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 2004-01-28, 20:14 #1 Neves   Jan 2004 616 Posts 51 problem Hello, I'm looking for somebody that know this problem: using four numbers 4 and the basic operations ( + - * / fatorial pow root ... ) find all number between 0 and 100 example: 0 = 4 + 4 - 4 - 4 1 = 44/44 2 = (4*4)/(4+4) . . . the really problem is to find the number 51. does anybody knows this problem and where can I found more about? sorry my english, I'm from Brazil.
 2004-01-29, 06:04 #2 gbvalor     Aug 2002 3·37 Posts Hello, Assuming you can use double factorial n!!=n*(n-2)*(n-4).... a solution is: 51 = 4! + 4! + 4!/4!! I have to admit it stuck me a while. Guillermo
 2004-01-29, 10:56 #3 Neves   Jan 2004 610 Posts Are you a Math? or just curious as me? I've never heard about double factorial. If it really exist, I'm sure it's is valid to use. Where can I found more about it? what I'm really trying to found is a formal description of this problem, and a math/algorithm solution to found the answers. congratulations due!!
2004-01-29, 12:02   #4
gbvalor

Aug 2002

1578 Posts

Quote:
 Are you Math? or just curious as me?
Just a Physicist (and Meteorologist) who likes math and programming.

About double factorial see

http://mathworld.wolfram.com/DoubleFactorial.html

Guillermo

 2004-01-29, 18:14 #5 michael   Dec 2003 Belgium 1018 Posts I see you can use sqrt(4)=41/2 can we use 42 too? -michael
 2004-01-29, 20:26 #6 Neves   Jan 2004 2·3 Posts 41/2 is only and diferent representatio of sqrt(4). if you found an other representation for any numbers but using only number 4, you can use it. as I said before I just listen about this problem, I'm looking for help to found the origem of this. The correct propose.
 2004-01-29, 20:38 #7 michael   Dec 2003 Belgium 5·13 Posts Then for instance F(sqrt(4)) * (4-4/4) =51 would be correct too? F(2) is second Fermat number (or 22[sup]2[/sup]+1). -michael Last fiddled with by michael on 2004-01-29 at 20:40
 2004-01-30, 11:13 #8 Maybeso     Aug 2002 Portland, OR USA 2×137 Posts I like the all-factorial solution!! But why stop at a paltry double factorial? Go for the tetrakaidecuple factorial!! (14-uple) tetrakaidecuple factorial = n!!!!!!!!!!!!!! = n*(n-14)*(n-28)... (sqrt(4)^4 + 4/4)!!!!!!!!!!!!!! = (2^4 + 1)!!!!!!!!!!!!!! = 17*3 = 51 You may want to limit the degree of factorial allowed, because too many numbers can be expressed with extreme factorials.
 2004-01-30, 11:23 #9 michael   Dec 2003 Belgium 5×13 Posts What about this one Pi(44-4!+4)=51 Pi(x) is amount of primes less or equal to x -michael
 2004-02-03, 06:58 #10 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 22·11·239 Posts According to a book that I got with my TI-35: Using 4 Fours and the following keys it is possible to build all whole numbers between 1 and 119. + - X / ( ) . ! ^2 =
 2004-02-03, 09:38 #11 michael   Dec 2003 Belgium 5·13 Posts Strange that they say up to 119 cause 120 is a very easy number being 5! (4+4/4)!=120 -michael

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