Algebra

[Achyut]

Do it if you like !!!

👍All the best👍

 

[Strange]

!

Moderator Note

Moved to Puzzles section

 

Spoilers:

https://oeis.org/A000043

https://www.wolframalpha.com/input/?i=factorise+(2^92+-+1)

[Sensei]

50 minutes ago, Strange said:

r

https://www.wolframalpha.com/input/?i=factorise+(2^92+-+1)

..but OP nowhere said it is sequence of binary ones..

 

[Strange]

51 minutes ago, Sensei said:

..but OP nowhere said it is sequence of binary ones..

 

Very good point!

[joigus]

4 hours ago, Sensei said:

..but OP nowhere said it is sequence of binary ones..

Right you are.

My doctor told me to stay away from factorizing numbers that are too high. 

[Sensei]

4 hours ago, joigus said:

My doctor told me to stay away from factorizing numbers that are too high. 

Why? Change doctor   It is easy.. If you have decimal e.g. 1111 factors will be 101 and 11. If it is e.g. 111111 factors will be 1001 and 111. And so on. You just have to answer question if number of decimal ones is dividable and has factors by itself.

91 is not prime. It is dividable by 7 and 13.

1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 

is dividable by 1111111.

[Strange]

So, as expected, the answer is the same whether we assume it is binary or decimal. 

[joigus]

40 minutes ago, Sensei said:

Why? Change doctor   It is easy.. If you have decimal e.g. 1111 factors will be 101 and 11. If it is e.g. 111111 factors will be 1001 and 111. And so on. You just have to answer question if number of decimal ones is dividable and has factors by itself.

91 is not prime. It is dividable by 7 and 13.

1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 

is dividable by 1111111.

Oh, yeah, divisibility by 11 criteria give you that, right? I'd forgotten about that. Must go over that again.

But don't get me into other numbers.

[John Cuthber]

https://en.wikipedia.org/wiki/Repunit