A nice math Olympiad problem |you should try to solve this

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Комментарии : 115

@aminimam5118 6 месяцев назад

b=1-a substitute for b in a^2+b^2=2. You will get 2*a^2-2*a-1=0 you can find a and b

@giuseppecaviggia4142 6 месяцев назад

Questo è il modo migliore, non quello del video, che non so dove porti.

@filipeoliveira7001 5 месяцев назад

Yeah, but it takes literally hours to raise it to the power of 11 and calculate🤦‍♂️🤦‍♂️

@shingshing01 2 месяца назад

The question is to find a solution to the differential equation, a" + b" =?. If it is understand that a and b can be functions then finding numerical values for a and b may be a part of the solution but will not determine what function a and function b are.

@shivkumaragarwal.1803 Месяц назад

@@shingshing01It's a¹¹+b¹¹ not a''+b''.

@turnandfacethedragon Месяц назад

This is what I did. Find a and b, square them for a^2 and b^2, square them for a^4 and b^4, square them for a^8 and b^8, multiply the ^4 by the ^8 and divide by the ^1's gives you a^11 and b^11. All you ever have are binomials over a numeric denominator and the a's and b's are the same except one is + while the other is -. Really wasn't complicated at all.

@dneary 4 месяца назад

There's a nice way using Viete's formulas - a,b are the roots of the equation x^2-(a+b)x + ab = 0, after finding ab=-1/2 you get 2x^2-2x-1 = 0 or x^(n+2)=x^(n+1) + (x^n)/2 - always true for roots a,b, giving the recursive relationship a^(n+2) + b^(n+2) = (a^(n+1) + b^(n+1)) + (a^n + b^n)/2.

@EasyMathSteps-SolveItEasy 3 месяца назад

Nice problem, nice solution!

@MegaSARITE 6 месяцев назад

Let a+b =S and ab=P. In the second equation a^2+b^2 =(a+b)^2-2ab=2 leads to ab=-1/2. If a and b are solutions of the quadratic equation X^2+PX+S=0, that give us a,b =(1±√3)/2.

@Ananya-lq9vq 4 месяца назад

What I was thinking tbh

@thunderpokemon2456 2 месяца назад

@@MegaSARITE nah ab is -1

@MegaSARITE 2 месяца назад

@@thunderpokemon2456 recalcute and verify again!

@thunderpokemon2456 2 месяца назад

@@MegaSARITE yeah sry it negative 1 by 2

@lifee2299 Месяц назад

2 equations, 2 parameters enough to solve. (a-b)^2 = a^2 + b^2 - 2ab (a-b)^2 = 2 - 2*(-1/2) = 3 a-b = sqrt(3) a+b = 1 Add above 2a = 1+sqrt(3) Hence a = (1+sqrt(3))/2 = 1.366 Hence b = 1-1.366 = -0.366 Therefore a^11 + b^11 = 31.

@karolissad.4270 5 месяцев назад

DUDE YOU NEED TO DRAW 1'S NOT LIKE STICKS CAUSE I THOUGHT YOU WERE TAKING THE DOUBLE DERIVATIVE OF A AND B!!!

@filipeoliveira7001 5 месяцев назад

Same😭

@gilbertodeoliveirafrota5345 4 месяца назад

not for gymnasium students

@thomasharding1838 4 месяца назад

YEAH! And in the answer it says g8g divided by 32. I never saw where the value of "g" was determined!!!

@filipeoliveira7001 4 месяца назад

@@thomasharding1838are u stupid lol it’s 989

@zyklos229 4 месяца назад

@@gilbertodeoliveirafrota5345hm? Not?

@MartijnLankhorst 4 месяца назад

Let Sn be a^n + b^n Then you can show Sn+1 = Sn + 1/2Sn-1 S11 can be calculated with primary school math starting from S1 and S2

@dneary 4 месяца назад

And S_0 = a^0 + b^0 too!

@prosenjitdas8269 Месяц назад

Step 2 ::: (a+b)^2= (a-b)^2+4ab =1 (a-b)^2=(a+b)^2-4ab=1 a-b=1 Step 3:::: a+b=1 a-b=1 By adding a=1 So b=0 Therefore, a^11+b^11=1^11+0^11=1

@Hero-00001 Месяц назад

😅😅bro u have mad the error in step-2 (a-b)= ±√3

@rasmiranjansahoo6607 4 месяца назад

Time taking and lengthy process. If you can find value of a-b by the formula (a-b)^2=(a+b)^2-4ab, after that find the value of a & b , finally you put the value get the answer

@ADthehawk 3 месяца назад

You could solve for a and b right after step 2. It would probably be easier (have to check). Also, this cannot be a math Olympiad question. This is something I was doing back in 9th grade or so for normal school exams.

@n.662 3 месяца назад

Music no need for this video. You can say) . Think you!

@satyanarayanakenguva3465 6 месяцев назад

Long proces

@QuiDocetDiscit Месяц назад

I grew a beard watching this...but still enjoyed it.

@mks-learning-easy Месяц назад

🙏🙏

@rajurkarj 4 месяца назад

Perfect solution..👍. Here there is no other shortcut trick.. What is the point to find value of a and b?

@zyklos229 4 месяца назад

The point is, that both solutions are equal ... technically each valid expression for "?" is a solution from math point of view.

@johnstanley5692 3 месяца назад

alternative? 2*x^2-2*x-1 =0 & x^11+y^11= (2048*x^22-1)/(2048*x^11) . Use synthetic division (by 2*x^2-2*x-1) on both numerator and denominator to obtain remainders. i.e. x^11+y^11 -> (1129438*x+413402)/ (36544*x+13376) = 989/32 .

@key_board_x 2 месяца назад

a + b = 1 ← this is the sum (S) (a + b)² = a² + 2ab + b² (a + b)² = (a² + b²) + 2ab 1² = (2) + 2ab 1 = 2 + 2ab 2ab = - 1 ab = - 1/2 ← this is the product (P) a & b are the solution of the equation: x² - Sx + P = 0 → where S is the sum and where P is the product x² - x - (1/2) = 0 x² - x + (1/4) - (1/4) - (1/2) = 0 x² - x + (1/4) = 3/4 [x - (1/2)]² = 3/4 x - (1/2) = ± (√3)/2 x = (1/2) ± (√3)/2 → the 2 solutions are: a = (1 + √3)/2 b = (1 - √3)/2 a² = (1 + √3)²/2² = (1 + 2√3 + 3)/4 = (4 + 2√3)/4 = (2 + √3)/2 b² = (1 - √3)²/2² = (1 - 2√3 + 3)/4 = (4 - 2√3)/4 = (2 - √3)/2 a³ = a * a² = [(1 + √3)/2] * [(2 + √3)/2] = (2 + √3 + 2√3 + 3)/4 = (5 + 3√3)/4 b³ = b * b² = [(1 - √3)/2] * [(2 - √3)/2] = (2 - √3 - 2√3 + 3)/4 = (5 - 3√3)/4 a⁴ = (a²)² = (2 + √3)²/2² = (4 + 4√3 + 3)/4 = (7 + 4√3)/4 b⁴ = (b²)² = (2 - √3)²/2² = (4 - 4√3 + 3)/4 = (7 - 4√3)/4 a⁸ = (a⁴)² = (7 + 4√3)²/4² = (49 + 56√3 + 48)/16 = (97 + 56√3)/16 b⁸ = (a⁴)² = (7 - 4√3)²/4² = (49 - 56√3 + 48)/16 = (97 - 56√3)/16 a¹¹ = a³ * a⁸ a¹¹ = [(5 + 3√3)/4] * [(97 + 56√3)/16] = (485 + 280√3 + 291√3 + 504)/64 = (989 + 571√3)/64 b¹¹ = b³ * b⁸ b¹¹ = [(5 - 3√3)/4] * [(97 - 56√3)/16] = (485 - 280√3 - 291√3 + 504)/64 = (989 - 571√3)/64 a¹¹ + b¹¹ = [(989 + 571√3)/64)] + [(989 - 571√3)/64] a¹¹ + b¹¹ = (989 + 571√3 + 989 - 571√3)/64 a¹¹ + b¹¹ = (2 * 989)/64 a¹¹ + b¹¹ = 989/32

@mks-learning-easy 2 месяца назад

Very nice ❤️❤️

@mohanpujar7403 Месяц назад

Brilliant mathematician 🎉 He rekindled my interest in mathematics at the age of 66 y

@mks-learning-easy Месяц назад

❤️❤️

@MathSync 4 месяца назад

i ❤ Mathematics

@Ramyleithy 5 месяцев назад

ليه المرار الطافح ده ، استخدم نظرية ذات الحدين واخلص

@jonathanwinesky1334 2 месяца назад

nice

@mohinkhan2503 4 месяца назад

@thunderpokemon2456 2 месяца назад

A+B=1 (A+B)^2=1^2 A^2+B^2+2AB=1 A^2+B^2=-2AB+1 1=-2AB AB=-1/2 A=-1/2b A^11+B^11= -1/2B^11+B^11=2B^121-1/2B^11

@mks-learning-easy 2 месяца назад

❤️❤️

@rumabhadra8941 6 месяцев назад

it is very easy to find the value of a-b

@shingshing01 2 месяца назад

But you are not trying to find values for a and b. You are trying to find what a" + b"equals.

@faranakhter007. 5 месяцев назад

2 ans

@carloheinz6465 2 месяца назад

If a+b=1 then 2×(a+b)=2=(a²+b²) 🤔 not so hard from here on.... 2a+2b-a²-b²=0 ....etc😊

@moh5463 5 месяцев назад

Solutions a=(1+3^1/2)/2 b=(1-3^1/2)/2 If you swap a and b, that is also a solution.

@prime423 3 месяца назад

You could use the a to the fifth equation to the same benefit.

@dailang2144 3 месяца назад

smart way but ... too "dump". Easy to find a, b => done :-)

@stones-qn2uj 2 месяца назад

Bro wrong answer cause at the end you have made a calculation mistake 247/8 + 1/32 = 988/32 = 247/8

@EpochRazael 2 месяца назад

Bruh. 247/8 = 988/32. 988/32 + 1/32 = 989/32. He's right.

@daddykhalil909 5 месяцев назад

1:07 too boring and complicated

@JJ_TheGreat 5 месяцев назад

11 (??)

@rexford9019 2 месяца назад

((sin(23π/12)·√2)^11)+ ((cos(23π/12)·√2)^11)=989/32

@michellewilson4217 4 месяца назад

Too long , attention span is 3 seconds max.🤣🤣🤣🧐🧐

@ugoc3300 2 месяца назад

And we never defined a or b. Interesting

@cogicube 5 месяцев назад

C’est vraiment passer derrière son genou pour se gratter la tête. 😂

@murvetaykac7041 4 месяца назад

2789/32 is result

@murvetaykac7041 4 месяца назад

1789/32.

@Rkssingh12121 Месяц назад

I would say 11😂

@girlfridayz5148 4 месяца назад

Algebra hates it

@olivier7660 3 месяца назад

Change music ?

@ramadevi7820 6 месяцев назад

What is 13multiplied by13,you did not check also

@JJ_TheGreat 5 месяцев назад

169

@tarek-md2mm 2 месяца назад

time wasted, too long, not the right way to solve this 😧

@rosslogan4154 2 месяца назад

The music is REALLY annoying

@nuvinpooliyadde7319 2 месяца назад

How would you solve this question for an exam? 😁

@niranjanchakraborty1139 2 месяца назад

Ans=2. , as a=1. & b=1. a^11+b^11=1^11+1^11=1+1=2.

@KaushikDutta-m4b Месяц назад

Wrong. That would give a+b=2. But here a+b=1.

@prosenjitdas8269 Месяц назад

Too much lengty.

@AllDogsAreGoodDogs 4 месяца назад

You never said anything about 2ab, thus you changed the terms of the puzzle. Plus you misspelled "equation" early on, and your 1's look horrid. The answer is a=1.5 and b=-0.5. This fits the two terms of the puzzle that was stated. Oh, and your "music' is repetitive.

@mks-learning-easy 4 месяца назад

Thank you for your feedback 😊

@AllDogsAreGoodDogs 4 месяца назад

@@mks-learning-easy You are welcome! The best of luck to you!

@atulbesra822 2 месяца назад

Childish maths problems are passed off as maths Olympiad problems for gullible viewers. Maths Olympiad, like the sporting Olympiads are tough and beyond the abilities of normal humans.

@jairamkhambadkone6423 2 месяца назад

Answer is 2

@thomaswalczak993 4 месяца назад

Besides, to multiply equations, you bring all terms to one side (e.g. a+b-1)=0 and then multiply them. You don't just multiply left terms with left terms and right terms with right terms. In other words, your solution is horse shit.

@prosenjitdas8269 4 месяца назад

@bornasarrafan4980 3 месяца назад

Dumb way to solve this, there is a much easier way to solve it

@dirk_t_wachter 3 месяца назад

Show it to us

@cd1168 2 месяца назад

Music is horrible

@GodIsEverPowerful 5 месяцев назад

RHS value = power of a or b= 11 Don't think too much

@anatolykatyshev9388 4 месяца назад

I am always puzzled when I see solutions like that. Who the hell need it? You have two equations with two variables. Easily find both of them as (1+=sqrt(3)/2 and after any of their combination. How twisted mind should people have to post 16 minutes video with crazy computations for that task?

@zyklos229 4 месяца назад

Now he got a result in Q where your result needs to be assumed to be in R ... unless you can get dont get rid of the "twisted mind" root in the term (x+sqrt(y))^11 + (x-sqrt(y))^11 ... whereas with binomial sum at least the odd exponents for sqrt(y) should cancel and the even are integer then 🤔

@anatolykatyshev9388 4 месяца назад

@@zyklos229Too complicated. In this particular case quadratic equation has 2 real roots. No other complex roots exists. To raise them in power 11 is not rocket science...

@MathSync 4 месяца назад

“Fantastic explanation! I love how you break down complex concepts. I do similar math tutorials and challenges on my channel. Check it out for a fresh perspective on math problems!”

@johnstanley5692 3 месяца назад

another solution: let d = (a+b-1=0), given g= (a^2+b^2-2=0) get remainder g/d = 2*b^2-2*b-1 =0 => b=1/2 +/- sqrt(3)/2. use synthetic division (a^11+b^11)/(a+b-1) and get remainder R(b): R(b)=11*b^10 - 55*b^9 + 165*b^8 - 330*b^7 + 462*b^6 - 462*b^5 + 330*b^4 - 165*b^3 + 55*b^2 - 11*b + 1. now use synthetic division R(b)/(b-(1/2+/-sqrt(3)/2) -> 989/32 => final result (a^11+b^11)=989/32)