51 problem

2004-01-28, 13:06

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.

koal · 2004-01-29, 11:48

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 ...

Jim · 2004-01-29, 13:32

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

Jim

Neves · 2004-01-29, 15:59

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 = ?
.
.
.

xilman · 2004-02-10, 13:26

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

Maybeso · 2004-02-10, 22:59

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.

2004-01-28, 13:06	#1
Neves 7,687 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.

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2004-01-29, 11:48	#2
koal Nov 2002 Vienna, Austria 41 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"((4^4SQRT(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 3² 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 110₂ 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, 22:59	#6
Maybeso 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.