At 4:10, there can be a = b = c = d = e = 0 but you said e is not 0 so I guess you don't count this one bcz it's not the maximum anyways, it's the minimum 🙂🙂

Naive

Thank you for your great teaching. Very well presented.

At last I found the meaning of this pi notation 😊

Guys, I am introducing a new constant for 3 digits i.e. 495 😊.

finally, ibp feels like a shortcut! u=4x^3 dv=(d/dx)^2((1-x^2)^5)•dx du=12x^2•dx v=d/dx((1-x^2)^5)=-10x(1-x^2)^4 I=120•int[0,1](x^3•(1-x^2)^4)dx ß(x,y)=2•int[0,pi/2](sin^(2x-1)(t)cos^(2y-1)(t))dt u=sin(t) ß(x,y)=2•int[0,1](u^(2x-1)•(1-u^2)^(y-1))du I=60•ß(2,5) ß(x,y)=Ř(x)Ř(y)/Ř(x+y) ß(2,5)=1•24/720=1/30 I=2

Ok. 2 to the power of 2 then to the power of 2. So 2 times 2 is 4 then 4 times by 2 is 8. My answer is 8. Now my question to you is. What is the largest known number to mankind? So your math get's into numbers no one learned in school. So how can anyone give you the right answer. Your 3/10 or 10 times 10 times 10. Is 1 with a billion zeros. Dude can you even tell me what that number is even called? So that's why we never learned tetraium in school

is zero a natural number?

Dang thats a good quote at the end

16 (Best way i can show tetration) 3 2 Is equal to 2^2^2. 2^2 = 2x2 = 4 2^4 is 2x2x2x2 which is equal to 16. 5:55

If a=b=c=d=e, then 0 or i.

³2= 2^2^2= 4^2= 16 🙋🤦🤷

16

n^3+11n = n^3 - n +12n = n(n+1)(n-1) + 12n The problem is solved because: n(n+1)(n-1) = (n-1+1)(n-1)(n-1+2) Think of (n-1) as a whole then you will get : n(n+1)(n-1) = (n-1)^3 + 3(n-1)^2+2(n-1) but 3(n-1)^2 has been able to be divided into three so look at : (n-1)^3 + 2(n-1) = (n-1)[(n-1)^2 +2] = (n-1)[(n-1)^2 - 1 + 3] = (n-1)(n)(n-2)+3(n-1) = (n-2)^3 + 3(n-2)^2 + 2(n-2) + 3(n-1)， then attention the (n-2)^3+2(n-2) ,do it such (n-1)^3 + 2(n-1) then you will find If you keep going on like this, the end this will be divisible by 3 sure i know that at least there is one number will be divisible by 3 , it's obviously but i wanna use the language of mathematics to prove why.

So it’s kinda like the exponent a number of times 3 and 2 is 2*2 then (2*2)*(2*2) which is 16

7289 requires 8 steps but no one wants to check my math. 7289 9872-2789 = 7083 (1) 8730-0378 = 8352 (2) 8532-2538 = 5994 (3) 9954-4599 = 5355 (4) 5553-3555 = 1998 (5) 9981-1899 = 8082 (6) 8820-0288 = 8532. (7) 8532-2358 = 6174 (8)

With any 4-digit number given, if you subtract the biggest one (abcd) and the smallest one (dcba) the product depends on a - d and b - c relations. Hereby you can count all the possible combinations for f(a-d)=x and f(b-c)=y for a>=b>=c>=d and connect the outcomes back to the corresponding combination until you find all the loop as ending in one at some point is inevitable for any case (after any iteration the outcome formula is 999x + 90y, which makes the outcomes finite set of 4 digit multiples of 9 and number of loops vary if we either change no. of digits or numeral system so we get different formula of the outcome).

sir you are great

Tried 1111

It is 16: 2 2 2

Why do I feel both educated and mildly aroused?

(2"2)2=16

I found an easier way to do it by remembering three things; (1) n! only accepts positive integers, (2) n! > n^3 for n > 5, and (3) 0 > a-b for a<b. (n-1)! + 1 = n^2 n! + n = n^3 n = n^3 - n! n^3 - n! must be negative for n > 5 by (2) and (3), therefore n < 6 by (1). bruteforce, and n = 5 is the only solution.

Hi Prime Newtons, I am an animator, and would love to collab with you to make a lego animated version of one of your old or upcoming videos. Would that be possible? - Dodo

Your a genius ❤

2,2,2,1,1...1,1,1,2,5...1,1,1,,3,3

This can be done without using the formula for Σi=(n)(n+1)/2 since If n=1, then 1=1, If n=2, then LHS=3, RHS=2 if n=3, RHS=LHS=6 And if n ≥4, 1+2+3+…+n <n+n+n+…+n=n*n=n^2, and n! ≥n(n-1)*2, and since (n-1)*2=n-1+(n-1)>n-1+1=n, n! ≥n(n-1)*2>n*n=n^2> 1+2+…+n.

1+1+1+1+1=5 1*1*1*1*1=1 1+1+1+1+2=6 1*1*1*1*2=2 1+1+1+1+n=4+n 1*1*1*1*n=n 4+n=n is impossible so the answer can't contain 4 ones 1+1+1+2+2=7 1*1*1*2*2=4 1+1+1+2+3=8 1*1*1*2*3=6 1+1+1+2+4=9 1*1*1*2*4=8 1+1+1+2+5=10 1*1*1*2*5=10 ------------------- solution 1+1+1+2+6=11 1*1*1*2*6=12 1+1+1+2+n=5+n 1*1*1*2*n=2n 5+n=2n only when n=5 1+1+1+3+3=9 1*1*1*3*3=9 -----------------------solution 1+1+1+3+4=10 1*1*1*3*4=12 1+1+1+3+5=11 1*1*1*3*5=15 1+1+1+3+n=6+n 1*1*1*3*n=3n 6+n=3n only when n=3 1+1+1+4+4=11 1*1*1*4*4=16 1+1+1+4+5=12 1*1*1*4*5=20 1+1+1+4+n=7+n 1*1*1*4*n=4n 7+n=4n only when n=7/3 which is not natural 1+1+1+5+5=13 1*1*1*5*5=25 1+1+1+5+6=14 1*1*1*5*6=30 1+1+1+5+n=8+n 1*1*1*5*n=5n 8+n=5n only when n=2 which is less than 5 1+1+1+x+n=3+x+n 1*1*1*x*n=xn 3+x+n=xn only when n=(x+3)/(x-1)=(x-1+4)/(x-1)=1+4/(x-1) which is natural only when x=2,3,5, but when x=5 n is less than x so we can't have any more solutions with 3 ones 1+1+2+2+2=8 1*1*2*2*2=8 ----------------------solition 1+1+2+2+3=9 1*1*2*2*3=12 1+1+2+2+4=10 1*1*2*2*4=16 1+1+2+2+n=6+n 1*1*2*2*n=4n 6+n=4n only when n=2 1+1+2+3+3=10 1*1*2*3*3=18 1+1+2+3+4=11 1*1*2*3*4=24 1+1+2+3+n=7+n 1*1*2*3*n=6n 7+n=6n only when n=7/5 which is not natural 1+1+2+x+n=4+x+n 1*1*2*x*n=2xn 4+x+n=2xn only when n=(x+4)/(2x-1)=(2x-x-1+5)/(2x-1)=1+(5-x)/(2x-1) which is only natural when x=1,2,5 but when x=5 n is less than x so we can't have any more solutions with 2 ones and 1 two 1+1+3+x+n=5+x+n 1*1*3*x*n=3xn 5+x+n=3xn only when n=(x+5)/(3x-1)=(3x-2x-1+6)/(3x-1)=1+(6-2x)/(3x-1) which is only natural when x=1,3 and n accordingly 3,1 which both disobey the generality 1+1+y+x+n=2+y+x+n 1*1*y*x*n=yxn 2+y+x+n=yxn only when n=(x+2+y)/(yx-1)=(yx-(y-1)x-1+3+y)/(yx-1)=1+(3+y-(y-1)x)/(yx-1) which only has natural combinations of x and n when y=1,2,3,5 but y=3(2,3; 3,2) and y=5(1,2; 2,1) combinations disobey generality i will leave the rest to the reader because i have no idea how to operate on the next level but it is supposed to look like this: 1+2+y+x+n=3+y+x+n 1*2*y*x*n=2yxn 1+3+y+x+n=4+y+x+n 1*3*y*x*n=3yxn 1+z+y+x+n=zxyn only when z=? y=? and natural x,n combinations are ... and they all(have to) disobey generality 2+z+y+x+n=2zyxn when ... 3+z+y+x+n=3zyxn when ... w+z+y+x+n=wzyxn when ... i understand a lot of this is unnecessary for getting intuition on the problem but we are nothing without vigor.

(a+b+c+d ) is more than or equal to 4

👍 t = x + 1/x = { 5/2 , 4 } 1) x + 1/x = 5/2 = 2 + 1/2 x = { 1/2 , 2 } 2) x + 1/x = 4 , x =/= 0 => x^2 - 4x + 1 = 0 D = 16 - 4 = 8 > 0 , 2 soluzioni reali x^2 - 4x + 4 - 3 = 0 ( x - 2 )^2 = 3 x - 2 = +/- sqrt( 3 ) x = 2 +/- sqrt( 3 ) 😊👍👋

First one. A=B=C=D=E=0

I didn’t think e was a variable lol

Your voice is so warm and calming! I'm a young guy, and number theory questions always sort of irritate me, but this delighted me very much! I'm going to be more open to these problems from now on! Thanks

I have never in my life had of tetration

This is amazing!!! 🔥🫶😜

0 is also a natural number, so (0,0,0,0,0) is also a solution. As the problem was originally stated (without a,b,c,d,e not necessarily in ascending order): (0,0,0,0,0) generates 1 solution (1,1,1,2,5) generates 20 solutions (1,1,1,3,3) generates 10 solutions (1,1,2,2,2) generates 10 solutions Totalling 41 solutions. Max. Possible max{a,b,c,d,e} (for (a,b,c,d,e) solution) is 5.

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