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 2004-01-28, 13:06 #1 Neves   41×79 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, 11:48 #2 koal   Nov 2002 Vienna, Austria 4110 Posts 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 ...
 2004-01-29, 13:32 #3 Jim   Feb 2003 118 Posts i would have that your other solution is illegal aswel, youre using 4x4 but also a 0. Jim
 2004-01-29, 15:59 #4 Neves   Jan 2004 2×3 Posts 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 = ? . . .
2004-02-10, 13:26   #5
xilman
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May 2003
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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

 2004-02-10, 22:59 #6 Maybeso     Aug 2002 Portland, OR USA 1000100102 Posts 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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