Who Wants To Be A Mathematician Final Round 2019
2019 ж. 18 Қаң.
288 037 Рет қаралды
The Final Round of AMS Who Wants To Be A Mathematician 2019. Congrats to Luke Robitaille for winning.
The Final Round of AMS Who Wants To Be A Mathematician 2019. Congrats to Luke Robitaille for winning.
Full credit to the host for being able to still read the questions straight after his coffee-drinking world record attempt.
Beautiful comment
proud of you luke, leave humanity an epic paper or conjecture with your name on it.
man just remembers questions, most likely photographic memory. doubt he will provide any new findings, just regurgitate answers
@@premierdota9128 jealous?
@@papesldjnsjkfjsn jealous of what? Im quite happy I dont have to store boring maths problems in my head, solved by millions before, and can use it for better stuff. If he provides any new research paper then he is using his brain otherwise its just a waste of memory space, like a chess player, keep repeating and remembering useless information. computers from the 1980s already made chess obsolete, spend your life on a useless thing
No.. luke is awsome ..he has won many maths championships..
yes, he already said he subscribed to many math magazines. Nice beef but great for a 15yo.
is it even possible for luke to lose a math competition lol
He lost in 2016 Raytheon semifinals. He also scored lower than some teammates at IMO. He's amazing, but not infallible. But I believe he's undefeated in videos posted by his relatives! :)
Yh he lost in 2016 but that wasn't his fault as the person he lost to was a 7th grader while he was a 6th grader,a champion among champions I call im
@@geses1 it wasn't the semis, it was the finals (he lost by 1 point).
@@geses1 He has 4 IMO golds. With world ranks of 18, 3, 4 and 9! Damnnn
That is not even math… As a mathematician its a shame that they call that ugly and meaningless shit math
Impressive speed, though they should've put more time into creating the questions I think e.g. the Ramanujan thing is very famous and is a free point to anyone who's seen it before, shouldn't belong on a quiz like this.
Also, Ramanujan came up with this identity by starting with the answer and then breaking it up. Nobody was going to get that one if they didn't already know the answer.
@@zanti4132 I didn't know it and found it by testing the answers, 2 rapidly showed incorrect so I tried 3, and by repeating squaring/substracting/dividing you notice a pattern 3, 4, 5 etc.
@@pierreardouin6441 Yeah, the way the answers are spaced out it should be possible to make an educated guess by doing a manual estimate from the first few terms. Even so, that takes time - there's no way it will beat someone who recognizes the identity, and you just had to figure Luke would! At the very least, the identity should have been tweaked in some way. For example, it could have been: √(1 + 1√(1 + 2√(1 + 3√(1 + 4√(1 + ... ) ) ) ) This looks like the Ramanujan identity, but the nested radicals start with a 1 coefficient instead of a 2 coefficient. This difference is enough to change the answer. For this modified problem, the participants would have to realize this is not the Ramamujan identity and not buzz in too quickly. The points would go to the one who sees how the Ramanujan identity can be applied to get the answer quickly.
@@zanti4132Would that change make the solution 2?
@@No-uu7wm Yep. You can substitute the known value of the Ramanujan identity into the identity I proposed to get √(1 + 3) = 2. That's what I figure Luke would have done in 0.1 seconds, so he still would have picked up the points.
This contest was HORRIBLY done!
Broo!!! So bad
And not the ‘who want’s to be a millionaire’ format at all.
I'm one of the worst human on earth at math and I'm watching this at 3am while drawing, my life is awesome.
Indeed
Same here
Hey, I just wanted to remind you that your drawing skills have probably not improved in the last 3 years. Have a good day though! 👍
May the force be with you, Luke..
Seems so clunky compared to the polished MathCounts competition! Still glad I got to see it.
Luke is litterally the main character
amazing development, he might be authistic?
Host did not take any public speaking classes in school
which is why he's so great
Host: Here's the next question *Contestants blink eyes* Luke: 1 B !
The contestants don't seem to get penalized for getting anything wrong, so it seems to me the best strategy is to just quickly guess the most likely possible answer without fully solving the equation.
Well, if you get it wrong the other contestants get to guess again and you don't, so you are guaranteed to not score any points for the round, whereas if you think it through and then give your answer you may score points. So guessing is not a good strategy, though educated guessing is a very good strategy that is being utilised and I think being able to give educated guesses quickly is a skill that is being tested.
@@mosesbejon6660 If you're watching a math competition video you should really see why you're wrong
@@timmyg72 Please enlighten us
@@mosesbejon6660 6 players. You've got a 1/5 chance of guessing. For those players that aren't above average, I'll let you figure out if guessing immediately is a good strategy
@@timmyg72 educated guesses don't care about probabilities though? As long as you understand the skeleton of the questions and which answers are viable quickly, you've got the best chance of winning here. So I don't get why you think he's wrong?, you can still get a buzzer in, without having to forfeit your round, losing points. And taking a bit of time to understand which answers are viable, comepletely changes your chances of winning. For instance the one question that someone got wrong and the guy who got points guessed on the fact that the other numbers just would've made sense.
Hello LUKE!
7:10 The trick is to rewrite 1/m + 1/p as (p+m)/pm. Then, since (m, n, p) is a geometric sequence, mnp = n^3 = 27 implies that n = 3, and mp = pm = 9. Also, since m+n+p = 12, m+p = p+m = 9. Finally, 1/m + 1/p = (p+m)/pm = 9/9 = 1. ✅
i have done the same thing 👍
But in this case wouldn't m be 1 and p would be 9 (geometric sequence being 1, 3, 9) but 1+3+9=13 not 12.
@@ggbasiQNot sure why you think m has to be 1. n HAS to be 3, due to the ‘geometric series’ condition combined with the triple product being 27. If you then try m = 1, you get: n + p = 11 p = 8 But then mnp = 1•3•8 = 24. In actuality, m = ~1.146 and p = ~7.854. The ratio of the geometric series (m, n, p) is ~2.618 (as opposed to 3, which is the ratio of the GS (1, 3, 9)).
@@quanta2770 you're totally right, idk why I thought they all had to be integers. 5am sleep brain ig
Luke answers, other contestants "how? I cant read that fast" 13:10 xD
heres me with dyslexia thinking i was good at math
For question 8, the only integer that works is 76. Here’s why: Each conditional puts the number into 3 different mutually exclusive ranges. Therefor Grace’s integer cannot fall into more than one category at a time. However these 3 conditions sufficiently cover all of the integers. Therefor Grace’s number must be in exactly one of the conditions (cannot be in none of them). Suppose Grace’s number is not a multiple of 4. This means that the other two conditions are false, therefor the number would be both a multiple of 6 and NOT a multiple of 3. However this is a contradiction because any multiple of 6 is a multiple of 3 as well. Therefor Grace’s number is not NOT a multiple of 4. We now know that Grace’s number is a multiple of 4. This means that it is either a multiple of 3 or it is NOT a multiple of 6 (which are mutually exclusive subsets anyways). Let’s look at them separately. Suppose her number is both a multiple of 3 and 4 (and thus 12). No integer in the range 50 and 59 inclusive will be a multiple of 12. Suppose her number is a multiple of 4 and not a multiple of 6. 72 and 76 are the only 2 numbers in the range that divide 4, however only 76 will NOT be divisible by 6 Grace is thinking of 76.
Duh, they have more difficult problems than that on Blue’s Clues
Great analysis, my way was a bit more brute force, but in the same way as you. If the number is a multiple of 3, then it will either be 51, 54 or 57. All of these numbers are NOT multiples of 4, therefore we cannot be in that 50-59 range. At the moment, we go to the 60-69 range. In that range, we would have to consider all the numbers that are not multiple of 3 or 4 (the range already tells us that it is not 4x, also if it's 3x, we would be in 50-59). The numbers that fit that are 61, 62, 65, 67, 68. We see that non of these are divisible by 6, therefore that would put us into the 70-79 range. In that range, we need numbers that are divisible by 4 (otherwise we fall into 60-69), that are not divisible by 3 (otherwise we go into 50-59) and by default not divisible by 6. Divisible by 4 are only 72 and 76 and out of those two, the only one not divisible by 3 and 6 is 76.
Luke's thinking on this one: "If the number is a multiple of 6, then it's a multiple of 3. But if it's a multiple of 3, it has to be 54, which isn't divisible by 4. Therefore it's a number in the 70 to 79 range that is a multiple of 4 and not a multiple of 6. Gotta be 76."
I made truth tables like a psychopath. Then I immediately got rid of ones where it was 6 T but either of four or 3 were not since 4 is hiding the two needed to prove six. Then I got rid of ones where both 6 and 4 were false since those couldn't coexist and that left FTF which leads to 70-79 and then you cut out the first bunch. I am proud since it has been a bit since I learned geometry.
i aint reading alltha
Luke looks a bit different from mathcounts
he seems better and chiller
I got the first question right while he was still explaining the question...i might have to enter this.
Go Brian!
Easier way to solve Question 7 (11:57): Every corner of the 100-d unit cube can be represented as some combination of 100 ones and zeros -- e.g., {1, 0, 0, 0, 1...}. The number of different distances to those corners (from an arbitrary starting corner) is the same as the total number of ones (or zeroes) in our vector. (For example, {1, 0, 0, 0, 0, 0...} and {0, 1, 0, 0, 0, 0...} are the same length.) There can be between 1 and 100 ones in the vector, inclusive, so the answer is 100.
Much harder* But yes I just noticed a square has 2, a cube has 3, then was confident enough to go with 100
tell that to Luke
I love keeping track of luke through different videos and championships
My simplest solution to Question 3 at 5:07: We have five non-gold medals to place on a 4x7 shelf that is horizontally symmetric. This gives 4 middle spots for unpaired medals and 12 outer spots for pairs. These are the only ways to fit our five medals: A) two pairs and one unpaired, or B) one pair and three unpaired. A) 12 spots for first pair, times 11 remaining spots for second pair, times 4 middle spots for unpaired. B) 12 spots for the pair times 16 arrangements for the middle (4 all-silver plus 4*3 mixed). So 12*11*4+12*16 = 12*60 = 720.
Howard Halim is Canadian and competed in the IMO for Canada
and he scored a 0 on this, use that as inspiration!
Luke always looks like he's going to have a stroke.
I was watching the Raytheon video and Luke was there too !
Luke and I are the same age and it makes me feel bad about myself but at the same time he’s a legend and I love him
Luke is acomeputer woow i wish that i have like his inteligince
Lykwise
Lukewise
Luke always with the hand in the Head
its a permanent p'card facepalm, if a person is that smart he's perpetually omg/wtf with the human race. the price paid for being 200+ iq
Luke is brilliant.
Luke is fast. Any one of those contestants could answer all of the questions just as easily, he just read them faster.
hes just done all the questions before
how so? I mean sure he seems to be reading hell a lot, but he didn't solve those answers he just knew the answers
@@Endrit719 he knows the answers probably because he's already solved them before and fully understands them lol, he is brilliant
@@crung he had seen them beford
Are you Lukes Father?
No, Anakin is.. Didn't you watch the movies?? Gosh!
that host really does act like the genius mathematician kid in the high school
And fast forward to today, Luke was probably the top score on the Putnam Exam.
Well, he won the MIT integration bee this year.
@@purplepinkclouds8882 There is a huge difference between Putnam and "integration" competitions. I'm not sure that someone who just can learn by heart is able to win Putnam (simply, not that level). Of course, I don't know Luke and all his abilities, so I'm not going to evaluate him here and I'm not claiming that his only ability is to learn by heart. Probably it is not Editted: Just checked. He won gold medal 4 times on MO, so he's really good
Standing affects the performance. It does for me anyway. I think better sitting.
At 4:50, can someone clarify what he meant? If we abstract the value 20 given in the problem, we get the following remainders (there are patterns, and the 3 appears a lot indeed) : 1-> 0 2-> 1 3-> 0 4-> 1 5-> 3 6-> 3 7-> 5 8-> 1 9-> 0 10-> 3 11-> 0 12-> 9 13-> 9 14-> 5 15-> 3 16-> 9 17-> 12 18-> 9 19-> 8 20-> 13 21-> 12 22-> 11 23-> 20 24-> 9 25-> 13 26-> 9 27-> 9 28-> 5 29-> 16 30-> 3 31-> 1 32-> 25 33-> 0 34-> 29 35-> 33 36-> 9 37-> 4 38-> 27 39-> 9 40-> 33 41-> 3 42-> 33 43-> 15 44-> 33 45-> 18 46-> 43 47-> 17 48-> 9 49-> 47 50-> 13 51-> 12 52-> 9 53-> 12 54-> 9 55-> 33 56-> 33 57-> 27 58-> 45 59-> 27 60-> 33 61-> 21 62-> 1 63-> 54 64-> 25 65-> 48 66-> 33 67-> 64 68-> 29 69-> 66 70-> 33 71-> 67 72-> 9 73-> 54 74-> 41 75-> 63 76-> 65 77-> 33 78-> 9 79-> 19 80-> 73 81-> 63 82-> 3 83-> 26 84-> 33 85-> 63 86-> 15 87-> 45 88-> 33 89-> 75 90-> 63 91-> 61 92-> 89 93-> 63 94-> 17 95-> 8 96-> 57 97-> 79 98-> 47 99-> 0
Every factorial above 4! is a multiple of 20, so they will give a remainder of 0 when divided by 20. 4! gives a remainder of 4, 3! gives a remainder of 6, 2! gives a remainder of 2, and 1! gives a remainder of 1. Add up all those remainders, and you get a total remainder of 13. As for the fact he mentions, every factorial above 4! is also a multiple of 10. So, for any n>4, the sum of any list of consecutive factorials 1!+...+n! will have a ones digit equivalent to the ones digit of 1!+2!+3!+4!. (All higher factorials will contribute nothing to the ones digit.) And 1!+2!+3!+4!=33, which of course has a ones digit of 3. So the sum of all higher factorials must also end in 3.
@@mriswith88 Thanks !
He said he "knew" that every sum of factorials 1! + 2! +3! ... ends in a 3. Therefore the remainder in question had to be a number ending in 3...and 13 was the only such choice. How on earth he "knew" the factoid about the 3's (true for any (n>=4) is beyond me. Turns out the last two digits are always 13 for any n>=4. For why it's true, @mriswith88 gave the perfect answer. The stuff these kids know "by inspection" is pretty amazing!
@@marcohanig176 He had probably just done a couple of similar problems previously and had remembered the fact.
@@mriswith88 your explenation actually makes sense unlike the one in the video which seemed to be a lucky guess.
That dude won the MIT integration Bee 2023....
Luke already saw the questions before the competition begins
anyone knows where i can find the answer paths to those questions? wanna know how to solve but struggle. Do they think their audience automatically knows? :D
Did the host do speed before the show?
Replace the host!
PLEASE
He's professor 🤣
He is the worst omg
Lol, maybe he’s acting normal for a mathematician. Us non mathematicians just might not understand.
@@fredirecko ..,mathematician or not we know how humans are
Explain me question number 4 please!!
It's perfect that that the duration of this video is 13:37. Luke is a LEET mathematician!
how on earth did i get the question with the medals right.. i havent been doin math soo long. i just crunched 7! divided by 7 in my calc out of intuition. i never learned or studied these kind of fields in math in school. was that random luck or what now hahaha
Andrew C good too
👏👏👏👏👏👏👏
I only got the first one
Same man :)
Holy shit Luke
5:32 his face turned red
Luke what the heck so good
"I'm sorry we ran out of time." HOW?
it took me around 10 mind to solve the question that's answer was 720 there are 2 major cases And then further 2 cases in One of these 2 cases so we have to solve three cases that's two much .How can they solve it so fast
superior genetics.
the host seems like a great guy lol
Full option by heated
I’m as good at this stuff as I would be if I’d never gone to school, that is, not at all.
Yup. School doesn't teach you this. Private mentoring, self teaching, circles, workshops do. School math will be boring for these guys, they would have finished calculus before high school.
Olive Zippered Jacket of Calculations +100 To Mathematical Ability
Luke might have finished all the questions on the graduation exam in his class, while the majority of his friends were still working on the first question.
Luke :-Its B) 3 Mod 24 Announcer:- Well theres no option like this Luke:- Its answer of next question. the ans of this is A) 2/3
Sorry to say but what happens here is not a proper representation of competitive mathematics. Lot of stuff done here is just practise some memory + Doing some tricks with the MCQ answer options.
So its more of a Math General Knowledge than Math solving competition?
They need a new MC
I didn’t know Chuck McGill moonlighted as a moderator.
I AM NOT CRAZY! I know he memorized those answers. B after B, as if I could ever make a mistake
Packed.
This ain't about math but memory
are calculators allowed?
Could GPT4 answer all of these problems in 5sec nowadays?
idk about gpt4 but gpt3.5 cant even do basic multiplication so doubtful. its a language model after all
Luke just became a Putnam fellow. Congrats man!
why Brian Liu here as MIT 2024 grand integrate not providing a score?
i definitely do not want to be a mathematician after watching this
Is Luke the real Good Will Hunting?
"Don't feel bad"
Luke Robutaille
At 4:48 what was his explanation I can't hear him
Host is extremely annoying!!
Is that todd Howard?
The others are just like we are living in luke world
these Chinese dudes are amazing bro fr
Am afraid luke will be a mad man before age 30. Looks like a confuse guy. He's dammed good.
he is already
:Host: What is the- Luke: I have figured an equation to get nukes. Host: How? Luke: 9:34
epic
Why is Luke not wearing the same shirt as the rest of them?
He's wearing the shirt underneath the jacket
Luke also said in an interview before the competition that the winter would be very cold for him there since he is from texas
is it bad to assume that 3s into the video... the guy wearing the hoodie with his hand to his head will clearly beat all the normies?
i got 10/9 for Q4 why is it wrong ?
Why would it be right? The contestant explained the solution
Wow! why did they make him as a Host?!
Good question!
@@Ace1King1 A harder question than any of the ones on the quiz
isn't the guy from MIT Integration Bee champion for 2023 ? the one in a hoodie
just watched luke winning mit integration bee and youtube recommended this...
Luke the skyWalker
Its awesome that you write who won in the description so that people who wants to see an exciting competition does not have to bother watching the video. Some people call that a SPOILER but honestly it saves a lot of time. Good thinking, you must be a math genius!
His son is!
Don't give options, they might try their luck by making mild calculated guess. Kind of gambling indeed..🤗
No idea why youtube recommended this to me but I have to say - the format is terrible! Just give everyone 30s to solve it (they will press the correct answer so you know the time when they had it) and then let them all show the answer and assign points based on speed... So Luke had it in 3s and gets 950p, someone else had it in 15s get 500p, etc... And also, don't use examples from books they might know.. just invest some time and prepare original questions.
They were running out of time? Didn't Luke save them a bunch of time? lol
they are guessing simple probabilty???
Yeah, its Luke from Raytheon again:D
Haha host just a little NERVOUS guys, but he’s having a fun time and so am I
Definition savant
Terrible MC
Oh you have 0 but don't feel bad about it. You can still be mathmagicians if you want to.
@@oneuniverse5687 You must be the fake account of the MC lol
100th like
No way, how there’s a math contest without an Indian. Lol
Is the host is nervous, his voice is freaking.
Ez. Grade 9 of Vietnamese schools
Ain't no way Luke wasn't cheating.
Who wants to be a computer
so they just remember answers
For the one on medal count I'm getting 528 (not one of the choices)... 12*11*4 Mistake?
You split the group of 3 into 1 and 2, but there's also 4 sets of permutations where the three medals are split across 3 rows (as the centre medal), which is 4 * 12 * 4 for the other 192 permutations. I did the same thing lol.
@@Degenx2 Ahh!! You're right, I am definitely missing those cases where 3 silvers are all in the middle slots. Except I think those only account for 4 * 12 = 48 permutations. Then there are the ones where the two bronzes are both in the middle. For those you get another (4 choose 2) * 2 * 12 = 144 That gets us to 720!
4(3*12+12+12*11)
I did this one by partitioning the non-gold medals using integer partitions of 5. Partition and config Count {5} 24 {4, 1} BBSS, S 72 {3, 2} SSS, BB 108 {3, 2} BSS, SS 108 {3, 1, 1} BBS, S, S 36 {3, 1, 1} BSS, B, S 72 {3, 1, 1} SSS, B, B 36 {2, 2, 1} BB, SS, S 216 {2, 1, 1, 1} BB, S, S, S 12 {2, 1, 1, 1} SS, B, B, S 36
We have five non-gold medals to place on a 4x7 shelf that is horizontally symmetric. This gives 4 middle spots for unpaired medals and 12 outer spots for pairs. These are the only ways to fit our five medals: A) two pairs and one unpaired, or B) one pair and three unpaired. A) 12 spots for first pair, times 11 remaining spots for second pair, times 4 middle spots for unpaired. B) 12 spots for the pair times 16 arrangements for the middle (4 all-silver plus 4*3 mixed). So 12*11*4+12*16 = 12*60 = 720.