don't read line by line, just look for the key steps

[low fiber]

is it fine i have no idea what the example is talking about

[pallid tundra]

evan writes these mostly for advanced olympiad-level problems but the same advice is still applicable to entry level computational contests

look for the key step from which you can finish the solution on your own

[low fiber]

does this guy post his olympiad training materials

[pallid tundra]

he has this whole paid program that he runs

https://web.evanchen.cc/

he has some resources on this website

[low fiber]

oh its paid

im to broke for allat

do u know any free ones?

[pallid tundra]

he has a few excerpts available for free

but you can probably find lots of student written resources (of varying levels of quality) on AoPS

the main draw of that site is their huge collection of actual past contest problems

[radiant jasper]

say say must explain

[pure mauve]

Er

Isn't this live?

While this isn't a qualification exam, I'm not sure whether this would count as cheating per se

For what it's worth, this doesn't mean you're violating anything; this is more a heads-up as to why someone might be reluctant to do this for you

If you have any work done on these, and you want someone to check your working out, I think people here would be more likely to help in that regard

[pallid tundra]

You may use resources such as textbooks, library books, calculators or computers. However, except for your partner, you may not seek help from other people, including people whom you could contact via the internet. Your teacher may discuss your progress with you from time to time.

<@&268886789983436800> potential cheating

[versed dagger]

The challenge is live from 3 March to 27 June of this year. This is cheating plain and simple.

[pallid tundra]

rip bozo

lmao he just dmed me calling me slurs

truly upstanding maturity

[pure mauve]

Dear fuxk

I honestly wanted to deescalate that

[pure mauve]

But that's the equivalent of breaking the volume dial

[pallid tundra]

then they shouldn't have tried to cheat in the first place 🤷‍♂️

[narrow isle]

Our future upstanding members of society

He's becoming a lawyer

Chatgpt help me defend my client ass

[pallid tundra]

unironically

one time i was sitting in my dorm lobby

and a buncha sleazy ass business students

were rehearsing a sales pitch for

an aI lEgAL wRitiNg tOOl

🤮

these people have no shame.

[pure mauve]

You guys seen that clip of "If it may please the court" and it did NOT please the court?

i.e. that obvious AI lawyer introductory argument

I think this was like last week or sth

[pure mauve]

In unrelated news, that one of the mod's names is "Omnipotent Entity" is major nominative determinism 🤣

(assuming he didn't change his name to that after becoming a mod lol)

[edgy vapor]

I consider $x+y-1 ,\vert, x^2+y^2-1 \implies x+y-1 ,\vert, 2xy$ then
$$ x+y+1 ,\vert, 2xy $$ and
$$ x+y-1 ,\vert, 2xy $$
I consider $ gcd(x+y-1,x+y+1) = 1,2 $
which case 1 is failed
case 2, I got
$$ x+y-1 = 2k $$ and
$$ x+y+1 = 2k+2 $$
I have pairs $(x,y) = (1,2),(2,1)$ for solutions but I don’t know if it’s just fakesolved

[gilded halo]

Sea

[edgy vapor]

||BMO2 2011 round 2|| for those who want the source

this question too

I given p is prime such that $gcd(a_i,p) = 1$ for $i=1,2,\dots,m$ and given f(n) for " "by FLT I got
$$ f(n+p-1) \equiv f(n) ,(mod ,p)$$
then if $f(n)$ is divisible by some $p$...
but I don't know how to show here

[gilded halo]

Sea

[edgy vapor]

for stronger bond I have $f(p-1) \equiv f(k(p-1)) \equiv 1 (mod ,p)$
thus $$f(n+k(p-1)) \equiv f(n) (mod, p)$$

[gilded halo]

Sea

[edgy vapor]

Stuck to prove f(n) con 0 (mod p) for some p ...

[edgy vapor]

It's a mock from my national MO, there's no solution yet

[spare bobcat]

Me probably but idk how far I'll survive lol

IOQM maybe and ISI

[chrome parrot]

Here is a hard riddle:

Riddle 3.0: Pythagoras's Revenge

A Queen, mage and a goblin are trapped in the top room of a very very high tower.

Below the window, there is a pulley with a rope around it covered in a shield so it cannot be touched.

There is an equal weighted basket at each end of this rope. They managed to escape by using this and a weight they found in the room. As well as a weight on the ground. The rope cannot be handled.

They cannot remove weights without getting in the basket, but they can drop them in.
**The weight on the ground, is in a secure unmovable box, only opened by the mage.**🧙‍♂️🔒

When the weights are equal, the forces exactly cancel out, due to the frictional lock.

With an imbalance, the extra weight moves the system fully, causing one to go down and the other one up.

It would have been extremely dangerous and too fast if any of them were lowered down weighing 15 lbs more than the contents of the other basket.

In summary (all you need):

One basket coming down brings the other up. Baskets weigh less than 5lb.

When the weights are equal it arrests the motion of the baskets.

Weighing upwards of 15lb more than than the other basket is dangerous. +15 = fine.

Dropping weights onto the ground and not a basket will smash them and spill their water weight everywhere. No removing weights without being in the basket either.

The 15lb can must be obtained by the mage first.

The rope cannot be climbed, nor anything else except climbing out the basket into the window. The mechanism cannot be stopped by jamming.

The Queen weighs 195lb, Mage 165lb, Goblin 90 lb,

The objects weigh [room] 75lb and [ground] 15lb, 🧙‍♂️ 🔓 but the 165lb Mage must unlock them first.

How did they do it? What steps did they take to all get down from the tower?

I can tell you these things:
The 75lb object was put into the upper basket. For some reasons it doesn't explode. Then the goblin goes into the new upper basket. He gets to the ground, still in the basket, he could leave or stay. <- It's a hint.

It would be helpful if you are able to solve and give me the solution how you did in the next 2-4 hours

[young heron]

If a basket is on the ground, can you throw weights or people in it from the top of the tower?

mb, thats a stupid qs, obv no

[chrome parrot]

No

[young heron]

Here's a question: Prove [(5^100) + (5^75) + (5^50)+(5^25) + 1] is composite.

[ornate blade]

it's a geometric series

[young heron]

yea bro, ik that its equal to [(5^125 - 1)/(5^25 - 1]

its a pth cyclotomic polynomial problem

[young heron]

+1

u forgot +1

[chrome parrot]

Oh right sorry

[young heron]

and u can't whip out wolfram alpha during an olympiad

you have to prove it

not necessarily factorise it

[chrome parrot]

Here you go buddy (Ik this is absolutely useless)

[young heron]

thanks, but yea

[ornate blade]

,w 3597750 prime factors

[gilded halo]

Results provided by WolframAlpha

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

Requested by sooouth.

[ornate blade]

RIP

[young heron]

Here's another one: Find all primes p and q s.t. p^2 + 7pq + q^2 is the square of an integer

[chrome parrot]

Finite answer?

[untold thunder]

In a functional equation, why can't you always assume $f(x)$ is constant and solve for that constant?

[gilded halo]

ProjectTime

[untold thunder]

I haven't heard anything of it in my textbook

[young heron]

Bcoz, functions are not necessarily constant? duh

[young heron]

yea, obv p = q is a solution for all integers, but are there any for p not equal to q

[soft vigil]

because its often times not???

you need to find all solutions to a functional equation

[untold thunder]

It definitely almost always says this but often times doesn't list the constant solutions

[outer mural]

constant solutions ought to count, in my opinion

[outer mural]

got it! || (3,11) ||

i think that's the only other one

hmm there's a problem with my argument, i'm missing the p=q solutions

oh i see i missed some cases

ok yeah there's no other solutions

|| (n+p+q)(n-p-q)=5pq ||
|| n+p+q | 5pq, so n+p+q ∈ {5,5p,5q,pq,5pq} (assume n>= 0 wlog) ||
|| case 5pq: so solutions ||
|| case pq: (3,11), (5,5), (11,3) ||
|| cases 5q,5p give p=q ||
|| case 5: no solutions ||

[young heron]

yeah, nice

[soft vigil]

Sure it does

[edgy vapor]

Here

[stable plank]

Quick question: is there a general form/fast way I can compute the possible moduli of $x^n \pmod{a}?$

For instance, $n^2 \equiv 0,1 \pmod{4}$ or $n^3 \equiv -1,0,1 \pmod{9}.$ I can compute the values using some algebra, but it is time consuming. Any answers appreciated, thank you!

[gilded halo]

Imkoma

[spark notch]

S'ABCD=121cm2 and S'AGC=82.5cm2 , S'EFGH=441cm2, FIND S of ABE

[vernal axle]

S(AEC)=S(AGC) since GE||AC. And S(ABC)=S(ABCD)/2. So, S(EFGH) is not necessary if ABCD and EFGH are known to be squares.

[spark notch]

wait im slow

[spark notch]

can you explain S(AEC)=S(AGC)

[vernal axle]

they have common base AC and equal heights because EG||AC. Therefore have equal areas.

[spark notch]

wait youre right

bro solved this with a single braincell

this makes so much sense

thx

[spark notch]

i have 1 more thing to ask

[waxen axle]

the quickest way i can think of is to just square 2...n-2 mod n

[brave dirge]

diophantine is killing me

[calm nymph]

hello, is there any roadmap for learning mathematics?

[ornate blade]

which level are you at now?

what are you studying / what have you previously studied?

[calm nymph]

I just realized this is supposed to be a college guy server but I am at grade 10, have completed everything of grade 10 with a lot of time to focus on maths. I was looking fo r a roadmap (I lack guidance)

[ornate blade]

nope! we do all levels of maths on this server

you're talking in the competition maths channel

do you want something for comp maths then?

[calm nymph]

yes sir, it will be great. I am sorry for my misconceptions.

[ornate blade]

#1347232691376095232 message

here is some general information about Indian olympiads

I am not familiar with those, so I'll refer you to the American olympiad instead

https://artofproblemsolving.com/store - here are some books you can buy

https://artofproblemsolving.com/wiki/index.php/AMC_10_Problems_and_Solutions

but there are a lot of free resources too on AoPS

https://artofproblemsolving.com/wiki/index.php/AoPS_Wiki:Competition_ratings

there's also this which has examples of questions by difficulty, from basic level to IMO

Evan Chen's EGMO is really good for Olympiad geometry

[calm nymph]

thank you, it is really helpful. I will surely go through it.

[ornate blade]

#book-recommendations message

I just searched up "road map"

[ornate blade]

no worries!

[calm nymph]

still better than whatever I was trying to do for half an hour. Thanks.

[torn galleon]

Hello, how does hmmt registration thingy work?

[reef hollow]

Some1 help I’m supposed to make an answer sheet for a competition that my non-profit is hosting but idk how to do one of the questions.

Freddy exists in the coordinate plane and is currently at the origin Freddie can make hops from (x,y) to (x+1,y) or from (x,y) to (x,y+1). If Freddie can never go above the line x=0.5x. How many ways can Freddie reach to point (10,5). Note that Freddie can touch the line y=0.5x

[pallid tundra]

you really couldn't have googled it?

https://lmgtfy2.com/s/AinPZO

[dawn agate]

Guys

If i post a problem in AoPS

Someone gives me a solution?

[pallid tundra]

if you’re trying to cheat on a homework problem

don’t.

[novel karma]

does anyone have free online resources specifically targetting intermediate to hard logartihm and exponential problems?
i would greatly apreciate ti

that does not require calculus to solve

[reef hollow]

Ty

[pallid tundra]

AIME often has a log problem early on

[ornate blade]

https://www.madasmaths.com/archive/maths_booklets/standard_topics/various/exponentials_logarithms_exam_questions.pdf

MadAsMaths is goated

[edgy vapor]

do you have China MO resources? or other is fine

[lapis crest]

You can find past questions on AoPS

[ornate blade]

nope

[livid star]

Hey is there anyone here who ever partcipated in IMO??

Could u gave me tips how to be Excel at math where do i have start ???

[molten rune]

,w 1 prime factors

[gilded halo]

Results provided by WolframAlpha

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

Requested by honestguy3674

[molten rune]

o

[ornate blade]

1 isn't prime lol

actually WA never said that 1 was prime if you read closely

[molten rune]

ohhh thanks for pointing out

[brave dirge]

im not gonna lie, how does one cheat on a homework

i mean technically its not a test

[near hill]

You can still cheat yourself out of the learning experience it would be to try to figure out a solution yourself.

(There are also places where homework solutions are graded and contribute to the student's academic scores).

[pure fjord]

Try

[grand iron]

prepping for amc 10, which aops book (intro to counting and probability, geometry, algebra, or number theory) should I start with

[soft vigil]

you can get through all by the time it rolls around

i would start with algebra tho

[grand iron]

true there are like 9 months left

[grand iron]

also if you dont mind would you also recommend which one I should do after algebra

[pallid tundra]

once you've gotten a baseline level of theory that you need the best practice is just drilling lots of problems

[soft vigil]

idk man

its been like half a decade since i did those intro books

js learn the stuff

you should be able to finish all of them

and then volume 1 of aops is good

[grand iron]

Alright tysm for the help

[ornate flare]

Yo

[dry idol]

topic channels usually aren't very active but you can post some problems here and spark some discussion

[ornate flare]

k

[maiden panther]

as long as you're exploring new problems and topics and you're enjoying it

you'll start to see some improvement quickly :)

[outer mural]

i would have no idea how to do this, if it didn't basically just follow the proof of wolstenholme's theorem

[soft vigil]

common denominator?

that would be the first step

you know that all denominator in the LHS do not have p as divisor

if you try an do it

it becomes something like $((p-1)!)^p\cdot\sum_{i=1}^{p-1}(i^{-1})^{p}$

[gilded halo]

buboblakistoni

[soft vigil]

and then use something with mod inverse maybe

yeah wait

you just need $\sum_{i=1}^{p-1}(i^{-1})^{p}\equiv 0\pmod {p^3}$

[gilded halo]

buboblakistoni

[outer mural]

yes

[soft vigil]

however

[outer mural]

you have inverses since gcd(i,p)=1 like you said

[soft vigil]

note that each term has a different mod inverse

so you now just need $\sum_{i=1}^{p-1}i^p\equiv 0\pmod {p^3}$

[gilded halo]

buboblakistoni

[outer mural]

i think the mod would need to be p for that to work

[soft vigil]

but

dont you need p^3|m

[outer mural]

yeah

[soft vigil]

and you know that p>3

im thinking that you can generalise it

playing around with small numbers i get:

$\sum_{i=1}^{p-1}i^p\equiv 0\pmod {p^{p-1}}$

[gilded halo]

buboblakistoni

[outer mural]

let p=3, then (1/1)^3+(1/2)^3 = 1+14^3 = 18 (mod 27) but 1^3 + 2^3 = 9 (mod 27)

so you can't change the sum of i^-p to a sum of i^p (mod p^3)

you can do it (mod p) and that is part of the solution later

first you need to || pair ith term with (p-i)th term ||

[modest marten]

Hey, I am trying to collect people that are interersted in joining a study group for the AIME. I am looking for people who live in the bay area and can meet every weekend. I am currently working on problems 1-6 on the AIME, and am looking for someone who is on my level or above so that we can pace and push each other. respond to this, or dm me if you are interested-bay area preferrably a high schooler as well please

[stable plank]

Should I study/know Legendre's symbol? For context, I won't be taking a number theory course anytime soon but I want to know if I should learn it for math competitions.

And quadratic reciprocity

[pure mauve]

dym "the Legendre symbol"?

You could learn it yh

[pure mauve]

If you are going to, it goes hand in hand with this

I've seldom seen applications of these in a comp problem tho

[stable plank]

Thank you

[lapis crest]

Theres is some ISL N9 that gets nuked by Legendre symbols I think

[radiant jasper]

Hi sir does anybody have good dpps or sheets for maths advanced level

[summer roost]

Any specific topic???

[outer mural]

Is N9 a difficulty rating?

[radiant jasper]

mod means modulo?

[outer mural]

yeah

[edgy vapor]

you meant to shortlist?

[outer mural]

Is that what ISL is?

Someone mentioned it in this channel

[soft vigil]

imo shortlist

[soft vigil]

9th number theory problem in isl

[outer mural]

oh thanks

[vague temple]

Find max |z|=4 of |(75+117i)z+(96+144)/z|.

[high goblet]

it's not nuked by legendre symbols, there just happens to be a 1 line soln using legendre symbols

IMOSL 2022 N8 is still insanely difficult tho

in recent years they are trying to avoid problems that are insta nuked by advanced theory

[high goblet]

it could be somewhat useful

i think in general learning a bit of in depth NT can be quite useful for developing intuition etc. for working mod p

normally you don't really need Q.R. for oly problems but this depends on the olympiad

for example, balkan MO and RMM tend to use more theory than IMO

you can read about legendre/jacobi symbols in evan chen's otis excerpts https://web.evanchen.cc/textbooks/OTIS-Excerpts.pdf

[lapis crest]

I agree

[lapis crest]

Is a one line solution not a nuke though even if it’s not the intended solution

[lapis crest]

I find that USAMO and USTSTST/TST also can be quite theory heavy but yeah RMM is the main one

[high goblet]

oh right that's what u meant

i think in my mind if a question is nuked by X then there's an easy to spot fast solution that uses X

[lapis crest]

I see

That’s a fair definition too

[edgy vapor]

the question look so innocent but …

[ornate blade]

yeah that should be the aim for all competitions

also hi LY!

[pallid tundra]

2016 USAMO hook length formula controversy

[ornate blade]

when you try your best but you don't succeed

[brave dirge]

Prove that the green shaded area is the same as the orange shaded area.

[dapper magnet]

k=4 is an counterexample lmfao

[outer mural]

i don't understand hint 66
why should || (p-2)^(p-1) not be congruent to 1 mod p^2 ? ||

[acoustic nova]

isn’t this the pizza theorem

[cerulean burrow]

I need to know other opinion about this problem: Given a square ABCD and we choose three random points XYZ on the the sides of the square, M-masscenter of triangle XYZ. We need to find the area of geometric place of M(assume that side of the square equals 1). I think that the answer is this kinda cross, but im not sure.

[outer mural]

is there a requirement that x,y,z are on different sides of the square?

[cerulean burrow]

yes,sorry, forgot to mention

I thought this way and assumed that by the symmetry we got the cross, but i didn't check it actually

[outer mural]

ah i found an explanation https://math.stackexchange.com/questions/4426368/prove-that-there-exists-1-le-a-le-p-2-such-that-p2-not-mid-a1p-1-1-an?rq=1

dang this problem is insane

[outer mural]

it seems right to me

[vapid crescent]

Nothing to prove just look at it and you will know

[maiden panther]

pay attention to the definition of A. If a number x is not in A, that means x^(p-1) is congruent to 1 mod p^2.
the hint is saying that p-2 and p-4 cannot both simultaneously not be in A. that is, at least one of them must be in A.

i can explain how you finish the problem from these hints if you would like

[outer mural]

yeah, that matches with the stackexchange post i found

[maiden panther]

hint: whenever you are adding two complex numbers of fixed magnitudes, the maximum will occur when they have the same argument. (it helps to think about it geometrically - this is just the triangle inequality)

[outer mural]

it says let (p-2)^(p-1) = 1 (mod p^2) and then simplify and square this equation
then let (p-4)^(p-1) = 1 (mod p^2)

is that what you had in mind?

or if you have another approach i'd be interested to hear it too

[maiden panther]

yes, i think so

you get something along the lines of 4^(p-1) * (p+1) = 1 (mod p^2)

this whole part of the solution is essentially black magic

[outer mural]

it feels that way to me too

[maiden panther]

i can try to explain some motivation if you would like

[outer mural]

yeah sure

[maiden panther]

let's say a number a is good if a^(p-1) = 1 mod p^2

hint 1 tells us that a and p-a cannot both be good

or in other words, if a is good, then p-a is bad

[outer mural]

yes

[maiden panther]

the goal of the problem is to find 2 consecutive bad numbers

so a good thought process is to proceed by contradiction

assume that from 1 to p-1, no 2 consecutive numbers are good

if a,a+1 are 2 consecutive good numbers, then p-a,p-a-1 are 2 consecutive bad numbers

so now we know no 2 consecutive numbers are good and no 2 consecutive numbers are bad

and since 1 is good, this means numbers alternate good-bad-good-bad-... from 1 to p-1

at this point, the "main part" of the solution is completed

starting from a relatively week assumption, we deduced exactly which numbers must be good and bad

and all we need to do is force a contradiction somehow

[outer mural]

I have to eat dinner but I'll check this later, thanks

[maiden panther]

now, which of "good" and "bad" is a stronger condition?

"good" is much stronger, since it tells you something is equal to 1 rather than not equal

so at this point it makes sense to focus on the good numbers

one property you can observe is that good numbers are multiplicative. that is, if a and b are good, then ab is good

this isn't very useful for small numbers, since we have 1,3,5,7,... are all good and multiplying 2 odd numbers just yields another odd number

but for numbers close to p-1, multiplying and expanding can yield nice identities

for example, p-2 is good --> (p-2)^2 is good --> 4p - 4 is good

and eventually you can use such identities to force a contradiction

[maiden panther]

also, because of how strong the condition here is, there is a wide variety of ways to finish the problem

for example, one strategy might be to multiply together odd numbers < p to produce an odd number slightly over p^2, which turns into an even number upon reduction modulo p

let me know if you have any other questions!

[outer mural]

thanks

[vague temple]

Thanks, the answer was quite ugly so I was suspicious

[hexed widget]

did u get an answer yet?

[cerulean burrow]

As i said, i think that answer is 5/9 for square with 1 side lenght

[hexed widget]

so ur finding the centroid of a triangle

[cerulean burrow]

yes

[hexed widget]

so what do u need help with

[cerulean burrow]

I dont know am i doing right or not

It seems like okay but i noticed that thing only in geogebra and not properly sure

[stable plank]

THANK YOU

[warm trail]

Hello, I have proved an inequality using the "tangent line" method (basically it's a plane), but I am not sure if my approach is correct given the fact the function i am using is not convex. I described by problem here: https://math.stackexchange.com/questions/5063308/proving-a-three-variable-cyclical-inequality-using-the-tangent-plane-method-e . Anyone can give an opinion is my reasoning is sound.

[midnight palm]

I have a question, still working on it.

[pallid ginkgo]

go back to jail!

[twin raft]

Show your work

[lapis crest]

Twin primes are like 27 and 29 right?

[slow granite]

indeed

[slender saddle]

No, 27 isn’t a prime

[lapis crest]

It’s a Tao prime

Like how 57 is a Grothendieck prime

[twin raft]

twin prime conjecture is about actual primes

so like 29, 31

[lapis crest]

@high goblet they don’t understand 🙁

[summer roost]

It is a joke

When Tao said 27 was prime

I think

[soft vigil]

https://media.discordapp.net/attachments/929079030588788789/1095374475341598751/maximum-tension.gif

[lapis crest]

Indeed

[chrome parrot]

Well well well, who do we have here?

[chrome parrot]

I think we know us in a parallel universe

[twin raft]

Yeah... are you from the crank server?

[chrome parrot]

Huh? No?

I might have confused you with someone else

[twin raft]

i mean the neo-pythagoreanism one

[chrome parrot]

No wtf is that

I only know the pythagorean theorem

[twin raft]

It looks kinda like this

[chrome parrot]

I don't remember joining the server... Yeah that has its reasoning

[pallid tundra]

has its reasoning???

what the fuck is that supposed to mean 😭

[lapis crest]

😭

[polar patio]

do you guys hate irrationals then? 🤔

[twin raft]

Not me, but they do

[outer mural]

the given solution has this sentence that doesn't seem obvious

[hushed charm]

It worked, the answer is indeed 40, but the time it took, would probably not be possible if i was taking the olympiad.
How could I solve this one faster? ( I used chatgpt only to translate the problem, the translation is correct )

My method was to see, with each one of the seven possibilities when n = 3, how would the winner team change the other teams depending on its results

[hushed charm]

which book is this from?

[outer mural]

modern olympiad number theory

[hushed charm]

Do you like it? I am eventually going to need some better books in the future

[outer mural]

i'm not sure

the problems are hard

[pallid tundra]

isn't that supposed to be a good thing

if you don't do hard problems you're not pushing your abilities

[outer mural]

i wonder if i'd learn faster by doing problems of intermediate difficulty

actually the problems in MONT have a range of difficulties

some are easy

[maiden panther]

i think it's a pretty good book

i learned a lot of my number theory by working through it

[maiden panther]

yes, it's indeed not obvious

at this point, the problem has been reduced from a number theory problem to a combinatorics problem. specifically, a graph theory problem

suppose you construct a graph in which you connect two pairs whenever they're congruent

then the vertices will be grouped into connected components for each remainder, and each connected component will moreover be a clique (although we won't actually need this ladder fact!)

so we just need to show that in a graph with p vertices and at most (p-1)/2 edges, there are at least (p+1)/2 connected components

and this is now a general graph theory fact that has nothing to do with the original number theoretic phrasing of the problem

in general, a graph on n vertices with k edges has at least n-k connected components

why? start out with the empty graph, and add in each edge one at a time. the number of connected components decreases by at most 1 for each edge, done.

[outer mural]

nice!

[outer mural]

that's elegant

[hushed charm]

does anyone have any idea on why i wouldn't read the entire solution, as said by evan chen? Maybe I solved the problem using a worst method, and by reading the entire solution, i would understand a possibly better method, wouldn't i?

https://blog.evanchen.cc/2017/03/06/on-reading-solutions/

i know that he said that remembering only the core of the solution is easier than remembering everything, but as i said, i might learn more knowing a better method told by the solution than sticking with the method i proposed, that might not be the best, right?

[outer mural]

well i think if you notice the key idea, that might be enough to see if they used a similar approach to you, or if it's a better method.

but i also don't think reading the whole solution is necessarily bad

[high goblet]

he's saying you shouldn't read line by line

if you're reading line by line then either:

• you didn't try the problem, in which case ur not really gaining anything - the steps you read will be unmotivated and you won't understand where they're coming from since you didn't attempt the problem
  or
• the problem is way too hard for you and you can't pick out the key details

there's nothing wrong with reading solutions to a problem after you've done the problem

you might have a worse method and it might shed some light on what could've been a better solution

but crucially, ur doing problems to improve & get better next time

after you've given a hard problem a proper attempt, you might then read the solution and it might say "we prove that ABCD is cyclic" or "we claim that quantity X is non-increasing"

for someone experienced, that should be all they need to finish the question

you should then think about how you could've come up with those steps yourself

hard questions are normally never literally packed with all hard ideas, usually they only have a few key results that you need to prove to finish the problem

[high goblet]

have you read his article in full? i thought it gave a pretty good explanation of what he meant with some illustrative examples

[hushed charm]

Yes I did, indeed his explanation was pretty good

There is this article here that is pretty interesting either https://blog.evanchen.cc/2014/07/27/what-leads-to-success-at-math-contests/

[limber vault]

help pls

[vague temple]

There are several approaches for this, you could coordbash, or use properties of median in right triangle + Stewart’s theorem on AGC

Here is coordbash:
||
Let B=(0,0), A=(a,0), C=(0,c). Then len BG=len G-B=len G.
By definition of centroid, G=(A+B+C)/3, so G=(a/3,c/3)=(a,c)/3. So our goal is sqrt(a^2+c^2)/3.
Connect C to midpoint M_AB=(a/2,0) similarly B to M_AC=(a/2,c/2) A to M_BC=(0,c/2)
We are given 3=len GC, so 3^2=9=len^2 GC. But GC=C-G=(0,c)-(a/3,c/3)=(-a/3,2c/3), then:
9=(a^2+4c^2)/9
Repeat with 4=len GA, so 16=len^2 GA=len^2 (a-a/3,c/3)=len^2 (2a/3,c/3)
16=(4a^2+c^2)/9
But now we can get len G by adding the equations:
5(a^2+c^2)/9 = 25
(a^2+c^2)/3^2=5
len G = sqrt(5).
In general,
len BG = sqrt((CG^2+AG^2)/5)
||

Here is Stewart’s:

||
Let x=BG. Note that the median from B has length 3x/2 (centroid) so AC has length 3x (right triangle median to hypotenuse).
Thus by Stewart’s theorem on AGC, man+dad=bmb+cnc
(3x)[(3x/2)^2+(1/2x)^2]=(3x/2)(3^2+4^2)
x^2 (9/4+1/4)=25/2
x=sqrt(5).
||

[scenic sun]

You can also notice that || the distance from one of the vertices to the centroid is 2/3 of that of the median coming through the same vertex. Then you have the lengths of two medians and have to find the third one which comes of vertex B. Use the formula for length of a median in terms of sides of the triangle, and then add the squares of all three medians of the triangle. Then use properties of a right triangle which are Pythagoras theorem and the fact that say BG extends to meet AC at X, that X is the circumcenter of triangle ABC. Then you can find the length of the median coming from vertex B and finally multiply back by 2/3 || to find what BG is, which is || sqrt5 ||

[ashen jacinth]

Putnam 2015 A2 ^

In many solution writeups i see ||By induction, a_n = ... but how do they get this form for a_n, do they use the characteristic equation thing for linear recurrences or are they just like doing it by inspection lmfao||

If anyone cna help ^ IT would be appreciated. Just about one small part of the solution not the whole thing

[ornate blade]

||yes, it is the characteristic equation thing||

[ashen jacinth]

crazy they omit these things in the proof lol i must just be that noob

[near hill]

Hmm, but those characteristic roots don't even look very promising for prime factors ...

Hmm, in the first writeup I could google, the Binet-like closed-form solution to the recurrence is the least opaque step.

Ah, there was one with a better-explained reasoning.

[pearl holly]

hi

is there a faster way or a technique when it comes to calculating prime fact?

[near hill]

What does "calculating prime fact" mean there?

[pearl holly]

like uhhh
wait I cannot rephrase my question lol

Okay

Imagine you have a really large number, let's say a million. Find the prime factors of 1 million.

[near hill]

2^6·5^6?

[pearl holly]

What I want to know here is what's the fastest way to calculate for the prime factors of 1 million (and some numbers)

[near hill]

I think you're trying to ask for efficient algorithms to factor arbitrary large numbers.

[pearl holly]

Yes

[near hill]

There's no (non-quantum) algorithm that's known to complete in polynomial time.
But it's an area of intense research and lots of work on getting up to several hundred digits in a feasible time (with computers), not least because the assumption that factorization is hard is the key assumption behind RSA-derived cryptosystems.

For numbers in competition problems, trial division is usually the best feasible way. (And have the prime factorization of the current year and a few numbers on each side memorized...)

[hexed sandal]

I need help with math

umhh??

[near hill]

!help

[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.

[ancient yoke]

Has anyone tried an AoPS course, what was it like, do you think they're worth it?

[limber vault]

lol ty

i js used pythagorean theorem tho

cuz i was scared to use it

i have a bad history with 2 varialbe squares

[limber vault]

tysm

[ashen jacinth]

what do you mean by opaque fact just curious

[near hill]

Hard to understand.

The solution I found first just derived the clised formula and then immediately observed, "because a2015/a5 is both rational and an algebraic integer ..."

That it is an algebraic integer may be an easy consequence of some standard ANT fact, but not one I'm actively aware of...

Later I did find a more accessible walkthrough pointing out that if we use the binomial theorem on each (2015/5)th power, the terms with sqrt3 in them would all cancel out pairwise.

[ashen jacinth]

i thought it was because the quotient must be of the form a + b sqrt(3) but since it is the quotient of two integers, it must be rational, so b must be 0 and the quotient must be an integer (this was my impression after looking at the solution)

[near hill]

That seems to assume you already know the quotient has the form a + b sqrt3 with integer a.

Without further, the most I can see is that it is in Q[sqrt3].

(And of course in Q itself, so that in redundant)

[ashen jacinth]

did you see the solution here https://kskedlaya.org/putnam-archive/2015s.pdf

[near hill]

No, that was not among the writeups I found.

[ashen jacinth]

ohh ok

i think this makes it maybe more clear for you not sure

i was just wondering because i myself didn’t see anything with the binomial theorem

[near hill]

Yes, that makes perfect sense.

The one with the binomial theorem went, roughly:
$$ a_{2015} = \frac{(\alpha^5)^{403}+(\beta^5)^{403}}{2} $$
with $\alpha^5 = 362+k\sqrt3$ and $\beta^5 = 362-k\sqrt3$ conjugate in $\mathbb Z[\sqrt3]$. If we use the binomial theorem on each of the to 403th powers, each of the terms with an $\emph{odd}$ power of $k\sqrt3$ is canceled by the corresponding term on the other side, and all the terms that are left has at least one factor of 362.

The one you link to is more straightforward, I think.

[gilded halo]

Troposphere

[ashen jacinth]

ah i see

[near hill]

(I suppose the term-for-term cancellation is not even strictly necessary here, because we know the terms with odd powers of sqrt3 must cancel out taken all together because a2015 is known to be an integer, hence rational).

[swift imp]

Is Putnam worth preparing for

[pallid tundra]

most of the MIT kids who dominate the putnam each year don't actually seriously study for it

they coast on their prior olympiad training

unless you're gunning for a top score probably not worth imo

[ashen jacinth]

aye im gonna try ong

problem solving makes my brain feel better

[iron terrace]

i took like 3-4 aops courses in the past 1-2 years

they're good but you just really need to actually spend time to study the topics on your own

or else you're just wasting money

[pallid tundra]

!status

[scarlet root]

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

[limber vault]

2

[pallid tundra]

hmm i'd suggest trying to draw in some extra lines and see if you can get anywhere

i'm trying this problem out myself too

ok hint: gonna need some trig

[limber vault]

rahhh

oh nvm i got it

25

[pallid tundra]

yeah that

took a bit of trig bashing

surely there's a more elegant method

[limber vault]

oh so draw the altitude of aeh to ae

and u get a 45 45 90 tirangle

and call the point x

so hxe is scongruent to edf

hx = 4 and ax = 4

xe - 7-4 = 3

3^2+4^2 = 25

[pallid tundra]

oh smart

i went full trig bash mode 😭 found cos GHC=sin AHE and then law of sines'd it

[ancient yoke]

Which ones, how were they?

[tribal owl]

I used to really good at math used to solve questions in my head when i was little now i can't even count
Really good to see people being competitive in math

[pallid tundra]

lowkey I forgot MATHCOUNTS is sponsored by RTX now after Raytheon dropped their sponsorship

(the national countdown is going on rn)

[minor frigate]

Raytheon is the subsidiary I think

It’s either that or they changed their name in 2023

[iron terrace]

c & p, number theory, pre alg B, and mathcounts/amc 8

they were good

[chrome parrot]

Which numbers have this fact?
Take the sum of the number. (x)
Take a divisor of that number. (y)
Do this: y - x. The result has to be a divisor of the number. Let's call them Wood Numbers!

[ashen jacinth]

what is the sum of a single number

it’s just that number?

[near hill]

Yes.

Letting it mean anything else would be confusing in the extreme.

[daring heart]

Anyone good at ap physics 1 that can help me study or give me tips

[pallid tundra]

no.

wrong channel

wrong server

and

!da2a

[scarlet root]

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

[daring heart]

Good math = good physics 😔 or so I thought

Physics is so humbling

[pallid tundra]

you might get better results asking in the physics server

invite link can be found in the #old-network channel

[sleek ivy]

true but nobody can help you unless you ask smthn

[radiant jasper]

anyone giving IMO?

[cerulean burrow]

Help me please to solve the problem, A* is reflection of A against IO, we need to proof that YDA* are collinear, i tried to do smth, but only figured out some inscribed quads and i tried to prove that D lie on radical axis YA* ,but stuck, also, i figured out that D is radical center of (ABC), (XYA*) and (BIC) circles

[snow elm]

Help

[ornate blade]

well all these points are at the same height: where FP cuts AG, where PE cuts GD, and where PH cuts GC

yeah it's super hard to find the exact boundaries of the region: this is my best guess

[snow elm]

I just stuck on finding the intercepted volume

[ornate blade]

I mean that's the entire question

you have to break it into smaller parts somehow

[snow elm]

Ik but the problem is how?

[pallid tundra]

similarity

[ornate blade]

ah consider the face BCHG: then you know that PH must cut GC

because B, C, H, G, P lie on the same face

similarly, B, D, E, G, P share the same face, so PE cuts GD

[snow elm]

I don’t think it apply to this question

[ornate blade]

it does!!

like the question is of course how exactly you use it

[snow elm]

We can’t prove that they are similar

[snow elm]

Then which part should we use

[pallid tundra]

yes we can

[ornate blade]

also another helpful thing is that y = 3 - x intersects y = x at (1.5, 1.5)

[pallid tundra]

(parallel lines/planes can make a lot of similar triangles appear)

[ornate blade]

yes then you can show that these three intersection points do in fact form a square

consider face by face as I said