Hardest Exam Question | Only 8% of students got this math question correct

[@guyhoghton399]

_x = ½(1 + √5) = ½( -(-1) + √( (-1)² - 4(1)(-1) )_ ∴ _x² + (-1)x + (-1) = 0_ ⇒ _x² = x + 1_ Multiply through by _x:_ _x³ = x² + x = (x + 1) + x = 2x + 1_ ⇒ _x⁶ = (2x + 1)² = 4x² + 4x + 1 = 4(x + 1) + 4x + 1 = 8x + 5_ ⇒ _x¹² = (8x + 5)² = 64x² + 80x + 25 = 64(x + 1) + 80x + 25 _ _= 144x + 89 = 72(1 + √5) + 89_ ∴ *_x¹² = 161 + 72√5_*

[@engjayah]

This is the simplest solution also derived independently

[@buffalobilly6046]

Yes anytime you work with the “magic number” you need to know how to simplify

[@oahuhawaii2141]

But you missed the coefficient of 2 simplifies things immensely: x³ = 2*x + 1 = √5 + 2 So, just square that twice, without having to deal with x in a symbolic manner, and substituting and simplifying yet again. x⁶ = 9 + 4*√5 x¹² = 161 + 72*√5

[@ranjithamarakoon8842]

Nice work. ❤

[@ayushdwivedi1154]

@@buffalobilly6046 can you tell me what is magic number?

[@twinclouds01]

Actually, it can be calculated in 3 steps: Since x^12={[(x^3)^2}^2, we first obtain [(1 + √5)/2]^3 = 2 + √5. Then (2 + √5)^2 = 9 + 4√5 and finally, (9 + 4√5)^2 =81 + 4*2*9√5 +80 = 161+72√5. Even no algebra is needed!

[@AdrienLegendre]

This is also how I would have done it. I agree this is better.

[@robertfitzjohn4755]

That's exactly how I did it. Much simpler than using algebra.

[@am-gobears5191]

Just excellent. And how do you calculate [(1 + √5)/2]^3 = 2 + √5 without algebra, may I ask?

[@UnwiseDolphin]

@@am-gobears5191 You do have to use FOIL to multiply it out. But you don't need to use any variables.

[@tranngochungdevwannabe]

@@UnwiseDolphinthe same anyway. Call it x or call it 5 that is the same. It just that with 5 u could come to answer right away. With x u write the formula down then slap the number in. But all the same

[@Ctrl_Alt_Sup]

φ = (1+√5)/2 where φ is the golden ratio φⁿ=Fₙ₋₁+Fₙφ where (Fn) denotes the Fibonacci sequence φ¹²=F₁₁+F₁₂φ with F₁₁=89 and F₁₂=144 φ¹² =89+144((1+√5)/2)=161+72√5

[@Viktor-m6p]

Exactly ❤❤❤

[@jpl569]

Another proof : noticing that (1+ √5)/2 = φ, and φ^2 = 1 + φ, we have the property φ^n = Un-2 + φ Un-1 for all n ≥ 2, where (Un) is the Fibonacci sequence. Then φ^12 = 89 + 144 φ = 161 + 72 √5. If you do not like finding the Fibonacci elements until n = 11, just take φ^6 = 5 + 8 φ, and square φ^6 : φ^12 = (φ^6)^2 = (5 + 8 φ)^2 = 89 + 144 φ = 161 + 72 √5. Thank you for your videos !

[@Escviitash]

I assume that you by Un-2 mean the (n-2)th Fibonacci number, or as it is usually written "F(n-2)" F(0) = 0 and F(1) = 1, and from there you get the rest of the Fibonacci numbers by F(n) = F(n-1) + F(n-2), and you can even go to in the negative direction by F(n) = F(n+2) - F(n+1). You will the get that F(11) = 89 and F(12) = 144. Compare this to the soltution and you will get that φ^x = F(n-1) + F(n)*φ, which you obviosusly got confused with the formula for finding the Nth Fibonacci number. A couple of examples where n is 0 or negative: Using F(n) = F(n+2) - F(n+1) you will get that F(-1) = 1, F(-2) = -1, F(-3) = 2, F(4) = -3 and so on, which is the same as the Fibonacci numbers in the positve direction, with the exception that they are negative if n is even. So if n = 0 the you get: φ^0 = F(-1) + F(0)*φ = 1 + 0*φ = 1 + 0 = 1 If n = -1 you get: φ^-1 (the reciprocal of φ) = F(-2) + F(-1)*φ = -1 + 1*φ = -1 + 1.618.. = 0.618... , which is indead the reciprocal of φ If n = -2 you get: φ^-2 (the reciprocal of φ^2) = F(-3) + F(-2)*φ = 2 + -1*φ = 2 - 1.618.. = 0.381... , which is indead the reciprocal of φ^2 This will hold true for any integer power of φ, not just for power of 2 or higher.

[@jpl569]

@@Escviitash OK... I didn't use the usual notation for Fibonacci numbers, in order to achieve a nice and easy solution with positive integers. Thanks for your remark !

[@robertveith6383]

​@@jpl569 -- In your first post, n - 2 and n - 1 needed to be inside grouping symbols, respectively.

[@jpl569]

@@robertveith6383 Yes, for sure, thanks !

[@ffggddss]

All good, except the index is off by one from the usual definition in parts of that. Yes, F(0)=0, F(1)=1; from which you eventually get F(11)=89, F(12)=144; but the rule for powers of phi is: φⁿ = F(n)φ + F(n-1) Fred

[@koenth2359]

We only have to calculate φ^2, φ^4, φ^8 by consecutive squaring and then φ^12 =φ^4×φ^8. It's just four lines.

[@ericlindholm9482]

That's what I did and it took maybe 90 seconds. Some of the other solutions in the comments look even faster.

[@redotto100]

There’s another elegant way to solve this problem by recognizing that (1+sqrt(5))/2 is the eigenvalue of the matrix A = [0,1; 1, 1] with largest magnitude. Taking this matrix to powers produces matrices with Fibonacci numbers as entries A^p = [f(p-1), f(p); f(p), f(p+1)], so A^12 = [89, 144; 144, 233]. The largest eigenvalue of A^12 is ((1+sqrt(5))/2)^12. Simplifying the characteristic equation for A^12 gives 0 = x^2 - 322*x + 1 and computing the larger root gives the answer.

[@wisdomokoro2255]

Just Brilliant!

[@CGoldthorpe]

Clearly this is more obscure, less intuitive, and less "pure"

[@vhalkola]

I did this rather similarly: let A = [1 1; 1 0]. Then the eigenvectors are [φ;1] and [ψ;1]. Putting these in a matrix S, we can diagonalise A = S^(-1)ΛS, and A^(n) = S^(-1)Λ^(12)S, so that Λ^12 = SA^(12)S^(-1). A^12 can be easily constructed using the Fibonacci numbers so that the only thing that requires some work is S^(-1).

[@vinny5004]

Yeah, this is simple… smh. Show off

[@ТимурГубаев-ж8ы]

To make it really elegant, you could’ve started with "Obviously, …" as most math textbooks do.

[@minchomilev]

(½(1 + √5))¹² = (¼(1 + 2√5 + 5))⁶ = (¼(6 + 2√5))⁶ = (½(3 + √5))⁶ = (¼(9 + 6√5 + 5))³ = (½(7 + 3√5))³ = ⅛(343 + 441√5 + 945 + 125√5) = 161 + 72√5 I can't get it why you made it so complicated?

[@gordonbraddock803]

I agree. I did it your way before watching the official method/explanation.

[@SeeJah87]

only 135√5, not 125√5

[@casey137]

I do like your method, but doing the cube first (followed by squaring twice) means we have no fractions to worry about for the rest of the calculation! ((1 + √5)/2)¹² = ((1 + 3√5 + 3(5) + 5√5)/8)⁴ = ((16 + 8√5)/8)⁴ = (2 + √5)⁴ = (4 + 4√5 + 5)² = (9 + 4√5)² = (81 + 72√5 + 80) = 161 + 72√5

[@slothbearanonymous]

@@casey137 Yes. I used this method also.

[@shadowfighter4341]

I have also solved this problem like this

[@jamesaitken4070]

I think it’s simpler just to expand out as ((((1+sqrt 5)/2)^2)^2)^3 as the expression squares out pretty simply each time

[@milan.matejka]

Well, probably just as complicated to calculate, but you definitely don't risk a dead end.

[@46swa]

I agree with you. It's much quicker this way.

[@user-gr5tx6rd4h]

I did so and got the exact same answer.

[@BotaTamas85]

In this particular case yes, but seeing how introducing an X for a part of the expression can simplify the problem is pretty valuable.

[@benjamingrunbaum3601]

I also did it this way. I wouldn't have even thought of the "X"

[@ChristopherBitti]

Let a be the number in parentheses. a is known as the golden ratio, and as such, it satisfies the equation x^2 - x - 1 = 0. Thus, a^2 = a + 1. Given this, a^12 = (a^2)^6 = (a + 1)^6 = ((a + 1)^2)^3 = (a^2 + 2a + 1)^3 = (3a + 2)^3 = (3a + 2)(9a^2 + 12a + 4) = (3a + 2)(21a + 13) = 63a^2 + 81a + 26 = 144a + 89 = 72(1 + sqrt(5)) + 89 = 72sqrt(5) + 161 as desired.

[@thomasrebotier1741]

It's the golden ratio. That said, knowing the golden ratio and the afferent properties (x=1+1/x) is more a matter of general math culture, and olympiads are for teenagers. some of them will know and some won't at all, this is NOT the spirit of the olympiads. Those problems should be based on problem-solving ability, not on random general culture.

[@ttt69420]

I mean.. look at the comments on these videos. It's always a flood of people saying just use x algorithm/theorem. Rote knowledge isn't intelligence and it isn't understanding, but it's all there is all through undergrad anyway (and beyond). The Olympiads would be boring if you had to wait 40 years to assemble 8 genius children to derive and prove what's already been done for hundreds of years anyway.

[@scottrackley4457]

While the golden ratio stuck out like a sore thumb to me, some haven't been introduced. I agree, the fact that's the first thing I see is kind of rote knowledge.

[@Escviitash]

φ^z = F(z-1) + F(z)*φ holds true for any integer power of φ, negative, zero or positive. F(z) is the Fibonacci numbers with index z. To find the Fibonacci numbers with negative index you can reverse the formula to F(z) = F(z+2) - F(z+1) to get the sequence F(-1)=1, F(-2)=-1, F(-3)=2, F(-4)=-3, F(-5)=5, F(-6)=-8 and so on, i.e the same as the positve indexed Fibonanacci numbers, but negative if the index is even. Plug in 12 for z and you will get: φ^12 = F(12-1) + F(12)*φ = F(12-1) + F(12)*φ = F(11) + F(12)*φ = 89 + 144*φ which can be simplified to F(z-1)+F(z)/2 + (F(z)/2)*sqrt(5) if F(z) is even.

[@pietergeerkens6324]

Such a great opportunity missed, to note in passing that [ (1 + √5) / 2 ]³ = (1 + 3√5 + 3*5 + 5√5) / 8 = (16 + 8√5) / 8 = 2 + √5; and thus give students insight into this instance (and others similar) of the Pell equation. Now the entire expression simplifies as [ (1 + √5) / 2 ]¹² = (2 + √5)⁴ = (4 + 4√5 + 5)² = 81 + 72√5 + 80 = 161 + 72√5

[@robertveith6383]

No, it does *not* simplify to that. Go backward to some steps: [4 + 4sqrt(5) + 5]^2 = [9 + 4sqrt(5)]^2 = 81 + 72sqrt(5) + 80 = 161 + 72sqrt(5) *Answer*

[@pietergeerkens6324]

@@robertveith6383 Oops! Thank you. I Copy-Pasted "²" twice instead of "√5". Now corrected.

[@aspenrebel]

Now that a good way to do it, start. Term to 3rd = 2 x sqrt of 5. Then sq, then sq =322

[@aspenrebel]

@@robertveith6383 that's what he did???

[@oahuhawaii2141]

@aspenrebel: OP made an error, which someone pointed out, so OP edited his first comment to fix the error. You saw the first comment after it was corrected.

[@florianbasier]

You complexified everything my friend. My idea was that we could detect a nice generalisation for Sn=x^n that would not force us to write Newton's formula with a coeff of 12. So I started looking at S1=(1+sqrt(5))/2, S2=(3+sqrt(5))/2, S3=2+sqrt(5). And now my idea of a nice formula vanished but S3 is so simple that it can be used to compute easily S6=S3²=9+4sqrt(5) and then S12=S6²=161+72sqrt(5)

[@tungyeeso3637]

Nice and neat.

[@46swa]

This solution is far too cumbersome. (((Term^2)^2)^3 is much faster

[@RexxSchneider]

You might alternately consider evaluating Term^4 = ((Term)^2)^2 as an intermediate step, then Term^8 = (Term^4)^2 and finally Term^12 = Term^8 * Term^4.

[@DandoPorsaco-ho1zs]

@@RexxSchneider That's how I did it.

[@billweihmillerjr9481]

Or x^12= (x^4)(x^4 )(x^4). Solve x^4 once, pull the power down to 1. Square, pull power down to one, then uae the third, power down ... I liked his deliberatw approach, skipping magic ratio, phi, and avoiding the temptation to go complex when algebra,does just fine.

[@billweihmillerjr9481]

Good problem to be VERY careful with.

[@oahuhawaii2141]

Nope, cubing simplifies the calculation by eliminating the denominator, so 2 successive squares can be done easily. ((√5+1)/2)³ = (√5+1)³/8 = (5*√5 + 3*5 + 3*√5 + 1)/8 = (8*√5 + 16)/8 = √5 + 2 Then, just square that twice.

[@robertbox5399]

I was screaming, 'GET ON WITH IT!', Lol.

[@AtanuKDey]

I downvoted the video because of that. The man takes many steps to explain why 1 plus 4 is 5 by adding 1 to 1 four times to get 5. Three minutes gone.

[@amunt3r]

Just evaluate the 2nd, 4th and 8th powers by squaring three times. Then multiply the 4th power by 8th power to get the 12th power. It's trivial arithmetic.

[@Pixiltation]

i don't think this video should be 11 minutes long

[@billyoung8118]

I suspect his ornate way of writing "x" accounted for about 1/3 of the video length

[@Philip-hv2kc]

I don't see why he needs an X . You could simply do the basic arithmetic. But now I see that √5 if multiplied by itself 12 times is going to give different answers depending on original number of decimal points .

[@souptikdam8424]

2x speed.

[@xl000]

this could have been a tweet... But this wouldn't be on KZbin

[@golastname7686]

Right. Also, in a timed test, students ought to be trained to initially skip questions that take 11 minutes to answer (or that cause drowsiness). Unanswered questions might not mean the question is hard, but that students strategize time elsewhere.

[@TristanBBBBB]

A good teacher doesn’t just show a series of steps. A good teacher explains the goal and explains why each step works toward the goal. Otherwise this is just rote learning and not knowledge.

[@ciprianteasca7823]

So, what's your point...?!

[@RexxSchneider]

@@ciprianteasca7823 The point is that his variable 'x' is actually the golden ratio, φ. You should know that it has the property φ^2 = φ+1. That means that any power of φ can be reduced to a linear expression in φ. The simplification that provides is the key to finding an easy to evaluate expression for any power of φ, rather than the slog of evaluating a large binomial expansion with increasing powers of a variable. You can then evaluate φ^12 as ((φ^2 * φ)^2)^2 or ((φ^2)^2)^2 * (φ^2)^2, etc. as a linear expression in φ.

[@echandler]

@@RexxSchneider *Any* simple radical expression for x can be cast as the root of some quadratic equation. φ is particularly nice!

[@nabuk3]

And almost all students will forget it an hour later, or right after the test.

[@RobertoStenger]

👍🏻

[@BN-hy1nd]

As a retired teacher of Maths in secondary schools, it is the/his step-by-step approach that makes his method interesting and less complex. Well done😄

[@whatwasisaying]

Interesting? Really.

[@KrytenKoro]

​@@whatwasisaying...yes, interesting. It's an education channel aimed at high schoolers

[@jerrygunning1449]

I was a software engineer for forty years. If this guy was working for me I would have fire him. I know this "guy" - he is far more interested in proving how smart is than actually getting any work done.

[@growleym504]

I would not just fire the guy who wasted 20 minutes answering the question with another problem. I would also fire the guy who came up with the question in the first place. When have you ever needed to raise anything to the 12th power in a real world situation? It takes about 30 seconds to type it into a python session or key it into a calculator, and get an actual usable numeric answer.

[@fredrodriguez6650]

Using some long hand and the OLD order of operations with a calculator is just 5 steps to an actual number. 5.402211e^9. Mathmagicians and their proofs.

[@michaeledwards2251]

@@growleym504 Unless you are performing aerodynamic calculations. 6d aeroballistics is full of tricks to reduce a multiple step, multiple term, equations to linear approximations. Examples of aerodynamic modelling ? Building wind shear, car cross wind stability, turbulence estimations in stadia, and effects of structures on wind speed. Both the problem presented, and aerodynamics, have a common theme : reducing the equation power to linear while controlling error. (Business calculations are inherently simple and standard. Modelling physical systems to ensure buildings won't fall down in a human life time is another matter. )

[@KrytenKoro]

So ...not a good engineer, then? Since this is a channel for explaining concepts, and not a "just use a calculator" channel?

[@KrytenKoro]

​@@growleym504good Christ, is your response to a mathematics education channel really "just use a calculator"?

[@italixgaming915]

You overcomplicated this poor little thing. Once you wrote that x²=x+1 you have: x^3=x².x=(x+1).x=x²+x=2x+1 x^6=(2x+1)²=4x²+4x+1=4.(x+1)+4x+1=8x+5 x^12=(8x+5)²=64x²+80x+25=64.(x+1)+80x+25=144x+89 So the value we are looking for is 72.(1+sqrt(5))+89=161+72.sqrt(5)

[@herbie_the_hillbillie_goat]

I don't see where your solution is any simpler, just different.

[@oahuhawaii2141]

By factoring 12 to 3*2*2, raising to the power of 12 can be done by cubing, followed by squaring twice. Let x = √5 + 1 We have x³ = 5*√5 + 3*5 + 3*√5 + 1 = 8*√5 + 16 Thus, ((√5 + 1)/2)¹² = (((x/2)³)²)² = ((x³/8)²)² = (((8*√5 + 16)/8)²)² = ((√5 + 2)²)² = (4*√5 + 9)² = 72*√5 + 161

[@whycantiremainanonymous8091]

Without watching: (1+5^½)/2 is the constant phi (sorry, no Greek keyboard here). Phi ^ n = Fn×phi + Fn-1, where Fn is the nth number in the Fibonacci series. Moreover, this is approximately equal to Fn+1 + Fn-1. So, for power 12, this would be 233+89=322. The actual number is more like 319.99.

[@literallydeadpool]

amazing, what a relation that is

[@Ringcaat]

Oh, impressive! Maybe some students knew that, but I doubt most did. I checked and it comes out right.

[@jontallen3319]

jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci

[@colinpountney333]

A more elegant approach would be to iterate. After noting the expression to be raised to the power of 12 is a root of x^2 - x - 1, it follows that x^2 = x + 1. So x^3 = x^2 + x = 2x + 1. And therefore by iteration any power of x can be expressed in terms of ax + b. If x^n = ax + b then x^(n+1) = (a+b)x + a. You can do the rest in your head.

[@twincities21]

I solved this questions in the following way. 12=2 x 2 x 3, so I converted (1 + √5/2)^12 to [{((1 + √5)/2)^2}^2]^3 {(1 + √5)/2}^2 = (3 + √5)/2 {(3 + √5)/2}^2 = (7 + 3√5)/2 {(7 + 3√5)/2}^3 = 161 + 72√5

[@ProfesSor-j2d]

Another way is to note that 12 = 8 + 4 and 4 = 2 + 2 so with φ=(1 + √5)/2 one have φ^12 = φ^8 * φ^4 and noticing that φ^8 = (φ^4)^2 and that φ^4=(φ^2)^2 one arrives at φ^12 = φ^8 * φ^4 = { (φ^2)^2 }^2 * { (φ^2)^2 } needing one square step to reach φ^2, another square to reach φ^4 and a final square to reach φ^8, with the multiplication of the previous two powers φ^8 * φ^4 to reach φ^12. All those ways leads to the same answer through some near routes. One of the most efficient ways to compute Powers is by BINARY Decomposition methods, so 12 in base 10 = 1100 in base 2 and the binary pattern of each power n leads to a simple implementation of a binary recursive power procedure

[@ProfesSor-j2d]

And if one notes that φ^2 = φ + 1 since ( (1 + √5)/2 ) ^ 2 = (3 + √5)/2 = (1 + √5)/2 + 1, one reaches for φ^12 = { (φ^2)^2 }^2 * { (φ^2)^2 } = { (φ + 1)^2 } ^ 2 * { (φ + 1)^2 } leading to a polynomial expansion just in terms of φ with φ^12 = { φ^2 + 2φ + 1} ^ 2 * { φ^2 + 2φ + 1 } = { 1 + 4φ + 6φ^2 + 4φ^3 + φ^4 } * { 1 + 2φ + φ^2 } = φ^6 + 6*φ^5 + 15*φ^4 + 20*φ^3 + 15*φ^2 + 6*φ + 1 with a much more cumbersome way to reach for φ^12 needing sucessive computations of powers of φ up to φ^6 and them combining in the characteristic polynomial just arrived...

[@baselinesweb]

It was good to see how you did it, but you can just square it, square it and cube it if you are in a hurry. Good to see the other comments two about the golden ratio and the eigenvalues. Thanks all.

[@alexisgs8800]

I love how every overinflated ego posts a different solution using more and more advanced mathematics, criticizing his approach, but no one cared to ask who this video was made for. You don't teach how to solve such a problem the same way when your target audience is high school students as you would if it were made for people pursuing a university degree in mathematics. Some of his videos have "Entrance exam" in the title, for different universities, which implies that this may be for people who don't have all the techniques some of you are talking about.

[@growleym504]

Cut to the chase. Calculate the solution. 322. The presumption is that there is a practical need for the solution, and there it is, now it is ready to be used for whatever purpose the question was conceived.

[@chrisbraun70]

Now that it has been more than 50 years since my university degree in mathematics (from Brown), I’m happy to just follow along 🙂.

[@Ghastrix9916]

The only skill required here is squaring binomials which I wouldn’t consider “college level”

[@donthuis]

Indeed the formula's being used are already known to students on secondary school level. There is no need to bring Fibonacci into the solution, if getting the answer without a calculator is the only requirement. The latter was implicit in the video IMO

[@donthuis]

On my later UNI, one professor of mine explaining mathemetical solutions to non-zero starting of electrical pulses using Laplace transforms first gave his own solution writing three boards full of formulas, followed by a three line solution found by one of his students during his aural exam. Of course both ways were correct. This thread with both long and short roads reminds me of such events.😅

[@paulortega5317]

Two Fibonacci series, each with different f(1) and f(2) values. 1,3 and 1,1. The latter can also be derived from (1/sqrt(5))*[((1+sqrt(5)/2)^n - ((1-sqrt(5)/2)^n]

[@jontallen3319]

jontalle.web.engr.illinois.edu/TEACH/AnInvitationToMathematicalPhys.pdf see page 71 or search for Fibonacci

[@kianooshboroojeni2551]

Want to have a shortcut? Fib(12) = 144. Considering the explicit form of Fib sequence, φ^12=144*sqrt(5) + ((1-sqrt(5))/2)^12 which is approximately equal to 144*sqrt(5) or 322.

[@DavidRTribble]

2:42 Instead of computing (x^2)^3 = (x+1)^3 the long way, could you just use the binomial expansion for (x+1)^3 directly?

[@hippophile]

Trying to keep it simple (to be safe) I just cubed the expression then squared it twice. Cubing gets rid of the denominator so it works out pretty nicely.

[@pietergeerkens6324]

And a passing reference that this happens to other Pell Equation solutions as well, such as [ ( 3 + √13) / 2 ]³ = ( 27 + 3*9√13 + 3*3*13 + 13√13 ) / 8 = ( 27 + 27√13 + 117 + 13√13 ) / 8 = ( 144 + 40√13 ) / 8 = 18 + 5√13 and [ ( 5 + √21 ) / 2 ]³ = 55 + 12√21 , might bee in order, as well as that this can only (but doesn't always) happen for √(4n + 1).

[@GlomgoldFlintheart]

I did the same. If only 8% of students got it right, that must be bad students.

[@fatroberto3012]

Modern students will not understand why the final solution is any simpler than the initial problem. This is understandable, because the point of solving problems without a calculator is debatable. If only 8% are going to get the right answer, then why do it? They are more likely to have a calculator than a book of Mathematical Tables or a slide rule, the use of which was the object of these exercises when I was at school fifty years ago.

[@michaeledwards2251]

The point of the question is to determine whether students are aware of the significance of reducing the power demanding evaluation. This is allows programming with minimal error growth. (In this instance the Harmonic ratio and Fibonacci sequence are useful, and may be part of a bonus, but the most significant factor is spotting ways to minimize error growth in numerical calculations. )

[@julianocg]

In this case should use any other number, not the golden ratio.

[@michaeledwards2251]

@@julianocg Why not use the golden ratio : it gives an opportunity for bonus insights.

[@michaeledwards2251]

@@julianocg An additional factor is the real world doesn't follow neat and tidy rules : numerical evaluation is an artform whose purpose is to achieve a resolution with minimum error. It is forgotten how important numerical evaluation is to the design of cars, houses, buildings, roads and much more.

[@RexxSchneider]

@@julianocg The whole point of picking the golden ration is that φ^2 = φ+1, which means that any power of φ can be successively reduced to a linear expression in φ. The key is that the student recognises that simplification provides a significant short-cut in this particular question.

[@julianocg]

@@RexxSchneider Just as I did.

[@cecilponsaing2749]

A beautiful solution. I did it without x at first, carefully, and got it right.After that I could finally understand your simpler approach , and I then did it that way too. Thank you.

[@fabrice9252]

Ce nombre est φ, le nombre d'or et on sait que φ est solution de l'équation x² - x - 1 = 0 donc: φ² - φ - 1 = 0 d'où: φ² = φ + 1 φ¹² = (φ²)⁶ = (φ + 1)⁶ = [ (φ + 1)²]³ = ( φ² + 2φ + 1)³ = (φ + 1 + 2φ + 1)³ = (3φ + 2)³ = (3φ + 2)² (3φ + 2) = (9φ² + 12 φ + 4) (3φ + 2) = (9φ + 9 + 12 φ + 4) (3φ + 2) = (21φ +13) (3φ + 2) = 63φ² + 42φ +39φ +26 = 63φ + 63 + 42φ +39φ +26 = 144φ + 89 = 144 (1 + √5)/2 + 89 = 72 (1 + √5) + 89 = 72 + 72√5 + 89 = 161 + 72√5

[@jriturria]

That equation (x²-x-1= 0), should have at least two answers. In fact, the equation passes by the points P₁ =(1.62 | 0) & P₂ = (-0.62 | 0).

[@ronaldnoll3247]

It is not clear to me what is simpler, the original equation or the calculated result.

[@zetvanzacharias9553]

exactly my thoughts.

[@Ridelto]

No need to use x. Just split the numerator and denominator, then square the numerator three times getting the common factor out every time simplifying with the denominator, and finally multiply one more time the numerator by 7+3sqrt(5) to get 644/4 and (288/4)sqr(5) or 161 + 72sqr(5). That simple.

[@panchostanza8712]

Here's what everyone is missing: the solution demonstrated would be doable by anyone who has done basic algebra

[@donthuis]

Not me, see my earlier remark

[@pinktoday5858]

(√5 + 1)/2 = φ - Golden Ratio φ² = φ + 1 φ³ = φ(φ + 1) = φ² + φ = 2φ + 1 φ⁴ = (φ²)² = φ² + 2φ + 1 = 3φ + 2 φ⁵ = φφ⁴ = φ(3φ + 2) = 5φ + 3 φ⁶ = φφ⁵ = 8φ + 5 ... Fibonacci Sequence as coeficientes Fₒ = 0 F₇ = 13 F₁₂ = 144 0 1 1 2 3 5 8 13 21 34 55 89 144 233 ... proceed to φ¹² = 144φ + 89 φⁿ = φFₙ + Fₙ₋₁ φ¹² = 144 (√5 + 1)/2 + 89 φ¹² = 161 + 72√5

[@chriscurtain1816]

Surely the 'answer' here is simply a different way of writing the question?

[@YoutubeHandlesSuckBalls]

Indeed.

[@nabuk3]

Right, after 11 minutes he still did not solve for x. I think he would have gotten it marked wrong on the test, if he did not run out of time.

[@Bill_Woo]

Small deduction for ambiguous "1" vs. "7" display. Yet a miracle occurs, that "c" and its inversion somehow produce "x", and it's clear and unambiguous. Every time I saw that I both winced and smiled.

[@ald6980]

phi = x=(1+sqrt(5))/2; phi^n=F(n)phi+F(n-1), where F(n) - the n-th Fibonacci number: 1,1,2,3,5,8,13,21,..... F(n)=F(n-1)+F(n-2). The same another way: x^0 = 1 = 1+0*sqrt(5) = (1;0) x^1 =(0.5;0.5) x^n=(a(n);b(n)) = (a(n-1);b(n-1))+ (a(n-2);b(n-2)).

[@ArtemKo___]

½(1 + √5) = x x^12 = (x^2)^6 x^2 = t x^12 = (t^2)^3 and just calculate

[@arekkrolak6320]

X is two straight lines crossing. What is this atrocity you are drawing? :)

[@blacksmith67]

In most mathematics, x is like a sine wave from zero to 2πr with a diagonal line through it. I have seen his style of x before and it bugs me too.

[@josepeixoto3384]

C'mon, it's so that you do not MIX it up with the x on 2x3=6

[@justliberty4072]

@@josepeixoto3384 but nobody doing algebra ever writes 2x3...

[@alexsam8554]

without Fibonacci: x^2=x+1, x^3=x^2+x=2x+1, (2x+1)^4=(4x^2+4x+1)^2=(8x+5)^2=64x^2+80x+25=144x+89

[@virtual-viking]

It's much better to firstly calculate x^3=2+√5 because it immediately gets rid of the /2. Then find the solution from (2+√5)^4 via two successive squarings.

[@oahuhawaii2141]

That's what I did.

[@justliberty4072]

What cultures write x with two curves rather that lines that cross?

[@jeannieheard1465]

DQ Culture

[@growleym504]

Yeah that really irritated me, too.

[@themulticourse2460]

​@@growleym504 that's is the best one to write with curve. I think you didn't have multiply symbol infront of x then you sould be confused if that's multiplication or x

[@Guidussify]

That was how I learned to write a lower-case x in school long, long ago in a galaxy far, far away.

[@HHHGeorge]

Very clever and very clearly explained. Thank you. When I initially looked at the problem I thought I couldn't do it but watching you I can see I could do it with the mathematical knowledge I do have.

[@r2boii882]

I'm part of the 8% Too bad for the rest of the 95% though :/

[@jonbondMPG]

I see what you did there 😏

[@MiloTsukroff]

A very elegant solution. (That is, if the year is before 1970 and no scientific calculators are available.) 161 + 72 * sqrt( 5 ) does indeed equal ( ( 1 + sqrt( 5 ) ) / 2 ) ^ 12

[@MarkEricson1]

Uh... yes, it does. I don't know about your calculator, but mine comes up with 321.99689438 (to 11 digits of precision) for both.

[@EyupSkydiver]

Something about writing “x” that way annoys me.

[@digbycrankshaft7572]

I learned to write it that way so it doesn't look like the multiplication sign. It's writing 9 like g that annoys me.

[@proletarian_unity]

I write the multiplication sign as a dot

[@jagjeetmule2271]

Isn't it simple ✓5 = 2.2360 then +1 and whole ÷ 2 and then 1.618 to the power 12 = approx 322

[@growleym504]

Duh. Exactly. Some people are so removed from the real world they would look at a micrometer and think it must be some sort of giant earring or something. We didn't land men on the moon by answering mathematical problems with more problems. We did it by calculating real world solutions and applying them.

[@nabuk3]

Right, it could have been solved in 30 seconds instead of partly solved in 11 minutes.

[@defevfefwow976]

Its math not physics. You cant do approximation because Its easier, you need to simplify as much as possible

[@SatishGhanekar-rm3qm]

Entire comments section is filled with toppers.

[@monstorgamer9635]

😂😂😂

[@johnblackledge4009]

8% of students got this right. 98% of the rest of humanity thought, "Why do we pay these people for this pointless gibberish?"

[@DeadlyBlaze]

98% of the rest of humanity can keep have that thought while working 996 unskilled labour till they're 70.

[@donaldasayers]

Revelling in his ignorance whilst posting using machines designed by people who understand mathematics, using mathematics.

[@echandler]

The 8% get to respond when the 98% ask "Do you want fries with that?" The problem itself is good. This explanation dwells too long on routine algebra that someone encountering this problem should have already mastered. The good part is that the operation of raising a number to a power can be reduced in several ways. First is by nesting it as a sequence of squaring and cubing operations. Second is by using a recurrence relation. Third is that replacing x squared to a linear expression in x using *recursion*.

[@stevealexander8010]

@@DeadlyBlaze Sorry dude - I have a math degree and there are any number of ppl who could never follow this vid yet who can buy & sell us. Bezos, Zuckerberg and Musk may be relative 'illiterati', but not "unskilled".

[@growleym504]

@@echandler The "problem" is made up out of thin air with no basis in the real world. Why would you ever need to know the solution? If it were based on an actual real world problem, a terminal window with python called up or else a $4.99 calculator are all that is needed. We don't have to re-invent mathematics every time we have a reason to solve something. Obviously the answer is calculated at approximately 322. Done. Now, get back to work. Play time is over.

[@jeffpudewell5884]

Awesome, thanks. Very helpful.

[@chriscurtain1816]

I just did it on my calculator. Then I got the same answer in less than 30 seconds. Except I came up with a figure - not just a different way of expressing the problem.

[@Liberty4Ever]

Three kinds of people enjoy math - accountants, engineers and mathematicians. Accountants use grade school math for practical purposes. Engineers use high school and university math for practical purposes. I'm an engineer and apparently you are too? Mathematicians abhor practical solutions because math is art, beauty, religion, et cetera. It's all about the journey and not the destination.

[@am-gobears5191]

Easily done a "peasant" way as follows. The square of (1+sqrt(5))/2 is (3+sqrt(5))/2 simply using the (a+b)^2 formula, so the original 12th power reduces to finding (3+sqrt(5))/2 to the 6th power. The same way, the square of the latter is (7+3sqrt(5))/2, so the problem further reduces to taking the cube of the latter. The square of that expression is (47+21sqrt(5))/2, so - by simply multiplying that by (7+3sqrt(5))/2 and collecting the terms - one gets 161+72sqrt(5). By the way, the value of that is pretty close to 322.

[@walcant5610]

Prop. Phi = x , x ^n = F(n) x + F(n-1) = F(n)/2 + F(n-1) + + F(n)/2 * 5^.5 F(1) = F(2) = 1 F(n) = F(n-1) + F(n-2) F(12) = 144, F(11) = 89

[@MikeBTek]

a = (1 + √5)/2 a^2 = (6 + 2√5)/4 = (3 + √5)/2 a^3 = (8 + 4√5)/4 = 2 + √5 (surprise!) a^6 = 9 + 4√5 a^12 = 81 + 72√5 + 80 = 161 + 72√5 Also note that ratio (1 + √5)/2 is called the golden ratio or phi.

[@michaelqi5202]

the way he wrote 1 is weird

[@和平和平-c4i]

Why ?? In what part of the world are you ?

[@xl000]

I' m ok with his 1's But the b are really weird. It's like a combination of two cursive styles

[@martinleitner8092]

His x ist weird. An x ist supposed to be two crossing lines.

[@xl000]

@@martinleitner8092 His x is ok. It's in cursive style.

[@yusufdenli9363]

This is golden ratio = φ We must find φ^12 You can use Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89,144,... φ^n = F(n) .φ + F(n-1) φ^12 = F(12).φ + F(11) = 144.(1+sqrt(5))/2 + 89 =161 + 72.sqrt(5)

[@reisele1980]

I noticed the 89 and 144 were in the Fibonacci sequence. Interesting thing is that the ratio between consecutive members of the Fibonacci sequence approaches phi. So for an astronomically large x, would the answer approximate to: (x)(1 + 1/phi) + (x/2).sqrt(5) ?

[@ВикторПоплевко-е2т]

11:04 this isn't an answer to the question

[@Helloworld-o5i]

It is 20^6

[@Hatifnote]

Why isn't?!

[@ВикторПоплевко-е2т]

@@Hatifnote think about it yourself I'm too lazy to explain it

[@stevet7522]

​@@Hatifnotebecause he didn't finish it. He just left it with another equation instead of an answer.

[@Hatifnote]

@@ВикторПоплевко-е2т Perhaps you saw or remak some thing I didn't do and as i think hé gived the answer your question had made me tell you why you think isn't an answer.. As you want not to tell me do thank you! Perhaps becaude at the end hé gived not thé symbole = ? Also he is perhaps lazy to finish something winch is easy

[@ProfesSor-j2d]

The most elegant solution is to note that φ=(1+sqrt(5))/2 = golden ratio = eigenvalue of Fibonacci matrix [[0,1][1,1]] and that φ^n = φ * Fib(n) + Fib(n-1) with Fib(n) = Fib(n-1)+Fib(n-2) denoting the n'th Fibonacci number, so ( (1+sqrt(5))/2 )^n = φ * Fib(n) + Fib(n-1). This shows the very deep interconnection of polynomial algebra, with matrices and notable constants like φ = golden ratio. Also noting that the n'th Powers of Fibonacci matrix [[0,1][1,1]] ^ n = [[Fib(n),Fib(n+1)][Fib(n+1),Fib(n+2)]] could be computed by FAST MATRIX MULTIPLICATION algorithms, taking NOTICE of BINARY decomposition of Powers, leading for the example in question φ^12 = φ^8 * φ^4 and that φ^8 = φ^4 * φ^4 and φ^4 = φ^2 * φ^2, so φ^12 = { ( φ^2 )^2 } * { ( φ^2 )^2 }^2 needing thus the calculation of the square of φ=(1+sqrt(5))/2 with φ^2 = (3+sqrt(5))/2 and φ^4 = (φ^2)^2 = (7+3*sqrt(5))/2 and φ^8 = (φ^4)^2 = (47+21*sqrt(5))/2 so φ^12 = φ^8 * φ^4 = (47+21*sqrt(5))/2 * (7+3*sqrt(5))/2 = 72 * sqrt(5) + 161. Note to have provided enough interesting relations to be explored by many with curiosity in all such intricacies of Matrix Powers with algebraic polynomials, characteristic and eigenvalues and the golden ratio constant. For those further interested φ = Phi constant there is a very captivating introductory book by Mario Livio - The Golden Ratio - The Story of PHI, the World's Most Astonishing Number - Broadway Books (2003). Taking cover of another interesting famous constant Gamma there is the introductory book by Julian Havil - Gamma - Exploring Euler's Constant - Princeton University Press (2003). For the ABSOLUTE REFERENCE on Mathematical CONSTANTS there are the books by Steven R. Finch - Mathematical Constants - Cambridge University Press (2003) and Mathematical Constants II - Cambridge University Press (2018). Noticing that 2003 was a rather inspiring year for 3 MUST READ books on some famous Mathematical Constants, appart from Pi and e the base of natural logarithms and sqrt(2) which have been much more extensively covered since the very dawn of ancient mathematics, most notably by greek exponents like Phytagoras, Euclides and Archimedes.

[@elgb5671]

x²-x-1=0 x²-x+1=2 x³+1=2x+2 x³=2x+1=2+sqrt5 x⁶=9+4sqrt5 x¹²=161+72sqrt5 수능 15번보다 쉬운듯

[@beatzedit1]

I understood it well till 2:25 but why do that all work just do (x^2)^6 = (x+1)^2(x+1)^2(x+1)^2 x^12 = x^2 +x + 1 + x^2 + x + 1 + x^2 +x + 1 x^12 = 3x^2 + 6x + 6 x^12 = ???? Whoops i made a mistake sry guys

[@FrancescoBalena]

Change pensil

[@oahuhawaii2141]

Pen

[@johnpool1111]

This is easy if you happen to recognize z = ½(1 + √5) as a root of x²-x-1, which gives z²=z+1. Now calculate z^12 stepwise and while doing so replace each occurence of z² by z+1. This gives z^12=144z+89. Substituting z = ½(1 + √5) gives 161 + 72√5.

[@nabuk3]

Right, so as I see it, it wasn't really solved, while several people here seemed to solve it well, faster, and more simply.

[@wadzanaikanongovere4798]

Stupid me was going to use Binomial Expansion/Pascal's Triangle. This solution you provided is elegant 🙏🙏🙏

[@mameahmed3758]

But their is simple method more than this u just use numerical method

[@robertveith6383]

Wrong. The host's solution is *not* "elegant."

[@seanbrendangarrette7644]

@@robertveith6383 What is the elegant solution?

[@TakuT.]

(1-√5)/2との和と積から解と係数の関係の逆からx^2-x-1=0は出せる。

[@zth809]

3:19 just use binomial theorem

[@guillemiscla3842]

This holds for n >= 1 P^n = p*(F(n)) + (F(n-1)) Where F(n) is the Fibonacci sequence starting with F(0) = 0, F(1) = 1

[@Gnowop3]

Hard because the question is incomplete. Some sort of result format specification needed. Decimals? Rational number format etc

[@jwm239]

...One could also 'break apart' that fraction into (1/2 + (\/5)/2) and use the Binomial Theorem to rapidly expand and collect terms.

[@JubeiKibagamiFez]

0:04 Is this suppose to be done without a calculator? If no, I did [(√5/2)+1]¹²=8150.7036132812426871779905887804. I used proper PEMDAS/PEDMSA. Now, improper PEMDAS gives us [(√5+1)/2]¹²=321.99689437998485814146050414865. Now, I don't know a third way to calculate this because the equation is very specific. If it were written √5+1/2¹²=X, then there's many more ways to get it wrong.

[@sanjogar]

Let a = ½(1 + √5); then is conjugate is a' = ½(1 - √5), which are the roots of the equation x² = x + 1. So a² =a + 1. Multiplying by a, a² =a + 1, a^3 =2a + 1; then a^6 =8a + 5. Finally a^(12) =144a + 89.

[@johnstanley5692]

Alternative? extract even and odd powers of expansion (1+x)^12 = p1(y) +x*p2(y) here x=sqrt(5), and y=x^2=5 then divide by 2^12. in this case p1(y)=y^6 + 66*y^5 + 495*y^4 + 924*y^3 + 495*y^2 + 66*y + 1 = 659456 -> 161. p2(y)=12*y^5 + 220*y^4 + 792*y^3 + 792*y^2 + 220*y + 12 =294912 -> 72=> (p1(y)+x*p2y)/2^12 = 161+72*sqrt(5)

[@stefantsarev4442]

Once you get to (x+1)^6, you can use the reverse Horner to make the polynomial.

[@albertomontori2863]

After you arrived at (3x + 2)^3 , It would be more obvious to just do the cubic binomial expansion (a+b)^3= a^3 + b^3 + 3a^2b + 3ab^2; also the Tartaglia triangle was a viable method to do that binomial expansion!

[@pythagorasaurusrex9853]

When you know some advanced geometry with the golden ratio, it is simple. (1+sqrt(5))/2 = phi, so phi^2=phi+1 square it: phi^4=phi^2+2phi+1=3phi+2 square again: phi^8=21phi+13 multiply last two results together: phi^12=phi^8*phi^4=(21phi+13)(3phi+2)=144phi+89=72*sqrt(5)+161. If you don't know the basic property of the golden ratio, then one will not get the first idea in this video. If you do, you can immediately start with step 2: x^2=x+1

[@olegalferov8125]

12=2*2*3. Notice that ((1+sqrt(5))/2)^3=2+sqrt(5) because of binomial: (1+a)^3=1+3a+3a^2+a^3. After that, square it twice by brute-force by formula (a+b)^2=a^2+2ab+b^2. This is it.

[@escapedlunatic27]

If this was set for a group of average schoolkids then I could believe only 8% of them got it right, but if it's aimed at Olympiad level or similar (i.e. a group who have already shown a considerable aptitude for mathematics) then I'd expect that almost all of them would. It's just simple multiplication - calculate the square, then the cube, then the sixth power, then the twelfth. You just have to be careful to avoid making mistakes when you're expanding the binomial then simplifying the result each time.

[@SALogics]

Very nice problem with very nice solution! ❤❤

[@BojanPeric-kq9et]

Simple way: do match by square, squaring and cubing or cubing, squaring, squaring without 1/2 to avoid bother with fractions. Divide when it is convenient and keep count of number of 2s used.

[@juandesalgado]

It's perhaps easier to figure out first the relation phi^n = F_n . phi + F_{n-1} where F_n is the n-th Fibonacci number; then compute phi^12 directly.

[@robertlunderwood]

I immediately set the golden ratio to x and jumped to x^2 - x - 1 = 0 => x^2 = x + 1. Multiply both sides by x, simplify, square, simplify, square again, simplify, substitute, and solve, abusing x^2 = x + 1 whenever possible.

[@Prypak]

Just know that phi²=phi+1 and boom, ez solution, just develop (phi+1)², replacing phi² by phi+1, and then do it with (a*phi+b)³ and you've got it

[@adamwhiteson6866]

I enjoyed the presentation of the technique. Very cool But I do agree it was way overkill for this problem.

[@MATEMÁTICAELEMENTAR_ALEXTSILVA]

Can you use a Pascal triangle???

[@rmsgrey]

Personally, I prefer to cube first, then square repeatedly - cubing is more difficult, on average, so doing it while the expression is simpler typically makes the most difficult step the easiest it's going to be. I immediately recognised phi, so was happy to work with the known property that phi satisfies x^2=x+1. Otherwise, I'd have just calculated it directly, starting by cubing, then squaring twice, knowing each step's result could be simplified to a rational plus a rational multiple of root five.

[@khuongduybui]

we can just split ^12 to ^2 ^2 ^3, calculate each step from inside out and minimize them after each step

[@Valdagast]

I don't know if I'm more impressed by the maths or the neat handwriting.

[@Sigma.Infinity]

Use the property of golden ratio: φⁿ = Fibonacci(n)φ + Fibonacci(n-1). So here, φ¹² = Fib(12)φ + Fib(11) = 144φ + 89 ≈ 321.997. This is the same as the final answer in the video. (Note that Fib(n) is zero based.)

[@amarjyoti7783]

Use the binomial theorem

[@GreatestPhysicistOfAllTime]

Why don't you just divide x^12 by (x^2-x-1) to obtain a first order remainder and calculate the value of it? It's nothing but a simple application of the remainder theorem.

[@aboutit9184]

If you are spending THAT much time on your "simple" way - you may have already calculated the initial equation (12=2*2*3, i.e. square it twice and put it into 3rd power once)...

[@lunstee]

You break down the 12th power into 2*2*3, taking x, squaring it twice, and then taking the cube. I would have taken the cube up front; x^3 = x^2*x = (x+1)*x = x^2+x = 2x+1, and then squared that twice instead. Or, observing that your x is the golden ratio φ, I would recognize that φ^n = (L(n) + F(n)*sqrt(5))/2 where L(n) and F(n) are the Lucas and Fibonacci series respectively. One can work out the series manually from the recursive definition, which for n=12 is feasible to do directly. For larger n, the methods for skipping ahead in these series are fundamentally the same as the manipulations shown in this video (multiplying two exponents).

[@oahuhawaii2141]

Note that the cube is 2*x + 1, which eliminates the denominator, so squaring twice is easy. No need to track all the coefficients symbolically as they get large. x³ = 2*x + 1 = √5 + 2 x⁶ = 9 + 4*√5 x¹² = 161 + 72*√5

[@julianocg]

If you consider that (1+(5)^.5)/2=Phi, the golden ratio, and the relation Phi^n=Phi[n]*Phi+Phi[n-1], where Phi[n] is the n-th number of the Fibonaci sequence. Thus, Phi^12=Phi[12]*Phi+Phi[11], and finally, Phi^12=144*Phi+89.

[@kidscartoon4446]

Slove using cube root of unity Omega = -1/2-√5/2 In question asked ( -omega^12) that is equal to Omega^12 And omega^12 = Omega^3 and that is equal to 1

[@doubop8021]

beautiful... it's look like tricky but they are more to deal with intermediary forms. there's only one number than can fit x^2=x+1. that's make phi so powerful.