Hardest math question

Sort:
Avatar of chessrohan2

also 99999999999999999999999999999999999999999999999999999999999999999999999999999999999 cubed times 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

Avatar of eulers_knot
slickQuickLiam wrote:

prove e^pi i = -1

If you'll grant me euler's formula, it is easy.

e^(ix)=cos(x)+i*sin(x) (eulers formula)

let x=pi

e^(i pi)= cos(pi)+i*sin(pi)

observe: cos (pi) = -1

               sin (pi) = 0

thus e^(i pi) = -1 + i*0 = -1.

 

Avatar of Superelevator

Name 3 euler theorems don’t look on the internet etc.

Avatar of Superelevator

Comment 22 answer: all those numbers ^ 4

Avatar of Superelevator

Nvm

Avatar of IJELLYBEANS

Prove that there exists a Hadamard matrix for every positive multiple of 4.

Avatar of Lucas_softhappycat

999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994089094508349505999999999999999999999999999999999999999999999999999999999999999309203902193093092109309329230923092130930392109231903120931290999999999999999^454343999999999999999999999999999999999999999999999999999999999999999999999999999999329808210391230832974837878479289202222222232322312139999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999*888

Avatar of hsv5609

Prove that there are infinitely many twin primes

Avatar of SciFiChess
chessrohan2 wrote:

also 99999999999999999999999999999999999999999999999999999999999999999999999999999999999 cubed times 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

=

9999999999999999999999999999999999999999999999999999999999999999999999999999999999699999999999999999999999999999999999999999999999999999999000000000000000000000000003000000000000000000000000000000000000000000000000000000029999999999999999999999999989999999999999999999999999999999999999999999999999999999700000000000000000000000000000000000000000000000000000000000000000000000000000000001

(388 digits)

To solve this, use a free online calculator like: Big Integer Calculator.

Here is a Quanta Magazine article about some impressive problems being solved in number theory, including work by Oxford mathematician and Fields Medal winner James Maynard.

Avatar of amrugg

https://projecteuler.net/ has some nice problems. https://projecteuler.net/problem=656 is a nice one.

Avatar of BPGHchess

Prove that the collatz conjecture holds for all integers.

Avatar of rsmab1

prove fermat's last theorem

Avatar of ThankfulBone
rsmab1 wrote:

prove fermat's last theorem

💀

Avatar of Abstract_Mindfield

If f(x)=xe^x then find the limit of (f^-1(x)/logx) when x->inf and domain of f(x) is (0,inf)

Avatar of Shri_Bo

what is the last digit of e (eulers number)

hahahahahahahahh Imbiciles

Avatar of Shri_Bo
ThankfulBone wrote:
rsmab1 wrote:

prove fermat's last theorem

💀

its already proved

https://www.scienzamedia.uniroma2.it/~eal/Wiles-Fermat.pdf

warning: google says its dangerous to visit so;

Avatar of Shri_Bo
Abstract_Mindfield wrote:

If f(x)=xe^x then find the limit of (f^-1(x)/logx) when x->inf and domain of f(x) is (0,inf)

1 my friend, 1

i was trying to use Lambert W function for xe^x but realized that would get me nowhere

Avatar of Future_AE_Major

2 x 50% of 9^ 284593559374584

Avatar of Future_AE_Major

Wait that’s just 9^284583559374584

Avatar of BPGHchess
youblundered_XD wrote:

Tell the ones place digit of the answer in 47^2020 (47 to the power 2020)

Nice, simple modular arithmetic. Ones place is just mod 10. Can use euler's totient function.
phi(10) = 10(1-1/2)(1-1/5) = 4
47^2020 (mod10) is congruent to 47^0 (mod10). Ans: 1