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Old 2004-01-28, 13:06   #1
Neves
 

24·373 Posts
Smile 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.
 
Old 2004-01-29, 11:48   #2
koal
 
Nov 2002
Vienna, Austria

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what do you mean with "pow" ? only exponents formed by "4"s (like "4^SQRT(4)"), or any power, like 4^0? Then it'l be 4!*SQRT(4)+4-4^0

If any "Function" is allowed, one could do the following

51 = "LEFT"((4^4*SQRT(4)),SQRT(4))

which is "LEFT"(512,2)
which is 51

but I'm sure, that's illegal ...
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Old 2004-01-29, 13:32   #3
Jim
 
Feb 2003

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i would have that your other solution is illegal aswel, youre using 4x4 but also a 0.

Jim
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Old 2004-01-29, 15:59   #4
Neves
 
Jan 2004

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There's an answer here
http://www.mersenneforum.org/showthread.php?t=2007

you can not use 0 only 4 and can not use functions.

now that 51 is know, I will put a little program that I did to found it, and insert the double fatorial in its expressions to use.
between 0 to 100 still less to found abou 30 numbers. I think double fatorial will help a lot cause it's the only way to reach the number 3 using only 2 fours numbers.
1 = 4/4
2 = sqrt(4)
3 = 4!/4!!
5 = ?
.
.
.
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Old 2004-02-10, 13:26   #5
xilman
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May 2003
Down not across

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Quote:
Originally Posted by Neves
There's an answer here
http://www.mersenneforum.org/showthread.php?t=2007

you can not use 0 only 4 and can not use functions.

now that 51 is know, I will put a little program that I did to found it, and insert the double fatorial in its expressions to use.
between 0 to 100 still less to found abou 30 numbers. I think double fatorial will help a lot cause it's the only way to reach the number 3 using only 2 fours numbers.
1 = 4/4
2 = sqrt(4)
3 = 4!/4!!
5 = ?
.
.
.
Correction: 2 = 4/sqrt(4)

You had only one 4 in your solution and the problem calls for two.


Paul
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Old 2004-02-10, 22:59   #6
Maybeso
 
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Aug 2002
Portland, OR USA

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Xilman,

I think what Neves is trying to do is make a list of numbers using the fewest 4's possible. He can pad them later to be legal, but he can also combine them to make other numbers. If 20 numbers can be found using 1 or 2 4's, then more difficult numbers can be expressed using those. If he doesn't have a "short" 5, he knows to avoid 5's in his equations.
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