Stanford math tournament algebra tiebreaker

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  • เผยแพร่เมื่อ 9 ธ.ค. 2024

ความคิดเห็น • 263

  • @arpitranasaria4878
    @arpitranasaria4878 2 ปีที่แล้ว +549

    Hi! I'm the organizer for SMT 2022! I'm glad you enjoyed our tiebreaker problems, and thank you for reviewing them!!
    For those of you who are high school students we encourage you to participate in this year's iteration of the contest, we're very excited :)

    • @blackpenredpen
      @blackpenredpen  2 ปีที่แล้ว +116

      Wow, great to have you here! As I said in both of my SMT videos, you guys have some really nice problems. Keep up the great work!

    • @theuserings
      @theuserings 2 ปีที่แล้ว +35

      @@blackpenredpen amogus

    • @kepler4192
      @kepler4192 2 ปีที่แล้ว +13

      @@theuserings no

    • @theuserings
      @theuserings 2 ปีที่แล้ว +6

      @@kepler4192 👽

    • @dibyojyotibhattacherjee897
      @dibyojyotibhattacherjee897 2 ปีที่แล้ว +5

      Yes, we would like to.

  • @cosmicvoidtree
    @cosmicvoidtree 2 ปีที่แล้ว +313

    For the second question, removing all of the high level notation you could say, how many numbers less than 40 are perfect squares but not perfect fourth powers.

    • @blackpenredpen
      @blackpenredpen  2 ปีที่แล้ว +145

      Yes!!! That’s a very simple way to interpret it. The hard part was definitely all the fancy math language.

    • @cosmicvoidtree
      @cosmicvoidtree 2 ปีที่แล้ว +11

      @@blackpenredpen Thanks. Also to finish the interpretation, it comes down to finding how many perfect squares there are below 40 and how many prefect fourth powers below forty and subtracting the later from the former

    • @poproporpo
      @poproporpo 2 ปีที่แล้ว +3

      @@cosmicvoidtree Alternatively, you can find/generate a list of squares of numbers that are not squares themselves. This works for n≤40 because we don't really have to deal with sixth powers and etc.

    • @kenhaley4
      @kenhaley4 2 ปีที่แล้ว +1

      Wonderful insight! Nice.

    • @mikehocksbig6800
      @mikehocksbig6800 2 ปีที่แล้ว +9

      Luckily there are no perfect 8th powers less than 40 or what you have said would not hold. Eg. 256 is 4^4 but it would satisfy the condition (f(256) can = 16 as f(4) = 2) perfect 16th powers don't hold as f(65536) != 256 since that is f(16). Will hold for 32nd powers but not 64th and so on.

  • @blackpenredpen
    @blackpenredpen  2 ปีที่แล้ว +21

    Another SMT problem (rotating y=x^2) : th-cam.com/video/gYAQg7xn-Xo/w-d-xo.html

    • @yodaimpostor4781
      @yodaimpostor4781 2 ปีที่แล้ว +1

      cant you just do the quardtaic formula for the last one and then plug in x?

    • @mckenzi4608
      @mckenzi4608 2 ปีที่แล้ว +1

      @@yodaimpostor4781 if I'm assuming what I understood from that
      I got (3+-√5) ÷ 2 for the first equation

  • @ASHISHKUMAR-vm3cx
    @ASHISHKUMAR-vm3cx 2 ปีที่แล้ว +30

    I am from India , i study in 12th standard. when i study integration, i have some questions having a problem then i search in TH-cam then i show your channels good working teachers. I really motivated your teaching skills , i love mathematics 🥰🥰🥰🥰🙏🙏🙏🙏🙏🙏🙏🙏

  • @kenhaley4
    @kenhaley4 2 ปีที่แล้ว +30

    That was a fun one!! It's so cool to see problems that I initially have no idea how to approach, solved with such a simple, easy-to-understand technique! Thanks!

  • @mastershooter64
    @mastershooter64 2 ปีที่แล้ว +93

    lol ever since i picked up dummit and foote the only things that come to mind whenever i see or hear "algebra" are rings, groups, fields, modules etc.

  • @EvanUnknown
    @EvanUnknown ปีที่แล้ว +2

    I wouldn’t be able to solve these in a half hour but you just make it looks so easy and make so much sense. Ty!

  • @andrew_owens7680
    @andrew_owens7680 2 ปีที่แล้ว +8

    I really enjoy your channel. I don't understand half of what you are saying, but it stretches my brain in the right direction.

  • @erikdegeorge866
    @erikdegeorge866 ปีที่แล้ว +5

    For the last one, I just multiplied both sides by x^2 + 3x + 1, which cancels out all the third and first degrees, and gives x^4 -7x^2 + 1 = 0. Rinse and repeat twice more. Basically the same process without dividing by x first.

  • @WestExplainsBest
    @WestExplainsBest 2 ปีที่แล้ว +106

    My 6th graders would probably wonder why lumber is involved in these 'log' math problems.

    • @thegoldlightning
      @thegoldlightning ปีที่แล้ว +3

      As a 6th grader, our math club took the questions from the Stanford-Math League Tournament. Even though our group was comprised of 6th, 7th and 8th graders, I ended up winning with a t-shirt which has “Stanford-Math League Tournament”. I’ve already learned logarithms and even some calculus.

    • @martinrosol7719
      @martinrosol7719 ปีที่แล้ว

      ​@@thegoldlightning 🏅

    • @zxlittle87xzexchernyap76
      @zxlittle87xzexchernyap76 ปีที่แล้ว

      ​@@thegoldlightning wut

    • @windowsxpmemesandstufflol
      @windowsxpmemesandstufflol ปีที่แล้ว

      ​@@thegoldlightningwait 8th graders don't learn log and calc yet?

    • @thegoldlightning
      @thegoldlightning ปีที่แล้ว +2

      @@windowsxpmemesandstufflol a lot of people there are in the top advanced program at our school. Some do even higher outside.

  • @stevesknee
    @stevesknee 2 ปีที่แล้ว +83

    I don’t know if I ever got one right but I remember seeing questions like the last one all the time in these extracurricular math competitions lol
    The idea of dividing by x to get a binomial where 2ab removes the variable… so simple yet so crazy to come up with if you’re not used to these types of approaches

  • @jaybacan650
    @jaybacan650 2 ปีที่แล้ว +2

    The best almost 15 mins. spent today. Thank you for the video.

    • @SQRTime
      @SQRTime 2 ปีที่แล้ว

      Hi Jay. If you want to see similar videos in math competitions, please consider
      th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊

  • @sphakamisozondi
    @sphakamisozondi 2 ปีที่แล้ว +40

    If i was in a math class and the 2nd question was in the test... I'd have passed out 😂
    Edit: how you tackled the 3rd problem was satisfying. No lie.

    • @cxpKSip
      @cxpKSip 2 ปีที่แล้ว +6

      That is the kinda problems I, as a math major, get asked as homework.

    • @bonjour7209
      @bonjour7209 2 ปีที่แล้ว +2

      the second one was by far the easiest imo

    • @idealized_
      @idealized_ ปีที่แล้ว +2

      @@bonjour7209 definitely.

  • @alexandros4990
    @alexandros4990 2 ปีที่แล้ว +4

    For the 3rd question, you can also notice that the k = sum of the roots of the equation. Then, sum the two roots of the staring equation (found via quadratic formula), each to the power of 8. This will be equal to k.

    • @sushant2664
      @sushant2664 2 ปีที่แล้ว +1

      Yes you can compute the sum of the 8th power of the roots using a similar method of squaring, adding/subtracting repeatedly. But, bprp's method is quite elegant in a way.

  • @sk8erJG95
    @sk8erJG95 2 ปีที่แล้ว +3

    Very fun!
    I did the first the same!
    For the second, I said that f(a)^2 = a can only be true if a is a perfect square, so a = 1,4,9,16,25,36.
    For a = 1, f(1) = 1^2, so we can get rid of that. And then 16 is the square of a square, so f(2) = 4 but f(16) = 4 breaks the bijectivity of f. No other is a 4th power so f(a) ≠ a^2 for a = 4, 9, 25, 36.
    For the third, I used the quadratic formula! Since x^2 - 3x + 1 = 0, we have x = (1/2)(3 ± sqrt(5)). And since x^16 - kx^8 + 1 = 0 is a quadratic in x^8, we get that x^8 = (1/2)(k ± sqrt(k^2 - 4)).
    raising (1/2)(3 ± sqrt(5)) to the 8th power (using the binomial theorem) gives (1/2)(2207 ± 987sqrt(5)) so
    (1/2)(2207 ± 987sqrt(5)) = (1/2)(k ± sqrt(k^2 ‐ 4)), which tells us k = 2207.

  • @draterdecriotz833
    @draterdecriotz833 2 ปีที่แล้ว +5

    this man explained the entire paper within 15 minutes time limit lol

  • @TheSafeSword
    @TheSafeSword 2 ปีที่แล้ว +14

    I love the first question, it helps my understanding for logs tats coming for my mocks, thanks!

  • @iHugoMMM
    @iHugoMMM 2 ปีที่แล้ว +2

    That 3rd problem was amazing!!

  • @Farhan-hp8kr
    @Farhan-hp8kr 2 ปีที่แล้ว +6

    I am glad that I was able to do the 3rd problem 😁

  • @ashirwadgarg174
    @ashirwadgarg174 2 ปีที่แล้ว +4

    Nice approach to solve problems 👍

  • @charadreemurr9337
    @charadreemurr9337 2 ปีที่แล้ว

    I tried doing it in 15 minutes and messed everything up with haste xD. Thank you for the solving methods exposed in this video.

  • @res5139
    @res5139 2 ปีที่แล้ว

    Your channel is truly awesome! I just wish you diversify the kind of problems you solve to include more stuff outside calculus :) Rock on!

  • @tambuwalmathsclass
    @tambuwalmathsclass 2 ปีที่แล้ว +1

    Amazing

  • @roberttelarket4934
    @roberttelarket4934 2 ปีที่แล้ว +2

    All three are really great problems!

  • @armanavagyan1876
    @armanavagyan1876 ปีที่แล้ว +1

    More math competition thanks Prof)

  • @gigachad3955
    @gigachad3955 2 ปีที่แล้ว +1

    In my highschool in Poland we did problems like the first one on daily basis when we had logarythms in 11th grade but instead of doing it by factoring it out we did t=log2(n) then solve delta and substitute t1 and t2 and then solve for log2(n)

  • @vijaykulhari_IITB
    @vijaykulhari_IITB 2 ปีที่แล้ว +2

    Wah new look 🤘🤘

  • @iSustainnn
    @iSustainnn 2 ปีที่แล้ว +2

    I don't know why I'm smiling on entire video, especially on the last part. Please help me lol.

  • @angelmendez-rivera351
    @angelmendez-rivera351 2 ปีที่แล้ว

    x^2 - 3·x + 1 = 0 x - 3 + 1/x = 0 x + 1/x = 3. We know that x^16 - k·x^8 + 1 = 0, so we want x^8 - k + 1/x^8 = 0, hence x^8 + 1/x^8 = k. The idea is to compute x^8 + 1/x^8 in terms of x + 1/x solely. This is doable, by considering the binomial theorem. This is the key to the question.

  • @kristian4243
    @kristian4243 2 ปีที่แล้ว +1

    Takk!

  • @Static_MK3Focus
    @Static_MK3Focus 2 ปีที่แล้ว +1

    I see that a lot of these, high school juniors could solve, If they had the knowledge, I tried the problem with all the solutions less than 40, I actually did it and it was more logic than anything in my opinion

  • @prakharjoshi2813
    @prakharjoshi2813 2 ปีที่แล้ว +2

    Satisfying explaination ..

  • @fergalmdaly
    @fergalmdaly ปีที่แล้ว

    The trick of dividing by x is very nice but you end up with something very similar by just doing
    x^2 = 3x -1
    square both sides
    x^4 = 9x^2 -2.3x + 1
    and notice that you have an equation for 3x above: 3x = x^2 + 1
    giving you x^4 = 7x^2 + 1
    Now do that 2 more times and you get the answer.
    It's less elegant but doesn't require as big a flash of insight, I think.

    • @iqwit
      @iqwit ปีที่แล้ว

      Quick tip you can use the asterisk symbol instead of decimal point for multiplication like 2*4

  • @rgarcia831
    @rgarcia831 2 ปีที่แล้ว

    Beautiful, that was quite fun. Well explained, and funny too

  • @jasonbowens8369
    @jasonbowens8369 2 ปีที่แล้ว

    Fiat Lux! As you were doing the question I thought in my head to make a common denominator on the left side and see that lon_2{n}=1. I was stunned the answer didn't line up. But then I saw you didn't switch the sign on when you moved 5log_2{n} to the other side. Maybe that will get my answer of n=1 and there is probably another one too. PS I made it to Cal because of you and Professor Leonard. TYSM

  • @lorddeath8696
    @lorddeath8696 2 ปีที่แล้ว +1

    BPRP: *Points at 9*
    Also BPRP: "Five"
    9:24

  • @AdityaKumar-gv4dj
    @AdityaKumar-gv4dj ปีที่แล้ว

    I liked the 3rd question and the 1st question a lot.

  • @petereziagor4604
    @petereziagor4604 2 ปีที่แล้ว +1

    Hahaha, I love the method you used in the third problem.
    Usually, I would do x^2= 3x-1, then square both side
    X^4= 9x^2 -6x +1, but x^2 = 3x-1
    X^4 = 9(3x-1) -6x +1 = 21x -8, square both sides again
    ...

    • @SQRTime
      @SQRTime 2 ปีที่แล้ว

      Hi Peter. If you want to see similar videos in math competitions, please consider
      th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 2 ปีที่แล้ว

      That won't give you x^8. However, I did it somehow similar, and that gave me x^8.
      x^4 = 9x^2 - 6x + 1 and 3x = x² + 1 ==> x^4 = 7x² - 1
      squaring gives x^8 = 49x^2 - 14x^2 + 1, and we have from the last stept 7x² = x^4 + 1 ==> x^8 = 47x^4 - 1
      squaring again...

  • @mathguruandscienceguru8893
    @mathguruandscienceguru8893 2 ปีที่แล้ว +3

    Proud of you sir i also want to be an mathematician like you👍👍👍👍🇮🇳🇮🇳🇮🇳

  • @petermhart
    @petermhart 2 ปีที่แล้ว +1

    Squaring 47 in your head. There is a cool way of working out squares of numbers between 41 and 59. 47 is 50-3 so 47^2 is (25-3)(100)+3^2=2209. Why does this work? (50+n)^2=2500+2(50n)+n^2= 2500+100n+n^2=(25+n)(100)+n^2. So for example 56^2=3136.

  • @CTJ2619
    @CTJ2619 2 ปีที่แล้ว +2

    UC Berkeley ?? That’s awesome

  • @littlejimmy5060
    @littlejimmy5060 ปีที่แล้ว

    that last question was satisfying

  • @BigOttomatic
    @BigOttomatic 2 ปีที่แล้ว

    I looked at the problem and thought u-sub by defining int(f(x)dx)=F(x). Pretty much the same thing you did but you get nice algebraic steps

  • @anggalol
    @anggalol 2 ปีที่แล้ว +25

    Alternative Solution to #3:
    Assume a and b are the root of x² - 3x + 1 = 0. Then a and b must also be a root of x¹⁶ - kx⁸ + 1 = 0.
    So,
    a¹⁶ - ka⁸ + 1 = 0
    b¹⁶ - kb⁸ + 1 = 0
    --------------------------- -
    (a⁸ + b⁸)(a⁸ - b⁸) - k(a⁸ - b⁸) = 0
    k = a⁸ + b⁸ = 2207

    • @blackpenredpen
      @blackpenredpen  2 ปีที่แล้ว +4

      That’s beautiful!

    • @topgearfan2596
      @topgearfan2596 2 ปีที่แล้ว

      How did you get 2207 in the last step? Do you use quadratic formula to find a and b?

    • @Prxwler
      @Prxwler 2 ปีที่แล้ว

      @@topgearfan2596 a and b were by assumption the roots of the first quadratic, so you just have to solve that one

    • @topgearfan2596
      @topgearfan2596 2 ปีที่แล้ว +3

      @@Prxwler Yeah I get it. I feel like calculating that is more tedious than the method presented in the video.

  • @JP-lz3vk
    @JP-lz3vk 2 ปีที่แล้ว +4

    Another excellent video! Thanks Professor!

  • @Goose____
    @Goose____ ปีที่แล้ว +1

    i love knowing every step used in the video but not being able to string them together to get a solution:))))))))))

  • @ltrizzle12
    @ltrizzle12 2 ปีที่แล้ว +11

    In the first problem:
    If you simply substitute “x” for “log (base 2) n”, you get (3/x) - (x/2) = (-5/2)
    Solve the quadratic and plug back in the 2 solutions (x = -1 & 6), setting them each = to “log (base 2) n”
    Same answer in the end. Just a less complex approach. You can actually solve it just by looking at it. It’s pretty clear what X is equal to. Regardless, more than one way to skin a cat. Per usual with the maths. 🤘😝🤘

    • @natebobdog23
      @natebobdog23 ปีที่แล้ว

      this isn't a different way to do the problem. it's literally the same exact thing except you let x = log_2(n). it doesn't save any time

    • @ltrizzle12
      @ltrizzle12 ปีที่แล้ว +2

      @@natebobdog23 “EXCEPT you let x = log_2 (n)”. THATS the difference. Same answer-yet the slightest of variety in approaching and yielding that same solution.
      Your comment just defeated itself. Congrats. 👏👏👏
      You also conveniently neglected the end of my original statement. Conveniently. No wonder you’re struggling over there.
      But hey, you’re learning! Good for you! I’ll be right here if u ever need me. ✌️

  • @MohitKumar-eu4pz
    @MohitKumar-eu4pz 2 ปีที่แล้ว

    I solved only first one log question and rest two no idea
    But i. Amazed 😮😲 by seeing solution of 3 rd one

  • @PXO005
    @PXO005 2 ปีที่แล้ว +4

    I did the first one another way, probably more complicated:
    () does not represent the base
    First, we can write the eqn as
    1/log2^3(n) + 1/logn(2^-2) = -5/2
    As logn(2^-2) =-2logn(2),,,
    And log2^3(n) =1/3log2(n)
    We get 3/log2(n) - 1/2logn(2) = -5/2
    Then
    As logn(2) =1/log2(n)
    6/log2(n) =log2(n) -5
    Which simplifies down to
    (log2(n)) ^2 -5log2(n) -6=0
    Which yields the solutions n=2^6
    And n= 1/2

    • @tenebrae711
      @tenebrae711 2 ปีที่แล้ว

      Solved this equation in a similar way to yours. Thought it'd be harder, but that's school grade maths

    • @tenebrae711
      @tenebrae711 2 ปีที่แล้ว

      But notice that solution n=1/2 is not integer, so this equation has only one solution n=2^6=64

    • @anshumanagrawal346
      @anshumanagrawal346 2 ปีที่แล้ว +1

      Not -1/2, but 1/2

  • @hxc7273
    @hxc7273 2 ปีที่แล้ว +2

    In the second problem, it's given that f is bijective. But I wonder if it makes sense to think about how you would prove such a function would actually be bijective.

    • @DrCorndog1
      @DrCorndog1 2 ปีที่แล้ว +2

      Sure. I imagine it would be a pretty basic proof by cases. "Suppose f(a) = b. Then either b = a^2 or b = sqrt(a)." Follow from there.

    • @adayah2933
      @adayah2933 ปีที่แล้ว

      Consider any sequence n, n², n⁴, n⁸..., where n itself is not a square. N is a disjoint union of such sequences, so it suffices to prove that f is a bijection on each sequence separately. On any such sequence the function is given by f(n^(2^k)) = n^(2^(k+1)) if k is even, n^(2^(k-1)) if k is odd. From this it is clear that f(f(x)) = x for each x, so f is its own inverse and it is bijective.

  • @Nishi-wc3zu
    @Nishi-wc3zu 2 ปีที่แล้ว

    First was nice bro.😍

  • @kaustubhnarkhede5166
    @kaustubhnarkhede5166 2 ปีที่แล้ว +2

    in the second question is there a generalized way to solve if, for example, it asked less than 500 or a larger number? how would we approach it? we surely can't go about counting every such pair...!?

    • @sharpnova2
      @sharpnova2 2 ปีที่แล้ว

      yeah just find numbers that have natural square roots but not natural fourth roots
      so take the number of squares and subtract the number of tesseracts

  • @shreyjain3197
    @shreyjain3197 ปีที่แล้ว +1

    for the 3rd question, cant you find out the value of x from the quadratic and then substitute in the 2nd eq?

  • @mathguruandscienceguru8893
    @mathguruandscienceguru8893 2 ปีที่แล้ว +1

    Love from india🇮🇳🇮🇳🇮🇳🇮🇳

  • @vibhupandya6103
    @vibhupandya6103 2 ปีที่แล้ว +8

    Hey i have a question for you. This is from an IITJEE prep book by RD Sharma. Lo and behold:
    y'•y'"=3y"²
    I solved it (somehow lol) but later thought, "well doesn't mean that y', sqrt3 y, and y"' are always in GP?" That didn't quite work out well though. I urge you to do this!

  • @ARKGAMING
    @ARKGAMING 2 ปีที่แล้ว

    I knew there had to be a more elegant solution to 3 than just solving the quadratic and making the second expression an equation for k

  • @nguyenkhoanam7386
    @nguyenkhoanam7386 2 ปีที่แล้ว

    You should try to differentiate the function f(x)=x!

  • @mathguruandscienceguru8893
    @mathguruandscienceguru8893 2 ปีที่แล้ว +1

    Very good

  • @shreejipatel2084
    @shreejipatel2084 2 ปีที่แล้ว

    Oh I am lovin' it!

    • @SQRTime
      @SQRTime 2 ปีที่แล้ว

      Hi Shreeji. If you want to see similar videos in math competitions, please consider
      th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊

  • @ministryoftruth8523
    @ministryoftruth8523 2 ปีที่แล้ว

    Awesome!

  • @aayushdhungana360
    @aayushdhungana360 2 ปีที่แล้ว +1

    8:24 this should be a meme! (The pause)

  • @lucyfer5887
    @lucyfer5887 2 ปีที่แล้ว +2

    hi i have a calculus question that i really hope you'll answer because it's annoying me so bad
    when you take the derivative of ln x or ln 2x or any natural log of nx it'll always be 1/x right? so why when we integrate 1/x do we just say the integral is ln x and not some ln ax because it can be any constant multiplied by that x
    please answer and ty

    • @blackpenredpen
      @blackpenredpen  2 ปีที่แล้ว +6

      bc of the +C

    • @DrCorndog1
      @DrCorndog1 2 ปีที่แล้ว +2

      ^ This. I was going to say, "We already do!"

    • @anshumanagrawal346
      @anshumanagrawal346 2 ปีที่แล้ว +2

      Let me ask you a simpler question, the derivative of x^2 is 2x, right? But so is the derivative of x^2+1, and x^2+2, so why then is the integral of 2x equal to x^2 and not x^2 +a, the answer? It is, that's we have the +C in indefinite Integrals, similarly as we have the integral of 1/x is lnx +C, where C is any real numbers, which is also equivalent to adding ln(a) for any positive a

  • @durgeshadurgesha8682
    @durgeshadurgesha8682 2 ปีที่แล้ว

    I am very glad while you change blackpenredpen

  • @jlp8573
    @jlp8573 ปีที่แล้ว

    let's compute the solutions of 1st equation : x1 = (3+sqrt(5))/2 and x2 = (3-sqrt(5))/2. Then k = x1^8 + x2^8 = 2207. (directly, or by Girard-Newton formula)

  • @knights_limit
    @knights_limit 2 ปีที่แล้ว +1

    #questions For the third problem, could you do a u-sub with let u = x^8? Also, I don’t get why we can’t just do the binomial expansion (x^2-3x)^8 = 1^8

    • @Erik-ij5jb
      @Erik-ij5jb 2 ปีที่แล้ว +1

      Because you have a ton of terms such as x^15, x^14 etc with a binomial expansion of your expression. Yeah you could set up a system so that all the unwanted coefficients equal zero but at best it will be too messy and time conseuming and at worst it may not even work (I’m not gonna give it a try).

  • @ATL45
    @ATL45 2 ปีที่แล้ว +1

    I did get all three, but made it harder than needed for #1 (took common log instead of log base 2, then had to substitute k=log4 and factor a multivariable quadratic expression before back-substituting) and for #3 (actually solved for x and found the 8th power--you don't get more and more terms, but you do get larger and larger coefficients--then compared that to the quadratic equation result for the 16th-degree polynomial).

  • @Predaking4ever
    @Predaking4ever 2 ปีที่แล้ว +1

    Cool way to do the last one.

  • @mahaprasadray2265
    @mahaprasadray2265 2 ปีที่แล้ว

    In second question 1 can be taken in both a² or √a form ...so the question should have mentioned least number of numbers less than 40

    • @adayah2933
      @adayah2933 ปีที่แล้ว

      The question is fine...

  • @Unique656
    @Unique656 2 ปีที่แล้ว +1

    Make a video on how to get the value of d/dx[erf(x)]

  • @adandap
    @adandap 2 ปีที่แล้ว +2

    To do 47^2 in your head quickly, try this: (50-3)^2 = 2500 - 300 + 9 =2209

  • @jeanroitelet51
    @jeanroitelet51 2 ปีที่แล้ว

    A very nice sol !

  • @anubhavlamsal5915
    @anubhavlamsal5915 ปีที่แล้ว

    3rd question lai can\t just ake the two roots of the first expression, alpha and beta. so you get
    alpha = (3 +sqrt5) / 2 and beta = (3-sqrt5)/2 use substitution y = x^4 so the sum of roots of y = -k or
    ((3+sqrt5) / 2 )^4 + ((3-sqrt5)/2)^4 = 2207 or k =-2207

  • @alejandrogarciaibanez1172
    @alejandrogarciaibanez1172 2 ปีที่แล้ว +1

    I didnt know Saitama took over this channel

    • @DrCorndog1
      @DrCorndog1 2 ปีที่แล้ว

      Busy man. He's been doing videos for CalisthenicMovement for a long time now.

  • @reidflemingworldstoughestm1394
    @reidflemingworldstoughestm1394 2 ปีที่แล้ว +1

    that was fun

  • @Craig31415
    @Craig31415 2 ปีที่แล้ว

    2:47 Shouldn't it be 5(log2(n)) instead of -5(log2(n)) because you're shifting to the other side?

    • @Crisprian
      @Crisprian 2 ปีที่แล้ว

      nah the -5log2(n) wasnt the one he was shifting, he shifted -(log2(n))² and 6

  • @gillrowley7264
    @gillrowley7264 2 ปีที่แล้ว

    You are amazing

    • @SQRTime
      @SQRTime 2 ปีที่แล้ว

      Hi Gill. If you want to see similar videos in math competitions, please consider
      th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊

  • @logiciananimal
    @logiciananimal 2 ปีที่แล้ว

    Who is the target participant for the SMT? I can see some good high school (grade 11 or so) students being able to do 1 and 3, but I don't think 2 would be suitable - maybe if "bijection" was replaced with "1-1 and onto".

  • @nimmira
    @nimmira 2 ปีที่แล้ว +1

    I like #3

  • @zohramartini9425
    @zohramartini9425 2 ปีที่แล้ว

    Sorry but 6*^ isn't it 36? I did not fully understand the step from logn2 to finding n. Thank you for your answer

  • @EE-ho1iz
    @EE-ho1iz 2 ปีที่แล้ว +1

    You look like a math master just came down from the mountain or something XD.

  • @javisings6960
    @javisings6960 2 ปีที่แล้ว +2

    Can you integrate (1/dx) ?

    • @sharpnova2
      @sharpnova2 2 ปีที่แล้ว

      that's not an integral. dx isn't a term to be wielded arbitrarily

  • @excelelmira
    @excelelmira 2 ปีที่แล้ว

    The first two problems are somewhat doable, but I would absolutely have no chance tackling the last problem.

  • @alexedrichlim828
    @alexedrichlim828 2 ปีที่แล้ว

    Tutorial on partial differentiation please🥺

  • @MathePhysikbyBasti
    @MathePhysikbyBasti 2 ปีที่แล้ว

    Nice one

  • @razvanfodor9203
    @razvanfodor9203 2 ปีที่แล้ว

    Try solving ln(i)

  • @GlorifiedTruth
    @GlorifiedTruth 2 ปีที่แล้ว

    But for the second question, can we rule out that f(1) =/= 1^2? Since it can go either way, isn't it a "maybe"? I would have written 4, and 5 if you count N = 1. But maybe I'm just stupid.

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 2 ปีที่แล้ว +1

      Huh? For a = 1, we obviously have f(a) = a². So this is not a number we have to count when looking for all numbers for which f(a) != a².

  • @Znwarp
    @Znwarp 2 ปีที่แล้ว +1

    For question 1, that was a 5? I thought that was an S, and figured they were asking about arithmetic progressions or something

  • @qwerty_____146
    @qwerty_____146 2 ปีที่แล้ว +1

    Thank you. guide me about: Integral (e^-x)/x Thank you

  • @ΣπύροςΦετσης-ν8η
    @ΣπύροςΦετσης-ν8η 2 ปีที่แล้ว +1

    can we use the quadratic formula on the 3rd question?

    • @sharpnova2
      @sharpnova2 2 ปีที่แล้ว +5

      yes but what a nightmare that would be

  • @와우-m1y
    @와우-m1y 2 ปีที่แล้ว +1

    best math sam a reader is convert is it

    • @와우-m1y
      @와우-m1y 2 ปีที่แล้ว +1

      sam math best

  • @PunmasterSTP
    @PunmasterSTP ปีที่แล้ว

    Dang, I was lost as to what that second question had been asking.

  • @xXJ4FARGAMERXx
    @xXJ4FARGAMERXx ปีที่แล้ว

    For the last one, x² - 3x + 1 = 0 => x¹⁶ - kx² + 1 = 0, I just said x = y⁸, which means
    (y⁸)² - 3(y⁸) + 1 = 0
    y¹⁶ - 3y⁸ + 1 = 0
    since the symbols arbitrary, we can just replace y with x
    x¹⁶ - 3x⁸ + 1 = 0
    k = 3

    • @edwardzhang1990
      @edwardzhang1990 ปีที่แล้ว

      The symbols are not arbitrary though, since we have specific values of x that solve the quadratic. You can check that the roots of the quadratic don't satisfy the new equation you wrote.

  • @adityaraajsingh4322
    @adityaraajsingh4322 10 หลายเดือนก่อน

    Is there an even quicker way to solve the 2nd one and if the 40 was replaced by 400 or something ........just wanted to ask ........pls reply

  • @iAzazelHD
    @iAzazelHD 2 ปีที่แล้ว

    i dont understand the bijection problem. U assigned 2 to 4, so the 16 couldnt have the 4 again. But why didnt u assign the 4 to the 16 (in blue) and then the 16 to the 4? You u will end up with 5 numbers, not 4

  • @ARKGAMING
    @ARKGAMING 2 ปีที่แล้ว

    I feel so dumb for not being able to solve 1🤦🏻‍♂️... I even though about change of bases but I was just like "nah I'll just give up and watch him solve it"

  • @Djake3tooth
    @Djake3tooth 2 ปีที่แล้ว +1

    6:17 why isn’t zero in there? 😢

  • @justinpark939
    @justinpark939 2 ปีที่แล้ว

    For the second question, can we omit 0 as a natural number?

    • @nurssa5991
      @nurssa5991 2 ปีที่แล้ว

      No 0 is not a natural number

    • @justinpark939
      @justinpark939 2 ปีที่แล้ว

      @@nurssa5991 if you consider the peano axioms, it is.

  • @EragonShadeslayer
    @EragonShadeslayer 2 ปีที่แล้ว

    Hello, I was wondering, when is it okay to divide by x? Because my teacher always tells us, never divide by x because it could be 0, and I’ve mentioned to her that you could just plug in 0 for x to find out whether it is before you divide, but she just told me not to do it.

    • @aayushdhungana360
      @aayushdhungana360 2 ปีที่แล้ว +2

      "she just told me not to do it" - that is obnoxious..... that's EXACTLY what you do.....say x²=4x is the equation.....x=0 is obviously a solution.....instead of dividing by x, you do:
      x²-4x=0
      x(x-4)=0
      x=0 or x-4=0 --> x=4
      Hence, x=0 or 4
      .....in equations like x²-5x+6=0 there is no way x=0 is a solution ....so if dividing by x helps you get a special useful form of the equation (like in this video), there is no reason in the world not to do it......
      BTW what grade are you in rn that your teacher is so restrictive?

    • @EragonShadeslayer
      @EragonShadeslayer 2 ปีที่แล้ว

      @@aayushdhungana360 Thank you! I am in 9th grade right now, however I am taking algebra 2 honors, so my classmates are almost all 10th graders who are slightly advanced, and the actual class is an 11th grade class. I hope that made sense and thank you for the explanation!

    • @neilgerace355
      @neilgerace355 2 ปีที่แล้ว +3

      In the third problem, you can divide by x because 0 doesn't satisfy the quadratic.

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 2 ปีที่แล้ว

      I think most teachers simply get desperate after teaching for years and decades and always having to correct errors like: x² = x is the same as x = 1. So they simply tell _all_ pupils that one should _never_ divide by x. And don't mention that there are cases where this _is_ allowed and even helpful...

  • @HershO.
    @HershO. 2 ปีที่แล้ว

    Could anyone suggest a method to prepare for this year's SMT? Mainly I struggle in combinatorics so would be nice if someone could suggest a good source.

    • @SQRTime
      @SQRTime 2 ปีที่แล้ว

      Hi Hesh. We have a lot of videos from SMT and other countries in our channel. please consider
      th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊

  • @shotdown5191
    @shotdown5191 ปีที่แล้ว

    why is it allowed to divide by x in the last question?