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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".

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"

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.

I agree, that solution was absolutely beautiful. It turned a complicated question into a problem that can be solved manually.

Flammable Quotes: “I don’t really give a shit about Pascal’s Triangle either” “we can tell the absolute value to fuck off basically”

That intro was absolutely adorable!

Papa got the schordingers cat

still waiting for a complex differential geometry video D:

This was so much fun!

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 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!

I love this maths channel and your personality;

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

Great video like always

10:28 when you canceled out the 2 i wasnt looking so i thought it was to the fourth power

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.

great video dude actually it was enjoyable for me

Such a wholesome intro!

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.

I really like this exam and I try to learn more .

damn first 3b1b uploads and a day after you upload! content everywhere, so nice

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).

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.

Yooo flammable math video.

I really did feel the binomial go brrr. Truly heartwarming.

Great video, enjoyed it a lot.

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

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.

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.

This is definitely a nice and fun problem to do after all the theoretical bs I do all day lmao

Are we just going to ignore the "2016 weeb MO"

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

0:10 "I'm German, don't know what the fsck Papa is saying."

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 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.

Binomial go "brrrr" got me

After not watching your videos for such a long time I now recall just how magnificent your accent is 😍

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

10:38 Feynman probably would've put it like that too.

Nice video ☺️☺️

yeah so even if i dont understand i still watch his videos at fucing 12pm

11:38 plz make a tutorial on how to make brrrr sound，its crucial for my next report

We need more cat intro!

go brrrrrrrrrrrrrrrttttttttttt "A-10 Warthog otw?"

Is it possible for you to explain why we need linear algebra and differential equations.

11:36 go BRRRRRRRRRRR *headbanging*

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

Nick cage reference nice

How would you integrate 1/(x^3 + c)^(1/3)??

That one doesn't seem too hard to me but still I couldn't solve from mind.

easy, use pascals triangle for powers of 11

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.

What about of negative integer? -11221 and +11221 by the way big fan 🖤🖤

plz what did he said at the intro? reversed one

Poor pascal's triangle, give it some love

Why don't you use Pascal's Triangle?

Are there any long term health concerns being around chalk so much?

Can some kind soul explain to me te joke on the shirt?

6:45 ah yes very kafkaesque

you could simply have used the identity a2+b2+c2 would have been much simplee

What is an intiger?

can anyone tell me why 65320 is not divisible by 80(8 x 10), when its divisible by 8 and 10?

You got me subbed at 10:39

3^x + 4^x =5^x Solve without knowing that the answer is 2 and also show all complex answers Please try it

0:50 That "111^4" looks like 1114 lol

0:27 What is Weeb, please?

such a cute cat omg

binomial go brrrrrrrrrrrrrrrr

the only thing i understood was the adorable cat at the start

Cat go purr, Binomial go brr

ALWAYS LOVE FROM BANGLADESH.

Your cat is so friking cute!!!!

1+1=2 Arithmetic is wild

Is your answer to be squared?

11:37 brrrrrrrrrrr

Solve for x and y x^2 +y^2 = a x^3+y^3 = b

Looks like Your Calico really Loves You. (this is not Shinja, this is Cristiella)

is she the cat from the vlog on your other channel?? anyways, she is very cute! what is her name??

9:45

that moment when you solve the problem but get 10221 because you suck at addition

We had a similar Olympiad ig, tho, not this level i think, 8=√(x+(√x+(√x+..... Infinity)))

Is this is of level 10 standard?

Binomial go brrrr 😂

Great video! Sorry for being abnoxious in the past comments.

definitely took me longer than 15 mins

cade

Math cat math cat math cat

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

So the fast way would be to go give up and let Jens do it for you.

Hättest du mal Pascals drei eck gelernt, würdest du bei Minute 9 sehen, dass du fertig bist😌

Just calculate it by hand ez

11221

This one was easy. :)

Meanwhile Endians: saar oi can solve dhis proablam in 0.1 milli seconds saar believe me saar ploxzzz😭🙏😭🙏😭

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

Weebs mathematics are indeed wild