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Tomasz Grysztar
It appears to be counting the amount of enclosed areas in the digit characters. Well, it has something to do with homotopy groups, so I'd argue it IS in a way related to higher mathematics.


18 Feb 2012, 16:07 

bubach
Meh..


18 Feb 2012, 16:29 

MHajduk
Well, bubach, I think that we must count number of "circles" in all those ciphers. It took me about 15 minutes to discover it.


18 Feb 2012, 16:38 

bubach
Yeah I discovered it too. Just took a while, maybe I'm too smart to see such a simple solution?


18 Feb 2012, 17:34 

MHajduk
bubach wrote: Yeah I discovered it too. Just took a while, maybe I'm too smart to see such a simple solution? 

18 Feb 2012, 17:44 

bubach
Heheh, yeah. I found the solution by trying out different values to each number, so I knew that 8=2 first but not why until later  hahha.


18 Feb 2012, 18:14 

cod3b453
yeah, no way I was gonna get that 

18 Feb 2012, 23:05 

revolution
The clue is "preschool" since all those squiggly lines will not have any type of mathematical meaning to such a child. My first though was the number of "tails" (unconnected endings) and then I tried the number of enclosed spaces and got the prize.


18 Feb 2012, 23:41 

edfed
ho shit, i scrolled too much...after 15 minutes wondering what a child would do with those signs.
i wondered about pure graphical things, like the apparent darkness of a line... or see a sort of global image by looking from far away... cool problem indeed, i'll try this on others at school. 

19 Feb 2012, 00:35 

Yardman
[ Post removed by author. ]
Last edited by Yardman on 04 Apr 2012, 04:47; edited 1 time in total 

19 Feb 2012, 01:53 

revolution
Notice that 4 is absent from the list. I presume that is because it can be drawn either with or without an enclosed space.
What about this?: ⓪⑧⑥⑨ = ? Or this: ❶❻❽❾ = ? Or even this: 〇一二三四五六七八九十 = ? 

19 Feb 2012, 02:23 

DOS386
http://lemeshko.blogspot.com/2009/01/greatestmathproblemever.html
Maybe the performance of well studied people would improve with this hack: Quote:
> Quote:
or Quote:
See also: http://board.flatassembler.net/topic.php?t=13987 EDIT : added link to t=13987 Last edited by DOS386 on 23 Feb 2012, 06:46; edited 1 time in total 

19 Feb 2012, 08:59 

MHajduk
I'm not sure if there is a sense to post here such a simple math problem because people on this forum are so smart... Okay, let me try...
Prove that the number presented below is composite: Code: 11...122...211...1 + 6 \____/\____/\____/ 14 7 14 

19 Feb 2012, 13:04 

revolution
Assuming it is base10 then it passes the simple divisionby3 test. Add up all digits and check that the result is divisible by 3.


19 Feb 2012, 13:25 

MHajduk
Very well.
So, how it goes, let N will be our number: N = 11111111111111222222211111111111111 + 6 = 11111111111111222222211111111111117 Sum of the number's digits is equal to S = 14*1 + 7*2 + 13*1 + 7 = 14 + 14 + 13 + 7 = 48 = 3*16 so is divisible by 3 and, consequently, the number is divisible by 3 (accordingly to the rule of the divisiblity by 3). 

19 Feb 2012, 13:29 

revolution
Full factorisation is given here.


19 Feb 2012, 13:32 

MHajduk
That's an interesting tool, thanks for the information. But I prefer calculations with a pencil and a sheet of paper.


19 Feb 2012, 13:35 

Tomasz Grysztar
MHajduk wrote: I'm not sure if there is a sense to post here such a simple math problem because people on this forum are so smart... Okay, let me try... Prove that the number presented below is composite: Code: 31313...1313 + 4 \_________/ 3*10^1313 

19 Feb 2012, 14:06 

MHajduk
Tomasz Grysztar wrote:
Let z = 31313...1313 + 4 ............\_________/ .............3*10^1313 So we can transcribe number z this way: z = 3*10^(3*10^1313) + 1313...1313 + 4 ....................................\_________/ .....................................3*10^1313 (I've just "separated" the first 3 from the rest here). Notice that the second component of the sum x = 1313...1313 ......\________/ .......3*10^1313 is "made of" the six digit cycles 131313 taken 5*10^1312 times: 3*10^1313 = 3*10*10^1312 = 3*2*5*10^1312 = 6*5*10^1312 so we can transcribe x this way x = 131313*1000001000001...1000001 ..................\____________________/ ........................3*10^1313  5 Factorization of the number 131313: 131313 = 3 * 7 * 13^2 * 37 So number x is divisible by 7. Now we should check divisibility by 7 of the number y = 3*10^(3*10^1313) + 4 (which is equal to z  x). We know from the Euler's theorem that 10^φ(7) ≡ 1 (mod 7) knowing that φ(7) = 7  1 = 6 we have that 10^6 ≡ 1 (mod 7) So 10^(3*10^1313) = 10 ^ (6 * 5 * 10^1312) = [10 ^ 6]^(5 * 10^1312) ≡ 1 ^(5 * 10^1312) = 1 (mod 7) and y = 3*10^(3*10^1313) + 4 ≡ 3*1 + 4 = 7 (mod 7) i.e. y = 3*10^(3*10^1313) + 4 ≡ 0 (mod 7) hence y is divisible by 7. Finally z = x + y as a sum of the numbers divisible by 7 is divisible by 7. I hope I haven't made any mistake here. 

19 Feb 2012, 16:29 

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