20040128, 20:14  #1 
Jan 2004
2·3 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. 
20040129, 06:04  #2 
Aug 2002
3×37 Posts 
Hello,
Assuming you can use double factorial n!!=n*(n2)*(n4).... a solution is: 51 = 4! + 4! + 4!/4!! I have to admit it stuck me a while. Guillermo 
20040129, 10:56  #3 
Jan 2004
2×3 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!! 
20040129, 12:02  #4  
Aug 2002
3·37 Posts 
Quote:
About double factorial see http://mathworld.wolfram.com/DoubleFactorial.html Guillermo 

20040129, 18:14  #5 
Dec 2003
Belgium
5·13 Posts 
I see you can use sqrt(4)=4^{1/2} can we use 4^{2} too?
michael 
20040129, 20:26  #6 
Jan 2004
110_{2} 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. 
20040129, 20:38  #7 
Dec 2003
Belgium
101_{8} Posts 
Then for instance F(sqrt(4)) * (44/4) =51 would be correct too?
F(2) is second Fermat number (or 2^{2[sup]2}[/sup]+1). michael Last fiddled with by michael on 20040129 at 20:40 
20040130, 11:13  #8 
Aug 2002
Portland, OR USA
2×137 Posts 
I like the allfactorial solution!!
But why stop at a paltry double factorial? Go for the tetrakaidecuple factorial!! (14uple) tetrakaidecuple factorial = n!!!!!!!!!!!!!! = n*(n14)*(n28)... (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. 
20040130, 11:23  #9 
Dec 2003
Belgium
5·13 Posts 
What about this one
Pi(4^{4}4!+4)=51 Pi(x) is amount of primes less or equal to x michael 
20040203, 06:58  #10 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2ACA_{16} Posts 
According to a book that I got with my TI35:
Using 4 Fours and the following keys it is possible to build all whole numbers between 1 and 119. +  X / ( ) . ! ^2 = 
20040203, 09:38  #11 
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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