#competition-math
1 messages · Page 16 of 1
Scratch paper exists for a reason lil bro
safhnaks;jdfghaopsihfalsidghl;ksafhpoadhnsgipsdufh'ansdfpik;asdhfo;asjdgf;/jsado;ghf;a
hehehehaw
whats crazy is i got 139.5 on B and 133.5 on A but got a B on my geometry test (basic school geo)
Bro I cannot relate cuz I sleep through all my math and science classes
i straight up dissing the teachers
i hate chinese characters ngl they all look the same (the chars)
erm actually
oml my teacher just throws packets at us
I sleep even though people snitch on me
i have a b+ somehow in my class
cuz me half dead and starving is still better than 99% of my school
the tests are harder than amc 12 ngl
it just doesnt make sense
im in precalc + ab mashup
my algebra teacher used to just throw me on desmos classroom and not teach anything
so i just like spent 30 out of 50 minutes of the class doing my own math homework
Bro the teachers let me sleep cuz they have no right to stop me; I'm literally better than them at their job 💀
oh the teachers at ur school not at uni right?
in my coding class some kid just took a few spinny chairs and started sleeping on them
even took his crocs off
valid
and I don't wear crocs
💀
blasting side angle side congruency
music
and walks into random normal LA classes
i like soh-cah-toa
do you have one of those happy-go-lucky math teachers?
lwk I kinda want to see that, I need a good laugh
typical coding class
Yes bro
Side Angle Side
Side Side Side
Angle Side Angle
Angle Angle Side
That's an easy way to memorize the reasons of congruent triangles!
cancer
LITERAL CANCER
WOMP WOMP MOMENTS
they even performed in front of the whole school
at assembly
watch the end of the video
they are doing it at the jazz concert
in front of the whole schooll
and my teacher is making us dance up there
what kind of school............
a weird school
Bro the weirdest thing my school has is electing an orchestral val
which is just stupid
there has been weirder
and then making said val play a 40-minute concerto in front of the entire school district with a youth orchestra 💀
My embouchure hasn't healed yet, and it's been a year 💀 RIP
And I wasn't even the guy playing the concerto
i was just there for support
what happeneed to your embouchoure bruh
tbh i should just study throughout the school
how does that even work
40 free minutes during my coding class
I couldn't bubble my mouth for a day and a half afterwards
💀
Bro and the guy was a trumpet player
i was forced to do either orchestra or band
I play bassoon; carrying that shit is literally an entire arm workout
Wait
let me show you what my section each had to play for a contest
https://www.youtube.com/watch?v=PYOPQuhdoQM (skip to 1 mintue marker)
Live concert in the wonderful Teatro Olimpico of Andrea Palladio in Vicenza, Italy
Aligi Voltan: bassoon
G.B.Rigon: Conductor
Orchestra del Teatro Olimpico di Vicenza
June 2004
I almost quit right there
yall play mozart?
my class just plays star wars and stuff
for the 5th graders apparently
Bro the arpeggios and fast runs killed my tongue
No, my youth orchestra
I feel choir can be more chill than orchestra though
why does that guys hair look like an M
💀
Hey don't disrespect my goat Voltan
i get sick way too often for choir
and also the choir sounds interesting...
too few voices
💀
im gonna go do some woot
you'll get there eventually
what do. you do in woot ?
Olympiad prep
i wouldnt say woot 1 is too hard
not for u 
its like aime level mostly
Yea it is
also why tf am i watching a video about group theory
im tweaking i havent even done calclus yet
once i looked at my brothers woot 2
you don't need calc for gt
yah but i should prolly just study calc so i can skip
and i also i keep failing my precalc tests so 🤷
Yea you should do that
aops
bruh for calc?
is it good tho?
idk
💀
i havent done a lot of calculus
ive heard of stewart and apostol
ye
What’s woot
Apostol is good but lengthy
If you're willing to really grind through it, then definitely do apostol
ok
huh
Is group theory useful for AIME? Should one learn it, and if so, where to start?
aops?
yeah k330-332
+1 just read over analysis
Does anyone know how to bulk download past contests from AoPS
Do u know the formula sinxsin60+xsin60-x
Yes use that and the multiplication formula for sin
alright
80?
sin (a+b)* Sin (a-b)
can also be used?
Anyone needs LaTex OCR API? I just purchased one, I can send you API Key if you want.
use double angle identity you get 80
Or you can solve for \alpha and subsitute
$\sin^210^\circ\times\left(\frac12\sin40^\circ\right)\times\left(\frac12\sin80^\circ\right)=\frac14\sin^210sin40sin80$
?
Aestusy
that identity is used to derive sin x sin (60-x) sin (60+x) = sin 3x /4
so yea same thing
Have you solved it yet?
does anyone know if reading napkin by evan chen will help w comp math
specifically pre-oly (like aime)
uh no lol
very very very maybe for olympiad maths
definitely not for pre-oly lmao
napkin by evan chen is meant as an exposition to university mathematics
fair enough
unless the maths competition syllabus has dramatically changed, knowing topology or galois theory is not gonna help on AIME lol
i heard group theory could help for the qs in 14-15
i'm not american so i've never done AIME
potentially some group theory could be helpful for some later qs
ah ic
but you almost certainly won't need anything more complicated than that
(sometimes some harder olympiad questions are immediately killed by advanced theory, i.e. 2003 IMO P6, but like those are quite rare)
(and will often involve lots and lots of theory)
interesting
basically read napkin if ur interested in uni mathematics, but don't read it specifically for maths competitions
ill take a look at 2003 p6
thanks for the response :)
r there any texts that helped u
a lot
nw!
ok so i'd like to preface by saying that i'm british
personally i didn't use that many theory books, i mostly did problems
i grinded all of the british maths olympiads R1 from like 1995 onwards and then also did a lot of BMO2
damnn i see
then i started grinding some IMO problems
to grind these problems
dont u need a base lvl of theory
or did u js pick it up
as u solved
there's like a few books that contain like a collection of just BMO1/2 problems & their solutions that i worked through
it's like "a mathematical olympiad primer" or smth like that
again, not american but like certainly for the UK
you don't need theory for the very start
u'll have to learn more theory as u get into like harder olympiads
but it's best to just start doing problems
sometimes there'll be a question that requires a tiny amount of i.e. modular arithmetic theory
but often ur meant to just pick it up as u solve it if u don't know the theory already
cus also if u do that it'll stick with u a lot better
thanks for the good advice :)
nw!
I have major ops on BMO 1 😂 Its next week
Sorry, I was busy doing smth else, stilll wiorking on it
Does anyone know where I can find combinatorics problems/handouts for AMC 12/AIME
I mean, just AoPS has like an absurd amount of problems, and youcould just search up like combo problems
Can someone explain why the red part is wrong
I know it causes arrangements of object as well but how
I think it was scrambled in the us too - I took the 12 and my question 11 was question 13, question 22 was question 20, etc.
my answers were scrambled too
I think they're just scrambling it for everyone to try to fight cheating ngl
does anyone have any recommendations for AoPS-style books that go beyond the AoPS curriculum?
yeah i saw some guy selling answer keys before amc10b in this server
to fight the people who just memorize the answers
does anyone here have any advice for the amc8
Don't stress (too much)
practice to fast
Only online I think, my physical copy for 12B has the exact same order as AoPS
Well tbf the physical copy is the “unscrambled/reference” version
Everything else is compared to that
Because the physical paper copy is what they release
i think its scrambled in china
because they take it earlier
and some people end up cheating
How do u apply for math competitions in middle school? I’m trying to get into some competitions because it sounds like it would really make my application to high schools sound good and I wanna be ahead the competition
I’m in the USA btw and im studying algebra 1
You don't need math comps to apply for high school; you don't apply for public high schools
maybe try local hs
Uh
some places do
Well, if your school supports it, you could try MATHCOUNTS or the AMC series
Normally, they'll tell you if they have it
check local university of email high schools near you
Some middle schools also give hs competitions
these are like goated ms tho
Like mine?
like top ms only really do that
Mine has absolutely no competitions am I cooked💔
Just email ur high school you'll be fine
Wym email ur highschool
like any highschool
I see those kids on a regular basis 😭
Or the one ur going to
any hs works but like start with the one that ur zoned to
email the school district highschool(s)
oh
probly not
brooo im so cooked for my highschool applications
What high school are you applying to?
my only thing going for me is me being in the gnt and having good grades
whats gnt?
Coltsneck highschool in New Jersey
bros trying to get into exeter 💀
Gifted nd talented
or somethign
What’s that
those who know
phonk plays
@tame burrow usmao awardees on a map
Nah... Exeter's overrated
💀 I can't do math so I get grilled on a regular basis; al I can do is physics
But seriously I’m stressing bc I’m in no clubs, my only thing going for me is straight A’s and taking advanced classes (gifted n talented)
am I cooked?
No competitions either
😞
Nah

if they're all public, why'd u have to apply?
I’m applying for like 3
You don't usually apply for a public high school
Oh then that's different
mhm
but u can’t just say “hi, I wanna join” in a top school
I’m also trying to apply for high technology high school which is top 10 highschool
Oh is it?
mhm
Then that's a totally different monster
coltsneck is like my happy school
like if I don’t make it huge then I can settle at coltsneck and Il be happy wit what I did
When you make Andover but can't pay the tuition
my parents make 200k+
I’m fine
I also got 2 dollars in the bank so I can pay for it 🤑
👍

💀
rich ass kid
💀
💀 not the ☠️ emoji
heheheha
1434
ok
WHAT'S andover
where did you find this?
Made it
I thought you were about to pull a Ramanujan
pull a what
1729 is the natural number following 1728 and preceding 1730. It is the first nontrivial taxicab number, expressed as the sum of two cubic numbers in two different ways. It is also known as the Ramanujan number or Hardy–Ramanujan number, named after G. H. Hardy and Srinivasa Ramanujan.
oh lol
Bro
I woke up at 5 am
I knew I could have done better
which is the most painful part
if I capped out at 99 I would be fine
What is the maximum score?
But the fact that I could have done better is sad
150
Have u taken amc10 before
how do you expect 102 if you already know how many questions you didnt answer
i thuoght i guessed them
i think that 0-0=0
hm'
Are u JMO
that probably explains why u think 102 is bad
😂
yes
no
Well well well
but im saying
Another jmo
3 point difference isnt much
there are too many
Oh
i got 12 point difference on 10a
Nah look
some people got like 30 point
if u got a 103.5 last year on the 10b
THEN a 3 point difference would have
donnnnnnnneeeeea lot
I have no clue
What does the colour of pointer mean
how do you do sigma notation
Idk
do you have a specific question
This calculus video tutorial provides a basic introduction into summation formulas and sigma notation. It explains how to find the sum using summation formulas for constants, i, i^2, and i^3.
Antiderivatives: https://www.youtube.com/watch?v=xaCPDMEkbig
Basic Integration Problems:
https://ww...
or are you happy with watching this org chem tutor video?
i learned it on tiktok but thx
RIP TikTok, but appreciate the thanks
ye
mathematicas puras
do a and b have to be integers?
or can they be real numbers
either way note that you must have $a^2 + b^2 = 13^2$ which is a circle
south, just south
ah it doesn't matter what a, b are then
cool
but how so
hopefully you see how this happened
||sub in a = x and b = y||
oh accha
||cause (x, y) = (a, b)||
exactly
once you come up with the equation yes the maximum distance has to be the diameter
Yeah
26 is right
For chicken McNugget problems of 3 and more container sizes does applying pairwise always work?
This person did it and it worked (luckuso in solution #4, on 2023 AMC 12B #16)
Say 6, 10, 15. So 6a + 10b + 15c.
First apply to 6 and 10, reduce to 3 and 5, so you can represent all combinations beyond (3-1)(5-1) + t for nonnegative integer t = 8 + t.
Then substitute into the original expression:
2(8 + t) + 15 c = 16 + 2t + 15c
Apply pairwise to 2t + 15c: you can form any (2-1)(15-1) + u = 14 + u
So you can form 16 + 14 + u hence 30 and above hence 29 is max
Or are there additional restrictions in some cases
dont memorize those though
chicken
math. is hard
chicken chicken chicken
Exactly
I like the video cause it tells you that working with summations is literally just working with addition
Good examples
wtf
there was a chicken nugget problem in amc 12?
The best strategy is to not assume
Check that 30, 31, 32, 33, 34, 35 are all possible
Then check 29 is impossible
Bro I can’t escape combinatorics, I went to watch a chess video and got hit with “how many possible configurations exist that allow a doubly disambiguated bishop move (i.e. 2 bishops lie within the same row and 2 bishops like within the same column and all of them can move to the same square, so they form an isosceles right triangle like in the picture)
Reasoning: consider the symmetries. First, light vs dark so x2. Now we see each square generates 4 choose 3 = 4 rotations of isosceles right triangles. Each square can only work if we have the same parity of rows/columns. Then, odd/even squares so x2. Looking at odd/even only tiles, of which we choose 2 hence 4 choose 2 squares. Hence
2*2*4*(4 choose 2)=96
Knights and queens
Actually if I am not mistake, knights is like
2*4*6*4 = 192
i dont think you should take making aime too seriously if you are hovering around the cutoff
you shouhld just focus on improving your own skills
Idgaf about aime
I mean doing well
I jsut wanna qual
And i have lots of time to improve anyway
Its j ust cause i woke up early so i did badly
yes but getting past a cutoff that doesnt really matter isnt something to stress about
you cant really do anything to make the situation better
just focus on improving yourself
not comparing yourself to others
@swift imp @radiant jasper @ornate coyote thx for the tips
bro math comp is easy
as long as you are smart and not like a neighborhood kid its easy to make comps
like just study around 3 hours a day for around 2 years
easiest 150 of your life
bro who put that
im not joking bruh
alr south
think about it
how many topics does a normal ap calc bc class cover?
around 10 or so?
you already have basics in algebra, geo, and number theory
if you take 1 hour a day with each
you can basically cover everything
What tips did i give you??
with time to spare for practice
You're proving yourself that it's hard
I dont remember lol
uhm practice tests or sum
bro
U got 150?
I’d say it’s practically impossible to get a 150 no matter how hard you practice
my friend did question 21 or sum from lat year in like 1 minute
and he studys only like 1 hour a week
Let alone have that much energy to spare for contest maths
bro im not gonna lie
There is a really wide ability range for maths yes
if you know how to do contest math
actual math tests r gonna be easy
plus
people actually studying for 150 probably dont struggle in their math class
ok gn gang
ill gangalank you later
Yeah it depends on what it is
I solved many aime questions under 2 minutes
No matter what there is always a 95 percent chance you will silly a problem or lose on time
Perfect scores are nearly impossiblr on difficult amcs
This and last year were exceptions
i don't think there's a clever trick
c)
the difference from 1 2 3... is 11 19 10 5 14 11 8 10 13 9 6 7 2 6 9 1 18 7 6 9
sort it 1 2 5 6 6 6 7 7 8 9 9 9 10 10 11 11 13 14 18 19
we can save 20 operations by targeting 2 3 4
the smallest diff is 1 again but now it saves 18 operations to increase the target by 1
then the smallest diff is 3, and it saves 3×16 operations
etc.
it looks clever once i wrote it lol
oh i misunderstood the problem
i thought an operation adds 1 stone, it adds a whole bunch
pretty neat actually
nah, study for 8 hours a day. dont sleep. you gotta put the hours in.

Wdym by targeting 2 3 4
?
Yeah looks like a good method
Anyone knows Mathematics SFTP server?
what's SFTP?
Ok so first thing is to convert the problem into a well-known and solved one of minimizing the MAD in a uniform distribution. So compute a new adjusted array A[i] = H[i] - (i - 1) (subtract 0-index). Now we end up with a new array A, for a it would be [2, 3, 4].
Our goal is to minimize the MAD between A and a target uniform distribution U of all values u, and this is done in the unconstrained case by setting u to the median of A. However, we have the additional constraint H[1] >= 1 meaning A[1] = H[1] - 0 >= 1.
But this implies that we cannot blindly pick the median in A to serve as the baseline but rather we require that u >= 1 as well. Hence, u must be chosen to be as either the median or if this median is nonpositive, then u must be as close as possible to the median, hence 1.
Example solution for (a):
H = [2,4,6]
A = [2,3,4] (already sorted)
u = 3
U = [3,3,3]
T = [3,4,5]
Calculating the MAD between the target distribution T and the original distribution H (equivalently, between A and U), we get |2-3|+|3-3|+|4-3| = 2.
(b):
A = [1, 1, 1, -2, -3, 0, -5, -4, -3, -6, 1, 0]
Sort A: [-6, -5, -4, -3, -3, -2, 0, 0, 1, 1, 1, 1]
In this case, we see the median of -1 is < 1, hence doesn’t work. We choose:
u = 1
Using Python brute force search, the answers are found as:
(a) 2
(b) 31
(c) 67
Hopefully this should match the manual way.
this is a quantative aptitude question please help me with the solution and explain
500
Distance is constant
Use d = vt
The numbers are somewhat unrealistic, I don't think anyone walks that slow compared to an escalator
Let
d be length of escalator (distance travelled, constant)
p be speed of person
e be speed of escalator
t be time to cover escalator in normal circumstances (walking and escalator)
We have the equations
Normal:
d = (p+e)t
Also, he walks 50 steps in normal circumstances.
pt = 50
Failure walk for 10s, then remaining is with escalator:
d = 10p + (t+9-10)(p+e) = 10p - (p+e) + t(p+e) = 9p - e + d
Hence
9p - e = 0 so e = 9p (unrealistic part but OK)
Thus, substitute back in:
d = (p+e)t = (p+9p)t = 10pt = 10(50) = 500 steps
Thank you sir
brother i had doubt in question number 6 is that question correct or not i am not aware but our sir told that option (a) is the correct answer
,rccw
he is right
the correct answer of question 6 is a
If you look at the graph on Desmos, you will see it works for a <= -d
wait can u do it in school?
idk since it's like 3.5hrs
Guys is it possible to do this without PIE, like recursion or something
Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10)
i noticed that the only way to have no people increasing is obviously decreasing order, so
6 5 4 3 2 1
and swapping any two consecutive creates a single pair with increasing order
6 5 4 3 1 2
while swapping two that are one apart creates an set of three, so cant do that
6 5 4 3 2 1 -> 4 5 6 3 2 1
and swapping one of the back ones for any one that is two or more apart creates two pairs
1 5 4 3 2 6
6 5 4 3 2 1 -> 3 5 4 6 2 1
6 5 4 3 2 1 -> 6 2 4 3 5 1
so maybe you can do something with these operations?
hello doers of math
I'm not even sure recursion works
I think the case n=6 adds complexity
Because you can have 2 consecutive triplets that are increasing
Or it could work but very complicated
How do I avoid getting beaten by combinatorics
It keeps happening, especially in competitions
To skip it
🤓
Idk how to get at comb too
Twin
Finger count: works every time, just gotta have patience. Trust. 👍
Find some books
It's like a google drive, or some cloud drive that hosts all the pdf files in one place
What's this channel for?
Idk what that is

Thanks man, great explanation, couldn't have found a better one on the internet(if i used a web browser or stuff)
I can
Can you give me solution
If you do trial and error method with options both option (a) and option (C) are getting same graph
gl!
yh i mean most ppl do it in school
Are you there?
Anybody help?
Pls follow my account.
I'm a new content creator, I really need your support.
Thank you.
https://www.instagram.com/sallys.universe?igsh=bXp2ZXhqOHR0cGV1&utm_source=qr
Hi I am here now
Do you use Desmos
I didn't find the answer in the option..... because I got -4038090 as answer.
Can anyone help
Thx
brotha do you not do anything other than math all day
What's that
online graphing calculator
Who uses aops here?
Matematicas puras
The answer is -(2010)(2011)/2
hey please dm me when you are free
then I will tell you clearly
How?
note that n^2-(n+1)^2=-n-(n+1)
1^2-2^2+3^2-4^2…2009^2-2010^2
(1-2)(1+2)+(3-4)(3+4)…(2009-2010)(2009+2010)=-
-1-2-3-4…-2009-2010=-2010(2011)/2
Lmk if you need a better explanation
I can't understand that,how 3rd line came from the 2nd line?
1-2 = -1, 2-3 = -1, … so on. So we have a factor of -1 we can pull out. The remaining terms is just (1+2)+(3+4)+…
In 2020 AMC 10A Problems/Problem 25, how did this person determine that the condition 2(a+b+c)>=21 has optimal probability by re-rolling 2 dice?
Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly 7. Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?
#math #amc10 #amc10a #mathtips #mathtricks #problemsolving #competition #mathcompetition
(1-2)(1+2)+(3-4)(3+4)…(2009-2010)(2009+2010)->
(-1)(1+2)+(-1)(3+4)…(-1)(2009+2010)->
1-2-3-4…-2009-2010->
-2010(2011)/2
does that makes sense?
Ya... thanks
hello everyone i am class 9th student in india i had recently give an olympiad and was not able to find the answer to the question please help me
Q if 3a=4b=4c and a+b+c=24sqr(29) then the value of sqrt(a^2+b^2+c^2)
a)3sqrt(29) b)81 c)87 d) 2sqrt(29)
Hint: Let $3a=4b=4c=k$ and solve for $a,b,c$ in terms of $k$
Civil Service Pigeon
i did do it like that but i couldnt get anywherre
!show
Show your work, and if possible, explain where you are stuck.
you should do this from the start of asking a question btw
so what did you get so far
a;b;c =4/3:1:1
let a=4k/3,b=k & c=k where k is proptional constant
4k/3+k+k=24sqrt(29)
10k/3+24sqrt(29)
10k=72sqrt(29)
k=72sqrt(29)/10
,w sqrt(a^2+b^2+c^2), a = 48/5 sqrt 29, b = 36/5 sqrt 29, c = 36/5 sqrt 29
the question legit doesn't have a correct answer 
yeah the 4, 3, 3 was screaming wtf to me
because 4^2 + 3^2 + 3^2 isn't a perfect square

P(a+P(a)) is divisible by P(a) for all a>=1. So, you can take n=a+P(a) which is infinite when a=1,2,....
Got cooked in bmo1 😔
I average 35-45 on past papers but I genuinely did so bad in this one
It’s so joever
no (public) discussion yet unfortunately
💀
Do you see me mentioning any of the questions?
i don't think you're allowed to say like how you found it either
no comment
that falls under embargo
math
math
math
where should i go for math questions arounud the late aime level to early usajmo
besides aime and usajmo
?
ARML
Hmmt pumac
Pumac is a good source ngl
Especially division b is a great source for that difficulty
Also all the problems are unoriginal which is good because that’s like aime
(pumac does have some errors floating around tho)
what you said is part of what is forbidden by the embargo
i suck at writing proofs so i still think early usajmo is harder than most of the aime
I'm so proud of myself
I guessed an AIME problem right
statistically you should get a perfect score if you guess on every problem

what are you saying
gamble

Hi nerd
That was quick ._.
this is from my notes, but I had trouble making it readable from the camera, so i typed it down on symbolab so you can get a quick observation
personally i had trouble on the equal part, so can somebody help me?
!help
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
I guess
is the c a typo, btw?
What competition is this from
Penn State University
A great college.
Chapter XIV in my textbook
your techincally paying to sit in a chair with a desk and be bored for many hours
Wait, what is the question
wait what's the question then
are you solving for c then
cause this is not true in general
ohhhhh
,w d/dx (sqrt(x) + pi^(-3/2))/(x^c + 2n) = 3n + 2 sqrt(x) sqrt(pi^(-4))
ok thanks 🙂
I think you're trolling us
wait a minute.
how?
I typed it into symbolab so it would be readable so instead i wouldent have to picture it directly on my textbook
it's just impossible to do algebraically
unless you have a solution that states otherwise
I'd say not to waste your time on this question
im just trying to make my professor proud
it doesn't even look like a contest maths problem
there is zero reason why you need to do that
if you really want to get into research and work with your supervisor very closely
then once you have a PhD you and your prof can seriously talk
and who knows maybe collaborate on something
like you go into higher education for your own learning and benefit
if you get a bad grade the only person who could be disappointed is yourself
or your family if they are that involved in your life
npnp then

I don’t understand
I don’t understand
play geometry dash to learn geometry
can someone tell me what the question is even talking about im so confused with what it's stating T.T
do you understand the examples?
{1, 2, 3, 6, 9} and {5}?
no
it says alternatively add and subtract
so shouldn't it be 9 + 6 -3 ...?
I think the question assumes you subtract first, yeah
ah
wait I'm so confused with how to approach to the question
if n is 7 what would the set even look like?
it could be {1, 2}
it could be {2, 4, 5}
it could be {1, 3, 6, 7}
basically the highest number must be 7 or fewer
and you can only choose numbers from 1, 2, 3, 4, 5, 6, 7
oh
where did you get that information from?
like how did u find that out
it says in the question
n = 7 so you choose from 1, 2, 3, 4, 5, 6, 7
the important part is
each of its non-empty subsets
which part?
oh
i learned math in korean so im unfamiliar with english terms 😦
do you understand the solution?
yeah, basically it's the same idea as the solution on the AoPS page
how is the total sum 12 when n is 3...?
if you choose a subset $S \in {1, 2, 3}$, say $S = {1, 2}$
take $S \cup {3}$, so ${1, 2, 3}$
when you add the alternating sums together, you get $(2 - 1) + (3 - 2 + 1) = 3$
I feel AIME was easier back then
anyways that's not relevant
oh
southlander!
basically because you have added on a positive term, 3, to the start of the alternating sum
the signs shift one place
so instead of 2 you get -2
instead of -1 you get +1
convince yourself this will be true for any list
i get it now
this is the key idea
are set questions commonly found in amc 12 nowadays?
also like to justify it, either your set contains 7, or it doesn't contain 7
i feel like im week with set and combination questions
right
so the number of sets in the original problem would be (number of subsets of {1, 2, 3, 4, 5, 6}) * 2
= 2^6 * 2 = 2^7
but then there are 2^7 sets
we want pairs of sets, so divide 2^7 by 2 to get 2^6
like each set A will have a counterpart A'
that's why it's 64 * 7
ah
7 being the sum of each pair
makes so much sense now
i was so confused with what the question was even talking about 😦
yeah it must be hard when English is not your first language, right
are set questions important in amc 12?
I don't see them a lot in the AMCs
i didn't see many in amc 10 but next year im taking 12 so
these sorts of counting problems seem more like AIME or higher
oh
possible but I think other types are more common
like some kind of geometry
probability
Sometimes there are set problems
or like yeah something algebraic is very possible for AMC12
really?
but there are like one or two questions right
yeah like try to qualify at least
easier said than done haha, I feel every year the problems get harder and harder for the same level
competitive maths didn't attract so many students 10 years ago
really it's to do with US uni admissions (and possibly other countries) being more and more selective each year I bet
yeah Evan Chen said something like this in his blog
But like 10 years ago a lot of the late problems were surprisingly easy
yeah and this problem was from 1983
Like one you just calculate a bunch of quadratic formula stuff i think 2011 aime ii problem 15
I suspect more and more people are going to move away from competitive maths
if it's just not worth the effort to get recognised
people have other interests
that's basically what happened to me
so many chinese students cheat on their exam
no offense to them or anything
I switched from my maths degree to sociology and boy, are first-year humanities such a breeze
but someone i know got 138 on amc by cheating
I know someone who got 90 on 10a but 150 on 10b
can't say anything for later years but yeah it just feels easier to try when there's not so much pressure on your shoulders
oh yeah it is a big problem
don't forget India also
they cheat as well?
from what i heard private chinese organization give the test papers to their students before the test day
China seems to be the most blatant though
they definitely cheat but it seems they don't for international competitions
mainly like the civil service exams for India
oh those stuff
More motivation to study harder ig
yessir
yes to some extent
but there's always a breaking point
it's like Hooke's law where you can only stretch a rubber band so much before it snaps
With cheaters the bar is raised but i think if you study really hard you can still make jmo
Its not like they are cheating on the jmo so like after you hit that bar of qualifying then the cheaters dont really matter that much
do you think competitive math is all about studying?
because i think until aime it is, but after that, i think you need to have like inborn mathematical brain
@ornate blade how do you find the question directly from aops website?
do you use the image search function? or
ok now that i've done drunk BMO1
i think i agree, this year's BMO1 was quite a lot harder than previous years
granted i was completely wasted by the time i got to Q4 and then didn't do any of Q4 or Q5 or Q6
(i read the question but like none of it was processing in my brain :(
but yeah Q1 i thought was significantly harder than the previous few years
i think we were expecting it to be an absolute piece of cake, but it stumped some of us for a bit longer than it should've
Q2 i thought was fine
Q3 i solved quite quickly, but i know some of my friends actually found it surprisingly hard
and then i didn't do Q4+ but i know my friends thought Q5 was quite hard asw
and Q6 was really hard
well ok i say i was too drunk to do any more maths problems but like my more sober friends were stuck on Q6 for quite a while until someone very drunk solves it perfectly 💀
when i have time today, i'll try to solve Q4 and Q5
except i'm currently hungover and i've still got tons of work to do
Searched the problem text up
i have a random question could you fit infinite bricks into a 1 by 1 foot box if they got smaller infinitely or is there a limit
yeah i think p1 and p3 should be swapped
absolutely! something like this is very possible
the infinite sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$
southlander!
the picture I sent also suggests $\frac{3}{4} + \frac{3}{16} + \frac{3}{64} + \cdots = 1$ as well
southlander!
I'll share my experiences with it, Q2 was definitely the easiest problem, think it shouldve gone 2,3,1,4,5,6. Got pretty much full answers for 1,2. I had like a billion other things in my mind so i just tried some stuff out for 3, got real close then said fuck this. Tried some random bs for 4,5,6 then just called it a day. Ended up only submitting 1 and 2, it is what it is.
Im really disappointed but life moves on
Im considering doing paid entry for BMO2 and actually studying, I started off pretty well with BMO1 prep but other priorities pushed it to the side
What about if each brick had to be smaller then the prior, you have 3 equally sized bricks every time you decrease the size
ok i've done q5 now, i thought q5 was kinda easy/standard but i've also done a lot of olympiads and i am no longer drunk
drunk me could not do q5 lol
England has about the world's worst drinking culture istg
aside from I guess eastern Europe
ahaha lol
Brits drunk on holiday
is enough of a horror story
sorry my mind just jumped to that
lol no that's fair
ok done 4 asw
wild guess but for question 1 are only n = 3, 4, 6, 12 possible
ah fuck
3 and all the evens
the 12 literally has 0 impact on the problem
I have the video solution
yep
how would you prove it?
show it's not possible for n=odd, n>3
like the UKMT video solution just used p, pq, q
then do a construction for n=even
oh like a contradiction I see
you can read my really really shitty write-up that i wrote yesterday lol
the video solutions don't tell you how to write it up formally
like just to clarify, this is an atrociously shitty write up
we were doing it for fun & were drinking so the write ups were never gonna be any good
i am expecting BMO2 grade boundaries to be lower than normal, but you can still pay to do bmo2
ngl i really regret not paying to do bmo2 when i was in y10
okay I've decided to think about q3 a bit
i was quite close to qualifying in y10, and i think if i paid to do it in y10, i would've revised a lot more and would've been a lot better when i did it in y11
I've figured out that if the current number is n = k^2 where k is any integer
the player could just choose a = b = k so that |a - b| = 0
so they will have won the game
has to be different numbers
oh shit
knew it
I think working backwards might help still
so if n = 1 that player loses
so if n = 2 that player wins
spoilers for Q3: ||i think drunk me took a little too long to realise it went R J R J R J R J lol||
and for n = p^2 where p is prime the next number has to be p^2 - 1 cause of what you said
oh wait
wow I just clicked the image
ah cause this must be even if p is an odd prime
so like (p + 1) - (p - 1) = 2 idk
yeah I think I'm overthinking it
just considering what happens when n = 2, then n = 3 and so on
suffices
tbh when i did it yesterday
i worked all the way up until 20
you start to see a pattern with how to deal with odd/even cases
&then that solves the problem
i mean a hint for the problem would be that in general, factorising numbers is a hard problem
(they gave u 10^6)
I feel these are just dirty tricks
so it'll be something clever, that you can do no matter what number it is
I'm still unconvinced
like on how for any even number, you are guaranteed to lose no matter what
I understand that for an odd number you can subtract 1 to get an even number, which is supposed to be a losing position
for n = 8 for example, can't you do a = 4 and b = 2
so you'd go down to n = 2
oh but then the second player would go to n = 1 and then the first player would be out of moves
ahhhhh so the first player chooses to go down from n = 8 to n = 7 to avoid this
yep
instead of n = 6 to n = 1
shouldn't that case be different then
as in if Rhian goes down from n = 6 to n = 1
Jack must be out of moves
ah I see so Rhian wins either way
so Rhian goes down from n = 6 to n = 4 to n = 2 on his turn
ah by making a table I see it
if you put R on the left column and J on the right column
both ways 1 ends up in the right column
so R wins either way if he goes first
so basically this relies on a parity argument of some kind
okay it makes a lot of sense
it just seems too simple to be true
like it doesn't make sense how the backward argument, considering how n reduces to n - 1 which we already have figured out
works at first sight








