*_Liked mah keht? Then sub to the channel and check out some of deez :^)_* Train your Number Theory Expertise by trying out Brilliant! =D brilliant.org/FlammableMaths Check out my newest video over on @Flammy's Wood ! =D ua-cam.com/video/WxoYVUKiHBA/v-deo.html Support the channel by checking out Deez Nutz over on stemerch.eu/products/deez-nuts-premium-3d-cutting-board?_pos=1&_sid=3ebd4ef06&_ss=r ! :3 Engi Watch: stemerch.com/products/the-incredibly-unrigorous-engineering-watch?variant=40377075728562
papa Flammy here is how I did it sqrt ((a^4+b^4+(a+b)^4) /2) I expanded it all out, I got sqrt(a^4+b^4+3(ab)^2 + 2b a^3+2a b^3 ) I just used the symmetric polynomials to motivate the rest, assume, s1=a+b=111(SIXTH OF EVIL) , s2=ab=1100 I and subtracted and added 3(ab)^2 + 2b a^3+2a b^3 again inside, I got sqrt(s1^4 -3s2^2 -2 s2 (s1^2 - 2 s2)) foiling it out, we get sqrt(s1^4 -3s2^2 -2 s2 s1^2 + 4 s2 ^2))= sqrt(s1 ^4 +s2 ^2 -2 s1^2 s2)= s1^2 -s2 = 111^2 - 1100 = 12321 - 1100=11221
And that works because 11 can be written as 10 + 1 = a + b, whose powers can then be expanded with the Binomial Theorem. You can extend the fun by using powers of 101, or even 1001. Fred
Even if you can't remember binomial coefficients, it takes seconds to write out Pascal's triangle and find the row 1 4 6 4 1 - and save a lot of chalk.
From the thumbnail, musing this thing, let's try this. Let a = 11, b = 100 [Note: Some aids in expanding/contracting powers of polynomials: 11⁴ = 14641; 111² = 12321] Then 11⁴ + 100⁴ + 111⁴ = a⁴ + b⁴ + (a+b)⁴ = 2a⁴ + 4a³b + 6a²b² + 4ab³ + 2b⁴ And ½(11⁴ + 100⁴ + 111⁴) = a⁴ + 2a³b + 3a²b² + 2ab³ + b⁴ = (a² + ab + b²)² = (121 + 1100 + 10000)² = 11221² So √[½(11⁴ + 100⁴ + 111⁴)] = 11221 Yeah, I think I can see my way clear to calling that wild! Fred
I don't really like using Pascal's triangle too. It gets more laborious the higher the exponent. I just like to think about the coefficients combinatorially. It's better because you can even think about general multinomials.
Brute force it's that bad: First observe that sqrt((11^4+100^4+111^4)/2) > sqrt((100^4+111^4)/2) > (100^2 + 111^2) / 2, hence the final answer is greater than 11160. In fact it should be very close to 11160 since 100 and 111 are close enough. Next, we need to calculate the last 2 digits of the number under square root. In mod 100 congruence classes, (11^4+100^4+111^4)/2 is equivalent to (11^4 + 111^4) / 2 is equivalent to (11^4 + 11^4) / 2 is equivalent to 11^4 is equivalent to 121^2 is equivalent to 41. Using the same trick we will be able to find last 4 digits are "0841" and 5th is either 1 or 6, this will be useful later. Since the last digit under the square root is 1, we know that the last digit of the final answer should be 9 or 1. Given the last digit of final answer is 9 or 1, the last-but-one digit under the sqare root is 4, we know that the last-but-one digit of the final answer should be 7 or 2. Now we know that the final answer is greater than and very close to 11160, last two digits are 21, 29, 71 or 79. We can start brute forcing: we will start from 11171 and go larger. We only need to check numbers whose last 2 digits are 21, 29, 71 or 79, and tests if the last 4 digit of its squre is "0841" and the 5th digit of its square is either 1 or 6. We should find the final answer 11221 in our 3rd guess.
Using the method where you complete the square twice for the x4 + y4 term converts it into a beautiful square of (x + y)^2 + x^2y^2. It also doesn't need you to have the knowledge of the binomial expansion for 4th power either. Took me a minute to solve it that way
Damn I thought I was shit at math but I made like 3 good moves and I just didn't notice the final factorisation. Now I have the copium to continue math.
I haven't looked at the video yet, but just use the identity: a^4 + b^4 + (a+b)^4 = 2(a^2 + ab + b^2)^2 Then the problem just ends up being to evaluate: a^2 + ab + b^2 at a=11 and b=100.
There is an even more elegant form of the answer with a bit more manipulation You have Sqrt (1/2*(a^4+b^4+(a+b)^4)) = a^2+b^2+ab =1/2*(a^2+ b^2+(a^2+b^2+2ab)) =1/2*(a^2+b^2+(a+b)^2) That is Sqrt (1/2*(a^4+b^4+(a+b)^4)) =1/2*(a^2+ b^2+(a+b)^2) Which is much more elegant imo
small comment: (a+b)^4 can be seen ass all the possible ways to draw 2 balls (red,blue) 4 times. Which means either all red a^4 or all blue b^4 or one red and 3 blue (4 possible ways to arrange) or 3 blue on red (4 times also, because of symmetry) or 2 blue 2 red (6 possible ways). edit: (if you need some help thinking about it, look at 3!, 2! 1! choices to arrange a color of your choice)
Well if you use some foresight you can see that (a+b)^4 consists of a^4 + b^4 plus the even-coefficient terms 4ab(a+b) + 6(ab)^2, so you know that can be used to get rid of the fraction right away. So, (a^4 + b^4 + (a+b)^4)/2 = a^4 + b^4 + 2ab(a^2 + b^2) + 3(ab)^2 = (a^2 + b^2)^2 + 2ab(a^2 + b^2) + (ab)^2 = (a^2 + b^2 + ab)^2
Factoring the polynomial under the radical can be done faster since you know it's a perfect square. a^4 and b^4 means it's (a^2 + k a b + b^2)^2 for some k, which clearly must be 1.
All those equations are slower than just do the math 11^4 = 14641 100^4 = 100000000 111^4 = 151807041 251821682÷2 = 125910841 And the complicated part is the √125910841 I think best way is by approximation So, You know the solution is close to 11200, so You try until You find it. It took me about 8 attemps, and about 6min
@@mantizshrimp Yeah. And we're assuming they can't use a calculator, wich for me has no sense at all. In a math olympiad the last thing You want to know is how good are the participants calculating stuff, what You really want to know is how good are they at solving problems, and turning those problems into math. If the olympiad is just problems like this one thats a math competition i'm not interested in watch or participate
I computed the square root by just plugging in a=10, b=1. Finding the square root of 12321 is pretty trivial (try 100^2, 110^2, 111^2), and that gives me a clue that the factorization would be 111^2 = (100 + 10 + 1)^2 = (a^2 + ab + b^2)^2 with a pretty high degree of certainty.
I figured it was intended to be a factoring problem due to the setup, but out of curiosity I simply computed the answer using traditional arithmetic algorithms. It took less than 4 minutes. The only multiplication that was even somewhat involved was calculating 111^2*111^2 or 12321*12321 and, even then, you could tell at a glance that there was no carrying in the multiplication part, so that took maybe 30 seconds to write down, then it was just an addition problem. All the multiples you had to find when doing the square root algorithm were either 1 or 2 with, again, no carrying making it trivial to do it in your head, so that was as easy as it gets. I get that brute force computation is against the spirit of the problem, but they should really make these problems more difficult if they want the student to have to resort to using factoring to simplify the problem. Any 6th grader should be able to do this armed with nothing more than the multiplication algorithm, the short division algorithm, and the square root algorithm.
Interestingly there 637 is the smallest number expressible in this way in three distinct ways (without permutation) i.e. (a,b) = (4,23); (7,21) and (12,17)
Yeah the "square root" symbol (or sqrt() in text) is really the "positive square root" symbol. For the negative square root you'd use -sqrt(). And since a²+b²+ab is always >=0 which can be easily checked as an exercise to the reader, we have sqrt( (a²+b²+ab)² ) = |a²+b²+ab| = a²+b²+ab, removing the absolute value we usually have in sqrt(X²) = |X|.
I am not the best at mathematics but I believe there might've been a mistake when factorizing a^4+ + b^4 + 3a^2b^2 as he might've seen the three coefficient as a two, therefore allowing him to factorize it normally. Of course I could always be wrong, since he is a a lot better at mathematics then I can ever aspire to be.
The oversimplified answer is physics. Linear algebra explains how vectors work, and pretty much all of physics is described in vectors. In addition, physics very commonly uses quantities like velocity and acceleration which are the derivatives of some other quantity, like position in this case. Since these are often related (as a simple example, take the force on a spring: f = -kx = ma is a second order differential equation) differential equations show up a lot.
After 20 years of teaching in high school I sometimes could hardly speak because of the chalk dust. But luckily then came the whiteboards with the alcohol based makers...
You switched divisibility laws : Let a,b,c be positive integers, if a | b and a | c then of course a | bc, but if a | c and b | c, ab | c if and only if a and b are coprimes but GCD(8;10)=2
i literally just took the square root of all the numbers that were ^4 which is surely that same number squared, added them and only then divided by two DO NOT ASK ME HOW *THAT* went right, but it did, and i came to solving this shit without using a single piece of sheet in under 30 seconds and most importantly, CORRECTLY you can try this at home, it'll surely work *it's not about getting the "right" path, it's about getting to the right destination* but seriously, this shit is so stupidly difficult you should probably just give finding the right answer up, and do ANYTHING just to get ANYWHERE and not go by this thing without answering it you might get it right, just like i did take care
Ah Pascal’s triangle……. I would prefer staying away from that side of math. Instinct keep telling me to raise everything to a 1/2 power and then multiply and divide out using exponent laws
@@ytbook9639 because I’m to use to dealing with multiplication in the root, it’s annoying when it’s addition. square root = being raised to the 1/2 power, but again since it’s addition inside of the root you can’t just do that. If you could then then the exponents would just be 4 times 1/2 or just 2.
*_Liked mah keht? Then sub to the channel and check out some of deez :^)_*
Train your Number Theory Expertise by trying out Brilliant! =D brilliant.org/FlammableMaths
Check out my newest video over on @Flammy's Wood ! =D ua-cam.com/video/WxoYVUKiHBA/v-deo.html
Support the channel by checking out Deez Nutz over on stemerch.eu/products/deez-nuts-premium-3d-cutting-board?_pos=1&_sid=3ebd4ef06&_ss=r ! :3
Engi Watch: stemerch.com/products/the-incredibly-unrigorous-engineering-watch?variant=40377075728562
deez what
@@e.b.1115 n u t z
papa Flammy here is how I did it
sqrt ((a^4+b^4+(a+b)^4) /2)
I expanded it all out,
I got sqrt(a^4+b^4+3(ab)^2 + 2b a^3+2a b^3 )
I just used the symmetric polynomials to motivate the rest,
assume, s1=a+b=111(SIXTH OF EVIL) , s2=ab=1100
I and subtracted and added 3(ab)^2 + 2b a^3+2a b^3 again inside,
I got sqrt(s1^4 -3s2^2 -2 s2 (s1^2 - 2 s2))
foiling it out, we get sqrt(s1^4 -3s2^2 -2 s2 s1^2 + 4 s2 ^2))= sqrt(s1 ^4 +s2 ^2 -2 s1^2 s2)= s1^2 -s2 = 111^2 - 1100 = 12321 - 1100=11221
@@RockHardWoodDaddy it's very pleasing to see how 11^2 + 100^2 + 11*100 = 111^2 - 11*100.
In which Flammy demonstrates that it's OK to love your pussy.
Fred
I did this in 6.9 seconds and 420 milliseconds by using a secret technique passed down through generations. It's called "the calculator methood".
You're pretty fast at typing on a calculator if it took you just 6.9 s + 420 ms for that
WOWOOWOWOWWO!!!!!! IA M SO IMPR ESSED but Math Olympiad doesn't allow you to take it inside the exam hall sire!!
@@psikoexe I did it in 4 mins using symmetric polynomials it is there as a reply in papa flammy's pinned comment if you are interested.
@@Nothingtonnobodson nope
@@Nothingtonnobodson not interested
Guys, did you know that 11^n corresponds to the n-th row of Pascal's triangle?
11^0 = 1
11^1 = 11
11^2 = 121
11^3 = 1331
11^4 = 14641
11^5 = 161051, whereas the row goes: "1 5 10 10 5 1"
Yes, and Flammy doesn't give a *BLEEP!* about Pascal's Triangle. An attitude that cost him much more work.
Fred
@@ffggddss Papa Flammy, say it isn't so!
wowwwww
And that works because 11 can be written as 10 + 1 = a + b, whose powers can then be expanded with the Binomial Theorem.
You can extend the fun by using powers of 101, or even 1001.
Fred
Wait that’s actually sick
Even if you can't remember binomial coefficients, it takes seconds to write out Pascal's triangle and find the row 1 4 6 4 1 - and save a lot of chalk.
He doesn't give a shit about Pascal's triangle or how much chalk he's using lol. The chalk is meant to be abused.
Flammable Quotes:
“I don’t really give a shit about Pascal’s Triangle either”
“we can tell the absolute value to fuck off basically”
missing "binomial go brr"
I agree, that solution was absolutely beautiful. It turned a complicated question into a problem that can be solved manually.
That intro was absolutely adorable!
hehe :)))
still waiting for a complex differential geometry video D:
Papa got the schordingers cat
From the thumbnail, musing this thing, let's try this.
Let a = 11, b = 100
[Note: Some aids in expanding/contracting powers of polynomials: 11⁴ = 14641; 111² = 12321]
Then
11⁴ + 100⁴ + 111⁴ = a⁴ + b⁴ + (a+b)⁴ = 2a⁴ + 4a³b + 6a²b² + 4ab³ + 2b⁴
And
½(11⁴ + 100⁴ + 111⁴) = a⁴ + 2a³b + 3a²b² + 2ab³ + b⁴ = (a² + ab + b²)² = (121 + 1100 + 10000)² = 11221²
So
√[½(11⁴ + 100⁴ + 111⁴)] = 11221
Yeah, I think I can see my way clear to calling that wild!
Fred
I don't really like using Pascal's triangle too. It gets more laborious the higher the exponent. I just like to think about the coefficients combinatorially. It's better because you can even think about general multinomials.
This was so much fun!
Brute force it's that bad:
First observe that sqrt((11^4+100^4+111^4)/2) > sqrt((100^4+111^4)/2) > (100^2 + 111^2) / 2, hence the final answer is greater than 11160. In fact it should be very close to 11160 since 100 and 111 are close enough.
Next, we need to calculate the last 2 digits of the number under square root. In mod 100 congruence classes, (11^4+100^4+111^4)/2 is equivalent to (11^4 + 111^4) / 2 is equivalent to (11^4 + 11^4) / 2 is equivalent to 11^4 is equivalent to 121^2 is equivalent to 41. Using the same trick we will be able to find last 4 digits are "0841" and 5th is either 1 or 6, this will be useful later.
Since the last digit under the square root is 1, we know that the last digit of the final answer should be 9 or 1. Given the last digit of final answer is 9 or 1, the last-but-one digit under the sqare root is 4, we know that the last-but-one digit of the final answer should be 7 or 2.
Now we know that the final answer is greater than and very close to 11160, last two digits are 21, 29, 71 or 79. We can start brute forcing: we will start from 11171 and go larger. We only need to check numbers whose last 2 digits are 21, 29, 71 or 79, and tests if the last 4 digit of its squre is "0841" and the 5th digit of its square is either 1 or 6. We should find the final answer 11221 in our 3rd guess.
Using the method where you complete the square twice for the x4 + y4 term converts it into a beautiful square of (x + y)^2 + x^2y^2. It also doesn't need you to have the knowledge of the binomial expansion for 4th power either.
Took me a minute to solve it that way
This was genuinely fun thanx papa!! Does brilliant do this, i'll sign up in an instant if brilliant is this fun!!
Mathematics is my passion and I enjoyed this video a lot…Really cool problem and really cool way to solve it!
quick lil' question, are you romanian?
@@TP-te4eb alright
Such a wholesome intro!
Damn I thought I was shit at math but I made like 3 good moves and I just didn't notice the final factorisation. Now I have the copium to continue math.
I haven't looked at the video yet, but just use the identity:
a^4 + b^4 + (a+b)^4 = 2(a^2 + ab + b^2)^2
Then the problem just ends up being to evaluate: a^2 + ab + b^2 at a=11 and b=100.
So 122100 then. This is what occurred to me personally too, except I was unaware of the equivalent RHS. It is not commonly taught in schools/colleges.
10000+1100+121 is 11221.
yep he did basically that
How do you turn the left hand side to the right hand side?
10:28 when you canceled out the 2 i wasnt looking so i thought it was to the fourth power
Great video, enjoyed it a lot.
same, weeb problems best problems
There is an even more elegant form of the answer with a bit more manipulation
You have
Sqrt (1/2*(a^4+b^4+(a+b)^4))
= a^2+b^2+ab
=1/2*(a^2+ b^2+(a^2+b^2+2ab))
=1/2*(a^2+b^2+(a+b)^2)
That is
Sqrt (1/2*(a^4+b^4+(a+b)^4))
=1/2*(a^2+ b^2+(a+b)^2)
Which is much more elegant imo
I really did feel the binomial go brrr. Truly heartwarming.
I love this maths channel and your personality;
small comment: (a+b)^4 can be seen ass all the possible ways to draw 2 balls (red,blue) 4 times. Which means either all red a^4 or all blue b^4 or one red and 3 blue (4 possible ways to arrange) or 3 blue on red (4 times also, because of symmetry) or 2 blue 2 red (6 possible ways).
edit: (if you need some help thinking about it, look at 3!, 2! 1! choices to arrange a color of your choice)
I think it is faster if I calculate using the standard method. Also, you can calculate it after you eliminated the numitor (the moment 8.00).
Are we just going to ignore the "2016 weeb MO"
yeye
I really like this exam and I try to learn more .
0:10 "I'm German, don't know what the fsck Papa is saying."
damn first 3b1b uploads and a day after you upload! content everywhere, so nice
11:38 plz make a tutorial on how to make brrrr sound，its crucial for my next report
Yooo flammable math video.
Well if you use some foresight you can see that (a+b)^4 consists of a^4 + b^4 plus the even-coefficient terms 4ab(a+b) + 6(ab)^2, so you know that can be used to get rid of the fraction right away.
So,
(a^4 + b^4 + (a+b)^4)/2 = a^4 + b^4 + 2ab(a^2 + b^2) + 3(ab)^2
= (a^2 + b^2)^2 + 2ab(a^2 + b^2) + (ab)^2
= (a^2 + b^2 + ab)^2
Factoring the polynomial under the radical can be done faster since you know it's a perfect square. a^4 and b^4 means it's (a^2 + k a b + b^2)^2 for some k, which clearly must be 1.
All those equations are slower than just do the math
11^4 = 14641
100^4 = 100000000
111^4 = 151807041
251821682÷2 = 125910841
And the complicated part is the √125910841
I think best way is by approximation
So, You know the solution is close to 11200, so You try until You find it. It took me about 8 attemps, and about 6min
what I was thinkin..
@@mantizshrimp Yeah. And we're assuming they can't use a calculator, wich for me has no sense at all. In a math olympiad the last thing You want to know is how good are the participants calculating stuff, what You really want to know is how good are they at solving problems, and turning those problems into math. If the olympiad is just problems like this one thats a math competition i'm not interested in watch or participate
Great video like always
I computed the square root by just plugging in a=10, b=1. Finding the square root of 12321 is pretty trivial (try 100^2, 110^2, 111^2), and that gives me a clue that the factorization would be 111^2 = (100 + 10 + 1)^2 = (a^2 + ab + b^2)^2 with a pretty high degree of certainty.
I figured it was intended to be a factoring problem due to the setup, but out of curiosity I simply computed the answer using traditional arithmetic algorithms. It took less than 4 minutes. The only multiplication that was even somewhat involved was calculating 111^2*111^2 or 12321*12321 and, even then, you could tell at a glance that there was no carrying in the multiplication part, so that took maybe 30 seconds to write down, then it was just an addition problem. All the multiples you had to find when doing the square root algorithm were either 1 or 2 with, again, no carrying making it trivial to do it in your head, so that was as easy as it gets.
I get that brute force computation is against the spirit of the problem, but they should really make these problems more difficult if they want the student to have to resort to using factoring to simplify the problem. Any 6th grader should be able to do this armed with nothing more than the multiplication algorithm, the short division algorithm, and the square root algorithm.
Nick cage reference nice
This is definitely a nice and fun problem to do after all the theoretical bs I do all day lmao
yeah so even if i dont understand i still watch his videos at fucing 12pm
great video dude
actually it was enjoyable for me
Interestingly there 637 is the smallest number expressible in this way in three distinct ways (without permutation) i.e. (a,b) = (4,23); (7,21) and (12,17)
bro i need to get those slidey chalkboards installed in my room
Binomial go "brrrr" got me
After not watching your videos for such a long time I now recall just how magnificent your accent is 😍
hehe :D
What about of negative integer?
-11221 and +11221
by the way big fan 🖤🖤
Well since it's a square root it can't be negative. Because sqrt(x) ≥ 0 always. Where x is any number
Yeah the "square root" symbol (or sqrt() in text) is really the "positive square root" symbol. For the negative square root you'd use -sqrt().
And since a²+b²+ab is always >=0 which can be easily checked as an exercise to the reader, we have sqrt( (a²+b²+ab)² ) = |a²+b²+ab| = a²+b²+ab, removing the absolute value we usually have in sqrt(X²) = |X|.
a and b aren't arbitrary as it is used to represent 11 and 100 so its not negative.
@@lars778 not in complex analysis.
go brrrrrrrrrrrrrrrttttttttttt "A-10 Warthog otw?"
We need more cat intro!
That one doesn't seem too hard to me but still I couldn't solve from mind.
10:38 Feynman probably would've put it like that too.
11:36 go BRRRRRRRRRRR
*headbanging*
I am not the best at mathematics but I believe there might've been a mistake when factorizing a^4+ + b^4 + 3a^2b^2 as he might've seen the three coefficient as a two, therefore allowing him to factorize it normally. Of course I could always be wrong, since he is a a lot better at mathematics then I can ever aspire to be.
No. He expanded it out first and then factored it. 3a^2b^2 = 2a^2b^2 + a^2b^2
that moment when you solve the problem but get 10221 because you suck at addition
Poor pascal's triangle, give it some love
easy, use pascals triangle for powers of 11
Can some kind soul explain to me te joke on the shirt?
Why don't you use
Pascal's Triangle?
plz what did he said at the intro? reversed one
0:27 What is Weeb, please?
It is a way of referring to those who are obsessed with Japan / anime, in this case it's used ironically to refer to the Japanese themselves.
How would you integrate 1/(x^3 + c)^(1/3)??
Nice video ☺️☺️
3^x + 4^x =5^x
Solve without knowing that the answer is 2 and also show all complex answers
Please try it
He already made a video with a similar problem. I don't remember the name of it but I'm sure you can find it.
Is it possible for you to explain why we need linear algebra and differential equations.
The oversimplified answer is physics. Linear algebra explains how vectors work, and pretty much all of physics is described in vectors. In addition, physics very commonly uses quantities like velocity and acceleration which are the derivatives of some other quantity, like position in this case. Since these are often related (as a simple example, take the force on a spring: f = -kx = ma is a second order differential equation) differential equations show up a lot.
is she the cat from the vlog on your other channel?? anyways, she is very cute! what is her name??
Nah, she's our pet, Hana :)
@@PapaFlammy69 wow
the only thing i understood was the adorable cat at the start
What is an intiger?
Are there any long term health concerns being around chalk so much?
After 20 years of teaching in high school I sometimes could hardly speak because of the chalk dust. But luckily then came the whiteboards with the alcohol based makers...
@@l.h.308 So you recovered? I have never used a chalk board for teaching!
you could simply have used the identity a2+b2+c2 would have been much simplee
such a cute cat omg
6:45 ah yes very kafkaesque
We had a similar Olympiad ig, tho, not this level i think,
8=√(x+(√x+(√x+..... Infinity)))
0:50 That "111^4" looks like 1114 lol
Looks like Your Calico really Loves You. (this is not Shinja, this is Cristiella)
You got me subbed at 10:39
ALWAYS LOVE FROM BANGLADESH.
Is this is of level 10 standard?
I mean to say question?
Yes
binomial go brrrrrrrrrrrrrrrr
can anyone tell me why 65320 is not divisible by 80(8 x 10), when its divisible by 8 and 10?
You switched divisibility laws :
Let a,b,c be positive integers, if a | b and a | c then of course a | bc, but if a | c and b | c, ab | c if and only if a and b are coprimes but GCD(8;10)=2
you used up a 2 twice
@@KS-lb9uc Thank you for clearing my doubt!
Your cat is so friking cute!!!!
Great video! Sorry for being abnoxious in the past comments.
definitely took me longer than 15 mins
1+1=2
Arithmetic is wild
Cat go purr, Binomial go brr
Solve for x and y
x^2 +y^2 = a
x^3+y^3 = b
x = y = a = b = 0
@@spaghettiking653 ✌😅
Is your answer to be squared?
11:37 brrrrrrrrrrr
So the fast way would be to go give up and let Jens do it for you.
Math cat math cat math cat
Just calculate it by hand ez
Hättest du mal Pascals drei eck gelernt, würdest du bei Minute 9 sehen, dass du fertig bist😌
i literally just took the square root of all the numbers that were ^4 which is surely that same number squared, added them and only then divided by two
DO NOT ASK ME HOW *THAT* went right, but it did, and i came to solving this shit without using a single piece of sheet in under 30 seconds and most importantly, CORRECTLY
you can try this at home, it'll surely work
*it's not about getting the "right" path, it's about getting to the right destination*
but seriously, this shit is so stupidly difficult you should probably just give finding the right answer up, and do ANYTHING just to get ANYWHERE and not go by this thing without answering it
you might get it right, just like i did
take care
This one was easy. :)
Binomial go brrrr 😂
🏵️🏵️
cade
Papa , really weeb Olympiad 😂
Weebs mathematics are indeed wild
Fuck yeah.
Ah Pascal’s triangle…….
I would prefer staying away from that side of math. Instinct keep telling me to raise everything to a 1/2 power and then multiply and divide out using exponent laws
Why
@@ytbook9639 because I’m to use to dealing with multiplication in the root, it’s annoying when it’s addition. square root = being raised to the 1/2 power, but again since it’s addition inside of the root you can’t just do that. If you could then then the exponents would just be 4 times 1/2 or just 2.
Meanwhile Endians: saar oi can solve dhis proablam in 0.1 milli seconds saar believe me saar ploxzzz😭🙏😭🙏😭
9:45