Nice Math Olympiad Problem a^11+b^11=?

Рет қаралды 5,639

Ай бұрын

The Math Olympiad is a prestigious competition that tests the mathematical prowess of high school students worldwide. Teams from various countries, including the USA Math Olympiad Team, compete annually, aiming for success at the International Math Olympiad.
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Пікірлер: 26

@roger7341 11 күн бұрын

I don't have any tricks up my sleeve, so I'll just try a=1/2+z and b=1/2-z. Substitute into the second equation: (1/2+z)^2+(1/2-z)^2=1/2+2z^2=2. Thus, z=±√3/2, a=1/2±√3/2 and b=1/2∓√3/2. It is now easy enough to raise a and b to higher powers and cancel out some terms, but I'll leave that as an exercise. I'm guessing that the answer is around 30+29/32 or so.

@glorfindel2740 8 күн бұрын

I have just 1 problem with the video: 989/32 is NOT strictly equal to 30.91. Its equal to 30.90625. And in math problems, you are required to give the exact, not approximated solution. I think you should have just left 989/32. Its fine.

@NadiehFan 17 күн бұрын

My favourite approach to solving these kinds of problems is using recurrence relations. Here is how I solved this problem. We are given that a + b = 1, a² + b² = 2 and are required to find the value of a¹¹ + b¹¹. As a preliminary remark, note that 2ab = (a + b)² − (a² + b²) = 1 − 2 = −1 so ab = −¹⁄₂. Therefore, (a − b)² = (a + b)² − 4ab = 1² − 4·(−¹⁄₂) = 3 so a and b are real. Let us define tₙ = aⁿ + bⁿ for any positive integer n, so we have t₁ = 1, t₂ = 2 and want to find the value of t₁₁. First, since (aⁿ + bⁿ)(a + b) = (aⁿ⁺¹ + bⁿ⁺¹) + ab(aⁿ⁻¹ + bⁿ⁻¹) we have tₙ·t₁ = tₙ₊₁ + ab·tₙ₋₁ and since t₁ = 1, ab = −¹⁄₂ this gives tₙ = tₙ₊₁ − ¹⁄₂·tₙ₋₁ and therefore (1) tₙ₊₁ = tₙ + ¹⁄₂·tₙ₋₁ so the sequence {tₙ} satisfies a second order linear homogeneous recurrence with constant coefficients. Secondly, we have a²ⁿ + b²ⁿ = (aⁿ + bⁿ)² − 2·aⁿbⁿ and therefore (2) t₂ₙ = tₙ² − 2·(−¹⁄₂)ⁿ Thirdly, we have a³ⁿ + b³ⁿ = (aⁿ + bⁿ)³ − 3·aⁿbⁿ·(aⁿ + bⁿ) and therefore (3) t₃ₙ = tₙ³ − 3·(−¹⁄₂)ⁿ·tₙ From (2) and the known value t₂ = 2 we get t₄ = 2² − 2·(−¹⁄₂)² = 4 − ¹⁄₂ = ⁷⁄₂ t₈ = (⁷⁄₂)² − 2·(−¹⁄₂)⁴ = ⁴⁹⁄₄ − ¹⁄₈ = ⁹⁷⁄₈ From (3) and the known value t₁ = 1 we get t₃ = 1³ − 3·(−¹⁄₂)·1 = 1 + ³⁄₂ = ⁵⁄₂ t₉ = (⁵⁄₂)³ − 3·(−¹⁄₂)³·(⁵⁄₂) = ¹²⁵⁄₈ + 3·(¹⁄₈)·(⁵⁄₂) = ²⁵⁰⁄₁₆ + ¹⁵⁄₁₆ = ²⁶⁵⁄₁₆ Finally, from (1) and using t₈ = ⁹⁷⁄₈, t₉ = ²⁶⁵⁄₁₆ we get t₁₀ = ²⁶⁵⁄₁₆ + (¹⁄₂)·(⁹⁷⁄₈) = ²⁶⁵⁄₁₆ + ⁹⁷⁄₁₆ = ³⁶²⁄₁₆ t₁₁ = ³⁶²⁄₁₆ + (¹⁄₂)·(²⁶⁵⁄₁₆) = ⁷²⁴⁄₃₂ + ²⁶⁵⁄₃₂ = ⁹⁸⁹⁄₃₂ Of course it is computationally somewhat more efficient if we consider that for m ≥ n ≥ 0 we also have (aᵐ + bᵐ)(aⁿ + bⁿ) = (aᵐ⁺ⁿ + bᵐ⁺ⁿ) + aⁿbⁿ(aᵐ⁻ⁿ + bᵐ⁻ⁿ) which gives tₘ·tₙ = tₘ₊ₙ + (−¹⁄₂)ⁿ·tₘ₋ₙ and therefore (4) tₘ₊ₙ = tₘ·tₙ − (−¹⁄₂)ⁿ·tₘ₋ₙ Using this we only need t₃, t₅, t₆ as stepping stones to calculate t₁₁. Note that (2) is actually a special case of (4) for m = n since t₀ = a⁰ + b⁰ = 1 + 1 = 2.

@grchauvet 12 күн бұрын

That would be my chosen method too

@zetutor 8 күн бұрын

Good Job

@sashagornostay2188 4 күн бұрын

God bless the graphs and geometry

@devanshgupta794 Ай бұрын

When you get a³ + b³, its best to cube them to get a⁹ + b⁹ and then calculate a⁷ + b⁷ by (a³ + b³)² (a+b) Then calc a¹¹ + b¹¹ easily Less calculation required imo

@medoesgames3145 4 күн бұрын

(a³+b³)³ is not a⁹+b⁹

@franciscook5819 6 күн бұрын

Calculate a, b and split the power seems easier ... a+b=1 …(i) a^2+b^2=2 …(ii) Squared => a^2+b^2+2ab=1^2 So 2+2ab=1 So b=-1/2a Sub into (i) gives two solutions a=(1/2) (1+√3) and b=(1/2) (1-√3) (or vice versa) Break the power 11 into 8 and 3: a^2=(1/4) (1+3+2√3) a^2=(1/2) (2+√3) and b^2=(1/2) (2-√3) a^3=(1/4) (2+3+(1+2) √3) a^3=(1/4) (5+3√3) and b^3=(1/4) (5-3√3) a^4=(1/4) (7+4√3) and b^4=(1/4) (7-4√3) a^8=(1/16) (49+4.4.3+2.4.7√3) a^8=(1/16) (97+56√3) and b^8=(1/16) (97-56√3) a^11=(1/64) (97+56√3)(5+3√3) a^11=(1/64) (989+571√3) and b^11=(1/64) (989-571√3) a^11+b^11=2 (1/64) 989+(1/64) 571√3-(1/64) 571√3 a^11+b^11=(1/32) 989

@zetutor 6 күн бұрын

That works

@Mecha_Math 26 күн бұрын

Nice🎉

@zetutor 13 күн бұрын

Thanks

@JeffryBezos-oy3nd Ай бұрын

Is there any other way to do this faster or is this the only way?

@liliancourjal4303 Ай бұрын

What i found is : a≠b as 2(1/2)^2≠2 (rember this it will be useful later) now, we Will be looking at a possibility for a so as a+b=1, a=1-b now, we solve the second equation by remplacing a by 1-b so (1-b)^2+b^2=2, i'll let you solve that as it would be way to long to solve it here. So you should end up with something like (1± sqrt3)/2. By solving 1-(1± sqrt3)/2, we find that a = (1± sqrt3)/2. As b≠a, one will be - and the other +. Now that we have a and b, we can do the calculus in the end easily. If you need me to clarify tell me because i tried to make it as short as i could but you have the main idea

@liliancourjal4303 Ай бұрын

My bad you need a calculator to do my version

@RONNIN667 Ай бұрын

you can turn the first equation into a = 1-b, then substitue that in the 2nd equation and do the equation and find the result

@echandler Ай бұрын

let f_n = a^n + b^n. This forms a 2nd order linear recurrence. (a+b)*f_n = a^(n+1) + b^(n+1) + a*b^n + b*a^n. Hence f_(n+1) = f_n * f_1 - f_(n-1) * (ab). We have f_0 = 2, f_1 = 1, f_2=2 giving (ab)=-1/2 So f_(n+1) = f_n + (1/2) f_(n-1). f_n = {2 1 2 5/2 7/2 19/4 13/2 71/8 97/8 265/16 181/8 989/32 ..} Depending on how fast you can do the arithmetic, this method may be faster.

@colt4667 Ай бұрын

@@RONNIN667 It's not easy. You get b=1+sqrt 3 and a=-sqrt 3 OR b= 1- sqrt3 and a= sqrt 3 Then you have to compute a^11 and b^11 (1+ sqrt 3)^11 yields 12 terms. For example the 8th term is 11*10*9*8*7*6*5*(11^4)*((sqrt 3)^7) / 7! Then collect all the like terms and simplify. Professor Zetutor shows us an easier way.