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bubach



Joined: 17 Sep 2004
Posts: 340
Location: Trollhättan, Sweden
bubach
See if you can figure this one out, took me like an hour or so, just like the text says. Feel kind of cheated, after going through lots of mathematics to try and find a pattern, when really it's not so much about the mathematics behind it... Rolling Eyes

Translated:
Quote:
This problem takes a pre-school child about 5-10minutes, a programmer about an hour, and people with higher education... well, try it yourself! Smile


Image






[edit]
do



not




scroll more if you don't want the answer

_________________
BOS homepage: http://bos.asmhackers.net/


Last edited by bubach on 18 Feb 2012, 17:38; edited 1 time in total
Post 18 Feb 2012, 15:49
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Tomasz Grysztar
Assembly Artist


Joined: 16 Jun 2003
Posts: 7724
Location: Kraków, Poland
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. Wink
Post 18 Feb 2012, 16:07
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bubach



Joined: 17 Sep 2004
Posts: 340
Location: Trollhättan, Sweden
bubach
Meh.. Razz
Post 18 Feb 2012, 16:29
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
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. Wink
Post 18 Feb 2012, 16:38
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bubach



Joined: 17 Sep 2004
Posts: 340
Location: Trollhättan, Sweden
bubach
Yeah I discovered it too. Just took a while, maybe I'm too smart to see such a simple solution? Wink
Post 18 Feb 2012, 17:34
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
bubach wrote:
Yeah I discovered it too. Just took a while, maybe I'm too smart to see such a simple solution? Wink
Hehe, but we aren't as smart as all those kids... what the entire education process made with our minds? Rolling Eyes Razz
Post 18 Feb 2012, 17:44
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bubach



Joined: 17 Sep 2004
Posts: 340
Location: Trollhättan, Sweden
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.
Post 18 Feb 2012, 18:14
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cod3b453



Joined: 25 Aug 2004
Posts: 619
cod3b453
Image
yeah, no way I was gonna get that Laughing
Post 18 Feb 2012, 23:05
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
The clue is "pre-school" 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.
Post 18 Feb 2012, 23:41
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edfed



Joined: 20 Feb 2006
Posts: 4237
Location: 2018
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. Smile
Post 19 Feb 2012, 00:35
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Yardman



Joined: 12 Apr 2005
Posts: 245
Location: US
Yardman
[ Post removed by author. ]


Last edited by Yardman on 04 Apr 2012, 04:47; edited 1 time in total
Post 19 Feb 2012, 01:53
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
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 = ?
Post 19 Feb 2012, 02:23
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DOS386



Joined: 08 Dec 2006
Posts: 1901
DOS386
http://lemeshko.blogspot.com/2009/01/greatest-math-problem-ever.html

Maybe the performance of well studied people would improve with this hack:

Quote:

8809=6
7111=0
2172=0


->

Quote:

8809->6
7111->0
2172->0


or

Quote:

f(8809)=6
f(7111)=0
f(2172)=0


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
Post 19 Feb 2012, 08:59
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
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... Wink Okay, let me try... Razz

Prove that the number presented below is composite:
Code:
11...122...211...1 + 6
\____/\____/\____/
  14    7     14
    
Very Happy
Post 19 Feb 2012, 13:04
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
Assuming it is base-10 then it passes the simple division-by-3 test. Add up all digits and check that the result is divisible by 3.
Post 19 Feb 2012, 13:25
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Very well. Very Happy

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).
Post 19 Feb 2012, 13:29
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revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
Full factorisation is given here.
Post 19 Feb 2012, 13:32
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
That's an interesting tool, thanks for the information. But I prefer calculations with a pencil and a sheet of paper. Wink
Post 19 Feb 2012, 13:35
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Tomasz Grysztar
Assembly Artist


Joined: 16 Jun 2003
Posts: 7724
Location: Kraków, Poland
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... Wink Okay, let me try... Razz
(...)
I made a modified version. Wink


Prove that the number presented below is composite:
Code:
31313...1313 + 4
 \_________/
  3*10^1313    
Post 19 Feb 2012, 14:06
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MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Tomasz Grysztar wrote:
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... Wink Okay, let me try... Razz
(...)
I made a modified version. Wink


Prove that the number presented below is composite:
Code:
31313...1313 + 4
 \_________/
  3*10^1313    
The given number is divisible by 7. Here you have how I calculated it.


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. Wink
Post 19 Feb 2012, 16:29
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