Japanese Math Olympiad Question | You should know this trick!
HTML-код
- Опубликовано: 1 сен 2023
- Hello My Dear Family😍😍😍
I hope you all are well 🤗🤗🤗
If you like this video about
How to solve √(11⁴+100⁴+111⁴)/2
please Like & Subscribe my channel as it helps me alot
🙏🙏🙏
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.
Well said!
Right
People are complaining how he skipped a step at 7.17 without explaining it.
@@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.
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
The last step can be further simplified:
a² + b² + ab = a² + 2ab + b² - ab = (a + b)² - ab = 111² - 11x100 = 12321 - 1100 = 11221 ...
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
Good idea
That works actually. I had to think about that negative number for a moment.
But it became even harder, how is it a simplification?
@@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 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
But a^2+b^2+ab is always positive
reread the last sentence please@@rssl5500
Use Pascal's triangle , for gods sake.
Or just the binomial theorem
@@14253689which has 1:1 connection to Pascal's triangle :/ (ong use potatoe! - or use spud.)
Delta it is sayed in my country
@@14253689that is what he used in the video
Tartaglia's Triangle
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 ! 🙂
Good teachnique for complex math solving
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.
I wondered why he didn't do that also.
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
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.
There is an algorithm for doing square roots by hand which makes the calculation very doable without guessing.
@@jonathansobieski2962gimme
You got that number mentally???
@@hajimehinata5854 Yes, I am gifted at mental math and can do mental computations to many millions.
The best i got at this mentally was 100159841+111^4
Wow
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
Just expand it according to (x+y)^2 = x^2 + y^2 + 2xy by letting x = a^2+b^2 and y = ab.
@@serbanudrea9429 Excellent; thank you for taking the time to respond. This was very helpful.
Not difficult.
Just the basics of
(a + b)² = a² + b² + 2ab
why didnt you recordnize the (A+B)^2 equality
@@huyminhha658 obviously because I need a lot more work. That's why I value this channel.
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
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 RUclips 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.
If 2+2 = 4, then 4 = 2+2 as well. It is the same for variables.
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.
Very nice and good idea,
You are the best!🎉
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...
Excellent video! Clear explanation using basic principles.
Glad it was helpful!
Thank you very much
Great, very clear!
Glad it was helpful! Thank you
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.
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
1000101
I think it would just be easier to multiply everything out
No not at all you need to multiply 11 six times. Then need to workout square root for huge number.
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.
This is defenitely 100% not a math Olympiad question.
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
I agree with your statement.
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
The important part is to see 111 = 100 + 11. The rest is a matter of "a little algebra yields..."
Yes
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
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!
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'
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.
An interesting exercise.
It is amazing how such a complex math problem after a few steps can equal a whole number and not a decimal.
The expression under the last rootsign should have been written [(a^2+b^2)+ab]^2. Would have been more clear.
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
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.
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.
You need to remember the algebra identities (a+b)^2 = a^2+b^2+2ab
etc
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;
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..
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
This man is a powerful wizard 👏
6:50 u said 2a^2b^2 but it was 3a^2b^2
THANK YOU!!!! I thought I was going crazy!
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
Thats my idea too.
121 +10000+12321/2
22442/2
11221
... and need not so much paper😊
You have done well in this long process
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.
GREAT
Thanks ❤️🙏❤️
nice
you could just use Pascal's triangle to find (a+b)^4 instead of going through 2 steps...
Solução elegante!!!
Thank you - this is a clever solution!
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
Are binomial theorem (of Newton) and squaring a sum of 3 terms such unheard things in Japan?
If you know how to calculate square roots by hand then this question is relatively easy to just do the calculation.
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.
I can't wait to use this.
Really really cool
Thanks 🙏❤️🙏
Nice.
Thanks!🙏❤️🙏
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.
Great job; crystal clear!!
Thank you! Thanks 🙏❤️🙏
Очень красивый пример! Моё решение немного отличается от вашего, но в общем всё одинаково.
Very good !!!
Thanks a lot! 😊👍
How to spend all your time on one problem and fail the test.
Be carefull when you simpify the square root of the square (this is a module definition).
Are you allowed to change (a^2)^2 + (b^2)^2 TO (a^2 + b^2)^2
excellently !!
Thank you very much!!
Better to expand (a+b)^4 all in one go, binomially?
Well, sorry to say, that's pure *grinding*, no trick at all. Olympiad-class students wouldn't and shouldn't do like this because that wasted time.
Wow
Thanks 👍☺️
I like maths
Красивое решение, понятное объяснение, Вы лучше всех объясняете на Ютубе
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.
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.
I have to admit, you completely lost me on this one. I love how you make algebra simple... usually. I watch your videos partly for fun. Partly to help my kids understand letters in math. I am going to have to rewatch this one
After rewatching I got it! I'm going to challenge my senior citizen Dad to this.
You can do it!
Great to hear 👍👍👍
@@learncommunolizer while I do understand the how part, I am still lost at the why you put the 111^4 to squared squared. It makes sense but I don't understand the why part.
@learncommunolizer my alternate method is rewrote 111 as 110 + 1 and take 4th power then 110 is same as 11×10 which you have factor of in jther terms..see what I meana?
@@learncommunolizerindont see why anyone would do the ma ovulation you do at 7:13..I don't see anyone thinking of that..would you agree?
Молодец, очень подробно изложено.
The technique explained is very lengthy and can be solved easier
Excellent explanation until about 8:16 when that step eluded me.
Easy peasy. 42
Knowing that root is power half, just divide all powers by 2. Then seperate them into partial Fractions.
(11^2/2)+(100^2/2)+ (111^2/2) = 11221
No Algebra Required.
You can't do that
And assume linearity?
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
I am almost the same as your video until "6:49" :p
It feels like a puzzle. You are stumped until you know the trick.
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?
That's easy : you don't know.
You just try and cross your fingers.
And if it doesn't work, you try something else!
These kinds of problems are designed to be simplified like this
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?😭😭😭
Cancel the square root by exponent-4, you're left with (11^2+100^2+111^2)/2
Then you solve the squares, you get (121+10,000+12321)/2
Then simply solve; 22442/2 = 11221
Please enlighten me because clearly this problem can't be this simple to solve considering I'm watching a guy expand this problem to this level
That gives the correct answer with these numbers, but exponents don't distribute over addition. sqrt(2^2+3^2)^(1/2) is not equal to 2+3.
How did you remove the square root from the denominator?
Hey Einstein, 11221 is not the correct answer. He omitted the calculation for 2ab(a sq +b sq.). It bothers me when math geniuses mess up. It makes me feel stupid
9:37 It's plot twist that you didn't do mental arithmatic for simple addings after all of these complex calculations..
It is Cardido indentity
Ecotú querido. Como te gusta complicarte.
Empirical thinking
we write that python and print answer. no calculator used
Not sure why in the minute 6.53 of the video you wrote 2 instead of 3 a^2 b^2.
He's splitting that out into 2a^2b^2+a^2b^2, because he knows that'll factorise neatly in the next step.
I have test rn, wish me gl
Only saw the thumbnail and calculated it in my head so I might have made a mistake, but my solution is 11221.
Rather laboured.
Den in nes step...
👍🏻
👍👍👍
I calculate all the problem in just 50 second
Calculation is easy than trick😂😂😂
Why those steps? Amazing
❤
11,221. I used a calculator.
At least this madness resulted in a single actual number and not some messy and unresolved quadratic equation solution!
Why didn't you use fourth power of sum formula?
Those are parenthesis, NOT brackets. This [ or ] is a bracket.
Um, yeah...I'll just go over here and hit this rock with another rock.
When I do the math on a calculator, I get an answer which is vastly different from the answer from the math. Also, I believe 111 to the fourth power is not equal to ( 100+11) to the fourth power. At least I am unable to get the same answers with a calculator. Either I am wrong or the answer is in error .
Your doing something wrong using a calculator you get the same answer albeit 10 mins quicker.
John, you must have made a mitsake somewhere because the expression resolves exactly to 11221.
Also 111^4 = (111)^4 = (100+11)^4.
You might have forgot to divide the inside of the root by 2… use parentheses on your calculator to combine the numerator, since it all has to be added first, and then divided by 2, and all that taken the square root of… just a suggestion…
The trap is, when someone tries to type an expression on a calculator because he doesn't master the expression, the typing is half the time wrong. Teacher's experience.
3:47 bro I think I skipped too far
111^2
I like when you say powa xd
I solve the same but using 111-11 in 100 ^4 bracket