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Old 2011-02-16, 19:37   #1
Puzzle-Peter
 
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Default Some four-play

Hi folks,

I've been trying to follow the rules of a little game I found in a footnote in Reader's Digest (no joke) and it turned out to be more fun than I expected. But now I'm stuck. Here are the rules:

Try to calculate the number 1, 2, 3, 4, ..... by using EXACTLY four instances of the number "four" and any combination of the following operations:

+ addition
- subtraction
* multiplacation
/ division
^ exponentiation
SQRT square root
! faculty

some examples:
1 = (4+4)/(4+4)
10 = 4*4 - 4!/4
20 = 4*(4!/4) - 4

Getting 1 to 10 is a matter of minutes, 11 to 20 takes substantially longer, then it gets difficult, especially for odd numbers. I tried this with fellow students, later with colleagues and we took this quite a distance. But there are still two gaps on the way to 50 (those being 39 and 43). How far can you take it?

Have fun! (and help me fill the last gaps please...)

Peter

Last fiddled with by Puzzle-Peter on 2011-02-16 at 19:38
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Old 2011-02-16, 19:51   #2
bsquared
 
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Slightly different rules, maybe, but see:

http://www.mersenneforum.org/showthread.php?t=4756

and

http://www.mersenneforum.org/showthread.php?t=7051
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Old 2011-02-16, 20:04   #3
xilman
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Quote:
Originally Posted by Puzzle-Peter View Post
Hi folks,

I've been trying to follow the rules of a little game I found in a footnote in Reader's Digest (no joke) and it turned out to be more fun than I expected. But now I'm stuck. Here are the rules:

Try to calculate the number 1, 2, 3, 4, ..... by using EXACTLY four instances of the number "four" and any combination of the following operations:

+ addition
- subtraction
* multiplacation
/ division
^ exponentiation
SQRT square root
! faculty

some examples:
1 = (4+4)/(4+4)
10 = 4*4 - 4!/4
20 = 4*(4!/4) - 4

Getting 1 to 10 is a matter of minutes, 11 to 20 takes substantially longer, then it gets difficult, especially for odd numbers. I tried this with fellow students, later with colleagues and we took this quite a distance. But there are still two gaps on the way to 50 (those being 39 and 43). How far can you take it?

Have fun! (and help me fill the last gaps please...)

Peter
Oh no, not again. This really is a mouldy oldie.

Note that if the log() function is allowed it is quite easy to produce all integers from four fours.

Paul

Last fiddled with by xilman on 2011-02-16 at 20:05 Reason: Fix tyypo
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Old 2011-02-16, 21:11   #4
science_man_88
 
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Quote:
Originally Posted by Puzzle-Peter View Post
Hi folks,

I've been trying to follow the rules of a little game I found in a footnote in Reader's Digest (no joke) and it turned out to be more fun than I expected. But now I'm stuck. Here are the rules:

Try to calculate the number 1, 2, 3, 4, ..... by using EXACTLY four instances of the number "four" and any combination of the following operations:

+ addition
- subtraction
* multiplacation
/ division
^ exponentiation
SQRT square root
! faculty

some examples:
1 = (4+4)/(4+4)
10 = 4*4 - 4!/4
20 = 4*(4!/4) - 4

Getting 1 to 10 is a matter of minutes, 11 to 20 takes substantially longer, then it gets difficult, especially for odd numbers. I tried this with fellow students, later with colleagues and we took this quite a distance. But there are still two gaps on the way to 50 (those being 39 and 43). How far can you take it?

Have fun! (and help me fill the last gaps please...)

Peter

number freak 1 to 200 pg. 187 has the answer to how far using simple expressions.
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Old 2011-02-16, 22:18   #5
Batalov
 
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Default Re: Some four-play

The natural thing to follow four-play is six.
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Old 2011-02-16, 22:27   #6
retina
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Quote:
Originally Posted by Batalov View Post
The natural thing to follow four-play is six.
Going fourth and multiplying?
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Old 2011-02-16, 23:13   #7
Batalov
 
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I am not going three elabornine...
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Old 2011-02-17, 14:20   #8
nuggetprime
 
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Solved it (without computer help) upto n=33.
I'm surprised you found a solution for it as I think it's not possible with the current allowed ops.
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Old 2011-02-17, 17:06   #9
Puzzle-Peter
 
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Quote:
Originally Posted by xilman View Post
Note that if the log() function is allowed it is quite easy to produce all integers from four fours.
Well, the rules are a bit more difficult (no square, no decimal point, no concatenation) than in the posts I was pointed at (thanks for looking them up, bsquared!) and I can't find the last missing solutions. If you can do them, please do so!

Quote:
Originally Posted by nuggetprime View Post
Solved it (without computer help) upto n=33.
I'm surprised you found a solution for it as I think it's not possible with the current allowed ops.
Not a computer, but a calculator helped find this:

33 = (SQRT(SQRT(SQRT(4^4!)))+SQRT(4))/(SQRT(4))
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Old 2011-02-17, 18:00   #10
nuggetprime
 
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thx
had not used exponentation in my calculations.
lets see if I cand find something for n=39
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Old 2011-02-17, 18:15   #11
xilman
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Quote:
Originally Posted by Batalov View Post
The natural thing to follow four-play is six.
An oldie, but I maintain the old ones are the best:

Q: what comes between fear and sex?

A: fünf

I love bilingual wordplay (but bilingual four-play I'll leave to your imagination). Ask me sometime about trilingual puns...


Paul

Last fiddled with by xilman on 2011-02-17 at 18:16 Reason: Add missing ltter in a typo
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