20040128, 13:06  #1 
3,697 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, 11:48  #2 
Nov 2002
Vienna, Austria
41_{10} 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)+44^0
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 ... 
20040129, 13:32  #3 
Feb 2003
3^{2} Posts 
i would have that your other solution is illegal aswel, youre using 4x4 but also a 0.
Jim 
20040129, 15:59  #4 
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 = ? . . . 
20040210, 13:26  #5  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
3^{2}×1,303 Posts 
Quote:
You had only one 4 in your solution and the problem calls for two. Paul 

20040210, 22:59  #6 
Aug 2002
Portland, OR USA
2·137 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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