Lilly's Technique
Lilly's Technique
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Nice Algebra Equation Solving | You should be able to solve this!#olympiad #maths
Nice Algebra Equation Solving | You should be able to solve this!#olympiad #maths
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Nice Algebra Equation Solving | You should be able to solve this!#olympiad #maths Hey there, amazing family! 🌈✨ I hope everyone is doing fantastic today! 🤗💖 If you like this video about how to solve this math problem! 🎉🔢 Please Like and Subscribe my channel. Your support means everything to me and keeps me motivated to create more educational content for all of us! 🙌 📚✏️ #matholympiad #algebra ...
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Nice Math Simplification Problem | You should be able to solve this !! #olympiad Hey there, amazing family! 🌈✨ I hope everyone is doing fantastic today! 🤗💖 If you like this video about how to solve this math problem! 🎉🔢 Please Like and Subscribe my channel. Your support means everything to me and keeps me motivated to create more educational content for all of us! 🙌 📚✏️ #matholympiad #algebra #...
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Math Olympiad Question | Nice Algebra Equation Solving | You should be able to solve this!#olympiad Hey there, amazing family! 🌈✨ I hope everyone is doing fantastic today! 🤗💖 If you like this video about how to solve this math problem! 🎉🔢 Please Like and Subscribe my channel. Your support means everything to me and keeps me motivated to create more educational content for all of us! 🙌 📚✏️ #math...
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Nice Math Simplification Problem | You should be able to solve this !! #olympiad #maths Hey there, amazing family! 🌈✨ I hope everyone is doing fantastic today! 🤗💖 If you like this video about how to solve this math problem! 🎉🔢 Please Like and Subscribe my channel. Your support means everything to me and keeps me motivated to create more educational content for all of us! 🙌 📚✏️ #matholympiad #al...
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ความคิดเห็น

  • @johnspathonis1078
    @johnspathonis1078 วันที่ผ่านมา

    ??? The quotient in the problem was 8 yet it mysteriously changed to 18 when working out the solution??

    • @LillysTechnique
      @LillysTechnique วันที่ผ่านมา

      Sorry for the mistake,,

    • @johnspathonis1078
      @johnspathonis1078 วันที่ผ่านมา

      @@LillysTechnique No worries. I only commented as I had spent time working on the problem as it was posted. Cheers.

    • @TruongNguyen-te3gy
      @TruongNguyen-te3gy 17 ชั่วโมงที่ผ่านมา

      Let’s call this “ fault advertising “

  • @MathEducation100M
    @MathEducation100M 10 วันที่ผ่านมา

    Nice

  • @ToalmKriuck
    @ToalmKriuck 13 วันที่ผ่านมา

    Ok but is no one gonna talk about how they over complicate it by making the numbers sqrt(x^2)

  • @unidentified7164
    @unidentified7164 13 วันที่ผ่านมา

    This sh*t cannot be a math olympiad question. Just stop fooling people into wasting time on fake olympiad questions.

  • @nikitaluzhbin8982
    @nikitaluzhbin8982 14 วันที่ผ่านมา

    b = -(x1 + x2) c = x1*x2 Vieta theorem

  • @halneufmille
    @halneufmille 15 วันที่ผ่านมา

    For those interested, I the four answers are: x=2, x=-11, x=(-9+√(-159))/2, x=(-9-√(-159))/2. I found the two real solutions using Gauss method.

  • @mathiq56
    @mathiq56 15 วันที่ผ่านมา

    So good solve

  • @deathwing3087
    @deathwing3087 16 วันที่ผ่านมา

    Other solution : Easy to see that we have to multiply 4 successive integers, so we expect the result to be more or less a power of 4 of the integers. It is then easy to test some integers. 6^4 = 36*36 = 1296, we are almost there ! Then we take the 4 numbers as slightly above a 6*6*6*6, let's try 5*6*7*8. It works :)

  • @genevaspring8974
    @genevaspring8974 16 วันที่ผ่านมา

    Practical solution I found. (x+6)!/(x+2)! = (x+6)(x+5)(x+4)(x+3). you work with factorials, the answer is expected to be integer. Decompose. 1680 = 2*2*2*2*3*5*7. Reorganise as a product of 4 following integers, with obviously 7 in them (you wont have one of the numbers begin 14 or 21), you find quicly 1680 = 5*6*7*8. x+3=5, so x=2.

  • @user-ee7nw2rx9s
    @user-ee7nw2rx9s 16 วันที่ผ่านมา

    Легко решить если учесть что 11-sqrt120=(sqrt 6-sqrt 5)^2 Тогда sqrt (sqrt 5+sqrt 6)*sqrt (sqrt 6-sqrt 5)=sqrt ((sqrt 6-sqrt 5)*(sqrt 6?sqrt 5))=sqrt (6-5)=1 30 секунд, а вы решаете 5 минут

  • @ulf-nicklassdegenhardt-mei3121
    @ulf-nicklassdegenhardt-mei3121 16 วันที่ผ่านมา

    After watching the video: Yes, you can do it the complicated way, but... Just a little bit thinking outside formulas would have been easier...

  • @ulf-nicklassdegenhardt-mei3121
    @ulf-nicklassdegenhardt-mei3121 16 วันที่ผ่านมา

    Well... This translates to: Find the 4 consecutive numbers, whose product is 1680, and substract 2 of the smallest to get X... Easy...

  • @alfredovargas3228
    @alfredovargas3228 16 วันที่ผ่านมา

    Una demostración clara de este problema.

  • @corentinb4894
    @corentinb4894 18 วันที่ผ่านมา

    Nicely done! I did this one a bit differently: I decided to do A=(2^333 * 111^333)/(3^222 * 111^222) and prove it is >1 It easily simplifies to ((8^111)/(9^111)) * 111^111 which I rearranged as ((11^111)/(9^111)) * 8^111. Now 11/9 > 1 => (11^111)/(9^111) > 1 and 8^111 > 1 => A > 1 and thus 222^333 > 333^222 It is quite similar in the end, but I think I prefer your technique :)

  • @JohnDoe-ti2np
    @JohnDoe-ti2np 18 วันที่ผ่านมา

    One can guess that 222^333 must be a lot bigger because the exponent has a much bigger effect than the base. To make this rigorous, note that 333^222 < 1000^222 = 10^666. On the other hand, 222^333 > 100^333 = 10^666.

  • @michaelmuller4356
    @michaelmuller4356 19 วันที่ผ่านมา

    Nice

  • @ANILDAS-ow1ii
    @ANILDAS-ow1ii 20 วันที่ผ่านมา

    9

  • @WPope
    @WPope 20 วันที่ผ่านมา

    Why does the x look like a n?

  • @zaidtalib4513
    @zaidtalib4513 22 วันที่ผ่านมา

    why u complicating this 1/0.5 is 2 so solve from there

  • @Heartsii_
    @Heartsii_ 23 วันที่ผ่านมา

    At step 5, where we say n^3 + n^2 - (3^3 + 3^2) = 0, why can't we move the (3^3 + 3^2) across the = to show n^3 + n^2 = 3^3 + 3^2? At that point, isn't it obvious that n=3?

    • @phaustho
      @phaustho 23 วันที่ผ่านมา

      It looks pretty obvious, I see your point. However, the equation has a cube power in it, which means that there could be up to 3 different solutions. Solving the problem implies finding all possible values for N (and a), not just the obvious one.

    • @sohamkumar2346
      @sohamkumar2346 22 วันที่ผ่านมา

      Yes, I was thinking the same thing...

  • @alejandropinto8130
    @alejandropinto8130 23 วันที่ผ่านมา

    Interesting

  • @LillysTechnique
    @LillysTechnique 24 วันที่ผ่านมา

    9

  • @ShaiyanKhan-ur6zc
    @ShaiyanKhan-ur6zc 25 วันที่ผ่านมา

    23

  • @savitrishah1955
    @savitrishah1955 25 วันที่ผ่านมา

    23

  • @savitrishah1955
    @savitrishah1955 25 วันที่ผ่านมา

    😊😊

  • @diptir3354
    @diptir3354 25 วันที่ผ่านมา

    9

  • @ThienVu3108
    @ThienVu3108 25 วันที่ผ่านมา

    you are making it complicated when it can be solved easily😂 5x/ √0.25 = 200. √0.25 = 0.5 200 x 0.5 = 5x 100 = 5x X = 100/5 = 20

  • @jayashreesarkar2236
    @jayashreesarkar2236 26 วันที่ผ่านมา

    13

  • @ParosGabriel63539
    @ParosGabriel63539 26 วันที่ผ่านมา

    13

  • @charlesokuom8747
    @charlesokuom8747 26 วันที่ผ่านมา

    Well evaluated

  • @user-cy7ui3ev7n
    @user-cy7ui3ev7n 26 วันที่ผ่านมา

    X=2

  • @jayadsilva67
    @jayadsilva67 27 วันที่ผ่านมา

    Is this the silliest question?

  • @RexxSchneider
    @RexxSchneider 27 วันที่ผ่านมา

    That formula only works when c = b + 1. Check out b + 1 = (n^2 - 1)/2 + 1 = (n^2 + 1)/2 = c. Note that a has to be an odd integer because the difference between consecutive squares is odd. If you're going to guess that that particular formula works, then you might as well guess b = c-1 = 612 and you find that a = √(613^2 - 612^2) = 35 immediately. You can make lots of formulae for Pythagorean triples by setting c=b+2 or c=b+3, etc. For example pick n to be even then a = n; b = (n^2 - 4)/4; and c = (n^2 + 4)/4 = b+2. If you want c=b+k, then a^2 = c^2-b^2 = 2bk+k^2. So you get a Pythagorean triple whenever you can pick k such that b = (a^2-k^2)/2k is an integer.

  • @danielduranloosli
    @danielduranloosli 27 วันที่ผ่านมา

    Wouldn't you need to state all pythagorean triplets have that form (2n; n^2-1; n^2+1) in order to be sure those are all the possible solutions?

    • @BabySarkar-tc1vt
      @BabySarkar-tc1vt 27 วันที่ผ่านมา

      Only for the positive integers solution,, I think this is the formula

  • @ThiruThiru-ed1ce
    @ThiruThiru-ed1ce 27 วันที่ผ่านมา

    2

  • @BabySarkar-tc1vt
    @BabySarkar-tc1vt 27 วันที่ผ่านมา

    13

  • @jayashreesarkar2236
    @jayashreesarkar2236 28 วันที่ผ่านมา

    27

  • @freedom8589
    @freedom8589 28 วันที่ผ่านมา

    b

  • @shreyaslokhande8010
    @shreyaslokhande8010 28 วันที่ผ่านมา

    bro...u just made the problem evn bigger :I

  • @adgod725
    @adgod725 29 วันที่ผ่านมา

    Hey, there's so many things wrong with this solution. First, it seems that you assume x, y, z are integers, otherwise your division thing at start wouldn't make any sense. But even if they are integers, (1 + 2^y/2^x + 2^z/2^x) can still be non integer, you have to assume x <=y and x <= z (it can be done because of symmetry) Even if you do so, that part can still be even if x=y, so x can be either 0, 1, 2, 3 (and you have to explain why they are not negative) and you have to check them all.

  • @namavenkatalakshmi9266
    @namavenkatalakshmi9266 หลายเดือนก่อน

    36

  • @user-hg7jp2ii9g
    @user-hg7jp2ii9g หลายเดือนก่อน

    59

  • @natarajanrajamanickam6602
    @natarajanrajamanickam6602 หลายเดือนก่อน

    23

  • @user-bn2ei9rk1x
    @user-bn2ei9rk1x หลายเดือนก่อน

    (10+20+24)÷2

  • @PElavarasi-ty5nu
    @PElavarasi-ty5nu หลายเดือนก่อน

    🐕 - 17, 😺 7 ,🐰_3.

  • @Learn_English_speaking-63.
    @Learn_English_speaking-63. หลายเดือนก่อน

    27

  • @BabySarkar-tc1vt
    @BabySarkar-tc1vt หลายเดือนก่อน

    Very nice

  • @andrewjknott
    @andrewjknott หลายเดือนก่อน

    Instead of changing base to 3, note that 3 = 9^0.5 and substitute that value for 3 in the denominator.