51 problem

Neves · 2004-01-28, 20:14

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.

gbvalor · 2004-01-29, 06:04

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

Neves · 2004-01-29, 10:56

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!!

gbvalor · 2004-01-29, 12:02

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

michael · 2004-01-29, 18:14

I see you can use sqrt(4)=4^1/2 can we use 4² too?

-michael

Neves · 2004-01-29, 20:26

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.

michael · 2004-01-29, 20:38

Then for instance F(sqrt(4)) * (4-4/4) =51 would be correct too?

F(2) is second Fermat number (or 2^2[sup]2[/sup]+1).

-michael

Maybeso · 2004-01-30, 11:13

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.

michael · 2004-01-30, 11:23

What about this one

Pi(4⁴-4!+4)=51

Pi(x) is amount of primes less or equal to x

-michael

Uncwilly · 2004-02-03, 06:58

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 =

michael · 2004-02-03, 09:38

Strange that they say up to 119 cause 120 is a very easy number being 5!
(4+4/4)!=120

-michael

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, 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 2^2[sup]2[/sup]+1). -michael Last fiddled with by michael on 2004-01-29 at 20:40

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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 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!!

2004-01-29, 18:14	#5
michael Dec 2003 Belgium 5·13 Posts	I see you can use sqrt(4)=4^1/2 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.

2004-01-30, 11:13	#8
Maybeso Aug 2002 Portland, OR USA 274₁₀ 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 2·5,477 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