Wow I'm shocked at the negative comments on this video. First there are those complaining the steps are not explained well even though he did the steps right down to basic arithmetic, along with one comment complaining the opposite (can't please all the people all the time and all that). Secondly, his solution is still a clever purely algebraic way of dealing with problem which assumes only basic maths. As for those droning on about Pascal's triangle, well why don't you make a video using this solution and see how long it takes you to first teach how the triangle comes about as a way of finding the exponents of a binomial expansion. Man there are a lot of haters out there. So let me just encourage the content creator and say well done on making a very carefully explained video of an elegant solution to the math olympiad problem.

@davidhowe6905

8 ай бұрын

Well said!

@BP-gn2cl

8 ай бұрын

Right

@Player-pj9kt

8 ай бұрын

People are complaining how he skipped a step at 7.17 without explaining it.

@rodbenson5879

8 ай бұрын

​@@Player-pj9ktin defence of the content creator, he had already expanded (a+b)^2 showing it to be equal to a^2 + b^2 + 2ab so someone watching carefully should be able to spot the refactorisation although I guess a quick sentence explaining he is reversing the process would have been helpful for those who know very little highschool algebra. However, this slight improvement does not justify the negativity of some of the comments.

@davidhowe6905

8 ай бұрын

It's the same procedure as expanding (x + y)^2 = (x + y)(x + y) = x^2 + y^2 + 2xy but with x=a^2, y=b^2@@Player-pj9kt

Good teachnique for complex math solving

I assume this problem is about right grouping and then using just 2 equations: 1st (a-b)^2=a^2-2ab+b^2 2nd a^2-b^2 =(a-b)*(a+b) This way it should be possible to avoid complicated calculstions. And 111-11=100 pressumes to use those equations.

If one wants to be extra rigorous, you need to make sure that the squared and rooted number is not negative before you cancel the power and the root. In this example it is obviously positive, but under any test-like conditions it's always better to show having considered such things

@rssl5500

2 ай бұрын

But a^2+b^2+ab is always positive

@kamilrichert8446

2 ай бұрын

reread the last sentence please@@rssl5500

Use Pascal's triangle , for gods sake.

@14253689

8 ай бұрын

Or just the binomial theorem

@HoSza1

8 ай бұрын

​@@14253689which has 1:1 connection to Pascal's triangle :/ (ong use potatoe! - or use spud.)

@ChristelleHilaire-lb6pu

8 ай бұрын

Delta it is sayed in my country

@brownie3454

8 ай бұрын

@@14253689that is what he used in the video

@lorenzolombardi1211

7 ай бұрын

Tartaglia's Triangle

Very nice and good idea,

The last step can be further simplified: a² + b² + ab = a² + 2ab + b² - ab = (a + b)² - ab = 111² - 11x100 = 12321 - 1100 = 11221 ...

@drinkchan4822

8 ай бұрын

the point was to not do 111^2 since it the whole point of this algebra was to make it possible to evaluate it in the easiest form when the algebra is converted back into numbers

@malaramesh8766

8 ай бұрын

Good idea

@user-qv2dd8ex8k

7 ай бұрын

That works actually. I had to think about that negative number for a moment.

@rakhatthenut3815

4 ай бұрын

But it became even harder, how is it a simplification?

@barttemolder3405

4 ай бұрын

@@rakhatthenut3815 You only have to do 2 multiplications and one subtraction this way. In the general case that would be easier. For this particular set of numbers it is not necessary as it is also trivial to calculate the result one step back, but the fun is in reducing the equation to the bare minimum.

I tried to do this one mentally. After awhile, I had to take the square root of 125,903,841, which I knew had to be roughly 11,xy1. Finding the digits of x and y was tough.

@jonathansobieski2962

6 ай бұрын

There is an algorithm for doing square roots by hand which makes the calculation very doable without guessing.

@RvVx7

5 ай бұрын

​@@jonathansobieski2962gimme

@hajimehinata5854

4 ай бұрын

You got that number mentally???

@DavidVonR

4 ай бұрын

@@hajimehinata5854 Yes, I am gifted at mental math and can do mental computations to many millions.

@austenmaster8981

2 ай бұрын

The best i got at this mentally was 100159841+111^4 Wow

An interesting exercise.

Well, the problem stands mainly upon the property : a^4 + b^4 + (a + b)^4 = 2 (a^2 + b^2 + ab)^2. Let us prove it with a far simpler calculation : We define P(X) = (X^2 + aX + a^2)^2 Then P’(X) = 2 (2X + a) (X^2 + aX + a^2) We define Q(X) = X^4 + (X + a)^4 + a^4 Then Q’(X) = 4 (X + a)^3 + 4 X^3 Using the well-known x^3 + y^3 = (x + y) (x^2 - xy + y^2), we obtain : Q’(X) = 4 (2X + a) [(X+a)^2 - X(X+a) + X^2) That is : Q’(X) = 4 (2X + a) (X^2 + aX + a^2). Obviously, Q’(X) = 2 P’(X), then Q(X) = 2 P(X) + k. Noticing that Q(0) = 2 P(0), we have k = 0 (qed). Thanks for your interesting videos ! 🙂

You are the best!🎉

It is amazing how such a complex math problem after a few steps can equal a whole number and not a decimal.

Thank you - this is a clever solution!

Great, very clear!

@learncommunolizer

9 ай бұрын

Glad it was helpful! Thank you

Excellent video! Clear explanation using basic principles.

@learncommunolizer

4 ай бұрын

Glad it was helpful! Thank you very much

nice

Using the "Pascal triangle" (in french) : 1, 1 1, 1 2 1, 1 3 3 1, 1 4 6 4 1, ... the developpement of (a+b)^4 is faster ;-) (NB This triangle is also used for Cn,p) Sorry for my english and thank you for the video;

When I was at school, decades ago, we configured "let" statements this way: "let a=b+1". In this solution, it's done like that. But many other maths problems here on KZread configure it as "let b+1=a". Is there a correct way, or does it just not matter? To my mind, the first way makes more sense, but I'm happy to be corrected if I'm wrong.

@hmwndp

3 ай бұрын

If 2+2 = 4, then 4 = 2+2 as well. It is the same for variables.

@AlaiMacErc

3 ай бұрын

Curious! I don't recall seeing the second style at all. I'd agree the first is much clearer, as it's saying which the new variable is, and it's giving a closed-form expression for its value. But one could argue any statement of the form "let P(x)" where P is some proposition that determines a value for its parameter is valid... if harder to follow.

this is of the form V(A/2) where A = x^4+y^4+(x+y)^4 expand the latter term and add corresponding terms so that A = 2x^4 +2y^4 + 4x³y + 4xy³ + 6x²y² hence A/2 = x^4 +y^4 + 2x³y+2xy³+3x²y² regroup this as (x^4 +y^4 + 2x²y²) + (2x³y+2xy³) + x²y² = (x²+y²)² +2xy (x²+y²)+ x²y² = [(x²+y²)+xy]² hence V(A/Z) = (x²+y²)+xy fill out x,y to find 10 000 + 121 + 1100 = 11221

This looks familiar from my school days what level of math is this? Algebra 3 and 4 ? Or pre calculus. Understanding those sub categories is where I lost it years ago..

GREAT

@learncommunolizer

8 ай бұрын

Thanks ❤️🙏❤️

The important part is to see 111 = 100 + 11. The rest is a matter of "a little algebra yields..."

@straightpool1964

7 ай бұрын

Yes

Let s = 5.5, t = 105.5 = 100+s, and let x = the answer. 2x² = 11⁴ + 100⁴ + 111⁴ = 16s⁴ + (t-s)⁴ + (t+s)⁴ = 2[8s⁴ + t⁴ + 6s²t² + s⁴] = 2(t² + 3s²)² x = (100+s)² + 3s² = 100² + 200s + 4s² = 10,000 + 1100 + 121 = 11,221

Are binomial theorem (of Newton) and squaring a sum of 3 terms such unheard things in Japan?

Great job; crystal clear!!

@learncommunolizer

8 ай бұрын

Thank you! Thanks 🙏❤️🙏

To find (a+b)^n expansion we can draw n+1 lines of pascal's triangle and the (n+1)th line gives the coefficients of all terms in the expansion. Then we can write the terms by decreasing the powers of a from n to 0 and increasing the powers of b from 0 to n along with the coefficients. And add all the terms. That is the expansion.

@josiahgibbs5697

5 ай бұрын

I wondered why he didn't do that also.

You have done well in this long process

Solução elegante!!!

I love the many ways to go about the solution. It is a shame that modern math ed does not encourage other processes for a soltion.

Nice.

@learncommunolizer

8 ай бұрын

Thanks!🙏❤️🙏

Fortunately, those of us not having to participate in the Math Olympiad can just punch it into our calculators. LOL It is still interesting to see how the answer is worked out using just pencil and paper. It is fascinating how many times a problem that looks complex as all get out can be simplified with the judicious use of substitutions.

@squirming_squirrels

8 ай бұрын

This is defenitely 100% not a math Olympiad question.

@andreibratosin1199

7 ай бұрын

Lol this is not olympiad level. Where I'm from we used to do these in 7th grade .. and much faster cus this method is archaic and lengthy

@tigistafine202

7 ай бұрын

I agree with your statement.

@aaronhansen706

6 ай бұрын

I am in my 50s. I got my GED at 17 and I think I aced it although it is nothing but pass/fail. I did a couple years at a community college and decided I would rather labor for a living. I do some algebra in my head as I walk around sometimes. Usually to figure out what the mindset of the civil engineer was thinking. I can only do it in my head when I am distracted by music. I need to be distracted to concentrate. My fav college course was actually a lifesaving class at De Anza college in Cupertino, CA. It covered some very... bad situations and gave a ton more info than the Boy Scout stuff I learned. I do love training my brain

Really really cool

@learncommunolizer

8 ай бұрын

Thanks 🙏❤️🙏

I can't wait to use this.

Usually you show us every step. This time for some reason you didn't show us the critical step. Why did you do that? It's extremely difficult to understand how to combine all of those terms into SQR of (a^2+b^2+ab)^2

@serbanudrea9429

9 ай бұрын

Just expand it according to (x+y)^2 = x^2 + y^2 + 2xy by letting x = a^2+b^2 and y = ab.

@xyz.ijk.

9 ай бұрын

​@@serbanudrea9429 Excellent; thank you for taking the time to respond. This was very helpful.

@hybridaccounts

8 ай бұрын

Not difficult. Just the basics of (a + b)² = a² + b² + 2ab

@huyminhha658

8 ай бұрын

why didnt you recordnize the (A+B)^2 equality

@xyz.ijk.

8 ай бұрын

@@huyminhha658 obviously because I need a lot more work. That's why I value this channel.

I think it would just be easier to multiply everything out

@malaramesh8766

8 ай бұрын

No not at all you need to multiply 11 six times. Then need to workout square root for huge number.

Better to expand (a+b)^4 all in one go, binomially?

Very good !!!

@learncommunolizer

7 ай бұрын

Thanks a lot! 😊👍

I have started in general approach; to solve (a^4+b^4+(a+b)^4)/2. which is actually: (a^2+b^2+ab)^2 and solution in general is a^2+b^2+ab...

Anyone would think of letting a=11 and b=100 and then using FOIL to simplify. I believe those who left negative comments on this video thought there should be more creative and easier way to solve this since he mentioned that there would be a 'trick'

excellently !!

@learncommunolizer

3 ай бұрын

Thank you very much!!

For getting the value of a² + b²+ab You just conver into (a+b)² -ab = (100+11)² -(100 x 11) =12321 -1100 =11221 OK.

👍🏻

@learncommunolizer

9 ай бұрын

👍👍👍

Wow

@learncommunolizer

7 ай бұрын

Thanks 👍☺️

The expression under the last rootsign should have been written [(a^2+b^2)+ab]^2. Would have been more clear.

@navi2710

5 ай бұрын

I just read it as : X = a2 + b2 Y = ab So (X + Y)2 = x2 + y2 +2xy = (a2 ×b2)2 + (ab)2 + 2ab(a2 + b2) Therefor (X + Y)2 is also (a2 + b2 + 2ab)2

The issue with those videos is that they are made with the solution and tactic already in mind, rather than experience it first hand. Although I understand your tactic , as others have already stated, you rushed it as the most critical point (which renders this whole process useless). Probably because you forgot the next step, but had the answer in mind.

Looks like you expanded it. Here is a trick: a^4 + b^4 +(a+b)^4 = a^4 + b^4 + 2a^2b^2 +(a+b)^4 - a^2b^2 -a^2b^2 =(a^2 +b^2)^2 - a^2b^2 +((a+b)^2 - ab)((a+b)^2 + ab) = (a^2 + b^2 + ab)(a^2+b^2-ab) + (a^2 +b^2+ab)(a^2+b^2+3ab) =2(a^2+b^2+ab)^2

@rakos1

7 ай бұрын

1000101

I like maths

I really enjoyed watching that and despite my never being that quite that good at expansions even at my best I still thought every step was clear and explained perfectly well and I had no difficulty following it at all. People sometime miss the fact that sometimes math can just be fun.

Красивое решение, понятное объяснение, Вы лучше всех объясняете на Ютубе

It feels like a puzzle. You are stumped until you know the trick.

❤

Очень красивый пример! Моё решение немного отличается от вашего, но в общем всё одинаково.

you could just use Pascal's triangle to find (a+b)^4 instead of going through 2 steps...

Молодец, очень подробно изложено.

If you know how to calculate square roots by hand then this question is relatively easy to just do the calculation.

Excellent explanation until about 8:16 when that step eluded me.

Easy peasy. 42

Beyond basic algebras 1 and 2, geometry, high school statistics, and precalc, math becomes so abstract. 😅😅😅 I suspect that being well grounded in the courses I've mentioned is enough to survive the business world. The exception is for those pursuing engineering and specific science endeavors

I like when you say powa xd

Are you allowed to change (a^2)^2 + (b^2)^2 TO (a^2 + b^2)^2

Good explanation as always. We can just use the binomial theorem. Could be solved in a jiffy. This is way too long and max time allowed for such a questions is 2 mins Max.

6:50 u said 2a^2b^2 but it was 3a^2b^2

@gyrlyninja

6 ай бұрын

THANK YOU!!!! I thought I was going crazy!

Still don't understand how to solve this. It seems like just random things get written down out of nowhere. So frustrating to be so bad at math and I try to learn with videos like this but I just get more confused and frustrated.

@logica_1989

6 ай бұрын

You should give up. Stop banging your head on something you were not made for. You're never going to be good at this. Pursue something else.

@funprog

2 ай бұрын

You need to remember the algebra identities (a+b)^2 = a^2+b^2+2ab etc

Be carefull when you simpify the square root of the square (this is a module definition).

Ecotú querido. Como te gusta complicarte.

Empirical thinking

The technique explained is very lengthy and can be solved easier

Most easy method i am telling you. If possible then note it down.... We know that in the language of exponent we define ^nroot a as a^1/n So root 11^4 becomes 11^4×^(1/2) root 100^4 becomes 100^4×^(1/2) And root 111^4×^(1/2) Now by fraction we know that a+b+c/2 = a/2 + b/2 + c/2 Thus, 11^4×^(1/2)/2 + 100^4×^(1/2)/2 + 111^4×^(1/2)/2 After cancellation we get =11^2/2 + 100^2/2 + 111^2/2 = 11×11/2 + 100×100/2 = 50 × 100 + 111×111/2 = 121/2 + 5000 + 12321/2 Using Associative law here Then, 5000 + 121/2 + 12321/2 = 5000 + 121 + 12321/2 = 5000 + 12442/2 = 5000 + 6221 = 11221. SIMPLE!

At least this madness resulted in a single actual number and not some messy and unresolved quadratic equation solution!

How to spend all your time on one problem and fail the test.

How did I get the same answer by taking the sqrt of the original problem and getting (11^2 + 100^2+111^2)/2. Which is essentially the same as the very last equation

@uwearnold3680

8 ай бұрын

Thats my idea too. 121 +10000+12321/2 22442/2 11221 ... and need not so much paper😊

Rather laboured.

9:37 It's plot twist that you didn't do mental arithmatic for simple addings after all of these complex calculations..

I have test rn, wish me gl

PAINFULLY slow - for the inner expression anyone doing this level maths can go straight to a^4 + b^4 + 2a^3b +2ab^3 +3 a^2 b^2 in a single step! Then (a^2+b^2)^2 + a^2b^2 +2 (...) then (a^2+b^2 +ab)^2 - 3 steps, and without the need for substitution with a,b

Um, yeah...I'll just go over here and hit this rock with another rock.

Den in nes step...

Why those steps? Amazing

Only saw the thumbnail and calculated it in my head so I might have made a mistake, but my solution is 11221.

My gush, you never heared from Pascal's triangle???

The question is "How did you know that the square root will cancel a square in the end?", did you saw the future, or just believing that every math question is always like that?

@thothorleboiteux9900

8 ай бұрын

That's easy : you don't know. You just try and cross your fingers. And if it doesn't work, you try something else!

@funprog

2 ай бұрын

These kinds of problems are designed to be simplified like this

I calculate all the problem in just 50 second Calculation is easy than trick😂😂😂

It is Cardido indentity

111^2

# 45 #

11,221. I used a calculator.

we write that python and print answer. no calculator used

The step before you reverted a and b back to 100 and 11, you didn't go over how you got root((a^2 + b^2 +ab)^2). I'm totally not seeing how you got to that point.

@billh5923

Ай бұрын

He doesn't know what cancel means. He used the term cancel when he was combining terms and crossing them out. This is just bookkeeping, it is not a cancel.

Elshadai, criador Elshadai, salvador Elshadai, o poderoso Elshadai, o grande eu sou

@RobiBue

8 ай бұрын

Você esqueceu El Shaddai, o matemático

I solve the same but using 111-11 in 100 ^4 bracket

This stresses me out so much. Maths has always stressed me out. It just 3 numbers 121+10000+12321 then divided in half. I don't understand all of this. No one ever explained to me why or what all the steps are for. It's literally 4 steps why is all of this necessary? Like what is the purpose? Even in school no one would ever explain to me why that was necessary. I always got in trouble because I couldn't show my work. But they never told me what work I was supposed to show, I just freaking gave up. Even on this why are they writing 47000 different numbers/letters? WHY are their letters? And where do the squares come from?😭😭😭

let a = 100,b=11 , =((2a^4+4a^3b+6a^2b^2+4ab^3+b^4)/2)^(1/2)=(a^4+2a^3b+3a^2b^2+2ab^3+b^4)^(1/2) =((a^2+b^2)^2+2ab(a^2+b^2)+(ab)^2)^(1/2) = (((a^2+b^2)+ab)^2)^(1/2)=(a^2+b^2)+ab = 10000+121+1100 = 11221

@TWJRPGGamming

9 ай бұрын

I am almost the same as your video until "6:49" :p

This can be done in more frequent way, If wanna know then...............

Простая школьная задача. Непонятно, что в ней такого сложного.

Not sure why in the minute 6.53 of the video you wrote 2 instead of 3 a^2 b^2.

@AlaiMacErc

3 ай бұрын

He's splitting that out into 2a^2b^2+a^2b^2, because he knows that'll factorise neatly in the next step.

😂😂😂 If we do the normal process, the solution takes 3-5 minutes, see video 10.

Didn't see anything ... calculating everything with brute force ... 111^4 = 100^4 + 4 100^3 11 + 6 100^2 11^2 + 4 100 11^3 + 11^4 (111^4 + 100^4 + 11^4)/2 = 100^4 + 2 100^3 11 + 3 100^2 11^2 + 2 100 11^3 + 11^4 11^2 = 121 ; 11^3 = 1331 ; 11^4 = 14641 = 10^8 + 22 10^6 + 363 10^4 + 2662 10^2 + 14641 = 100000000 + 22000000 + 3630000 + 266200 + 14641 22000000 3630000 266200 14641 125910841 square root 1.25.91.08.41 1 25 1 21.1 = 21 4 91 2 222.2 = 444 47 08 2 2242.2 = 4484 2 24 41 1 22441.1 = ok Answer = 11221

@andreibratosin1199

7 ай бұрын

This man is a powerful wizard 👏

3:47 bro I think I skipped too far

A=11cm ; B=100cm

直接按計算機比較快。