The four fours problem

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Wed May 02, 2007 8:36 am UTC

51: -~(44+sqrt(4)+4)

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 8:41 am UTC

52: 4! + 4! + √(4) + √(4)
53: 4! + 4! + 4 + φ(√(4))
54: 4! + 4! + 4 + √(4)
55: 4! + 4! + Γ(4) + φ(√(4))
56: 4! + 4! + 4 + 4
57: 4! + 4! + 4/(.4...)
... meaning repeating of 4 (that is 4/0.444... = 4/(4/9) = 9)
58: 4! + 4! + Γ(4) + 4
59: (Γ(4)! / 4!) * √(4) - φ(φ(4))
60: 44 + 4^(√(4))

EDIT: unicode ftw! and changed 58: 4! + 4! + 4/.4 to 4! + 4! + Γ(4) + 4
Last edited by poizan42 on Wed May 02, 2007 6:44 pm UTC, edited 1 time in total.

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Postby skeptical scientist » Wed May 02, 2007 12:33 pm UTC

I thought φ was only defined for integer inputs? I think this is illegal. You could do floor(sqrt(4)), but that's really ugly. Elsewhere you have φ(φ(4)), which is better.

61=(gamma(4)! / 4!) * sqrt(4) + φ(φ(4))
62=(gamma(4)! / 4!) * sqrt(4) + sqrt(4)
63=(4^4)/4-φ(φ(4))
64=(4^4)/(sqrt(4)*sqrt(4))
65=(4^4)/4+φ(φ(4))
66=(4^4)/4+sqrt(4)
67=<4/4,4/.4>
68=(4^4)/4+4
69=<4-φ(φ(4)),4+4>
70=<4,4/.44...>
Last edited by skeptical scientist on Wed May 02, 2007 12:46 pm UTC, edited 1 time in total.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

PaulT
Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

Postby PaulT » Wed May 02, 2007 12:44 pm UTC

By the by, way back at 3, isn't (4+4+4)/4 much neater than messing around with sqrts? And (4*4 + 4)/4 for 5. Also for 7, 44/4 - 4. How far can you get only using the basic +-*/ functions? I can get to 12.

Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Wed May 02, 2007 1:14 pm UTC

71: 4!*~-4-4/4
72: 4!*(4-4/4)
73: 4!*~-4+4/4
74: 4!+4!+4!+sqrt(4)
75: 4!+4!+4!+~-4
76: 4!+4!+4!+4
77: 4!+4!+4!-~4
78: 4!*~-4+sqrt(4)+4
79: 4!*~-4+4+~-4
80: 4!*4-4*4
81: (~-4)**(~-4)+4-4

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 1:46 pm UTC

skeptical scientist wrote:I thought φ was only defined for integer inputs? I think this is illegal. You could do floor(sqrt(4)), but that's really ugly. Elsewhere you have φ(φ(4)), which is better.


What?? Since when wasn't sqrt(4) = 2 an integer??

PaulT wrote:Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?


sqrt(4) = 2
4*4 = 4^(sqrt(4)) = 4^2 = 16
44 + 16 = 60
where's the problem?

EDIT: merged posts

EDIT2:
hotaru: what does that ~ mean? and ** is that exponentiation?
Last edited by poizan42 on Wed May 02, 2007 1:50 pm UTC, edited 1 time in total.

User avatar
Gelsamel
Lame and emo
Posts: 8237
Joined: Thu Oct 05, 2006 10:49 am UTC
Location: Melbourne, Victoria, Australia

Postby Gelsamel » Wed May 02, 2007 2:02 pm UTC

I did this upto 40 or 44 or so In year 9 for some reward at school or something.

I only used Factorial/Multiply/Divide/Add/Subtract/Power and I might have used forth root, but I can't remember.

The teacher had to cheat and use 44 or .4 for some of them (at least, I thought that was cheating).
"Give up here?"
- > No
"Do you accept defeat?"
- > No
"Do you think games are silly little things?"
- > No
"Is it all pointless?"
- > No
"Do you admit there is no meaning to this world?"
- > No

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 2:16 pm UTC

hmm no one said the successor function succ(x) (or S(x)) wasn't allowed :/
But well... that's not very fun :)

82 = sqrt(4)^(gamma(4))+4!-gamma(4)
Last edited by poizan42 on Wed May 02, 2007 2:33 pm UTC, edited 1 time in total.

User avatar
mattmacf
Posts: 90
Joined: Sat Sep 02, 2006 11:19 pm UTC
Contact:

Postby mattmacf » Wed May 02, 2007 2:26 pm UTC

82 = 4^(5/2) / .4 + sqrt(4)

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 2:32 pm UTC

mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

1. too slow!
2. you aren't allowed to do 4^(5/2)
3. you lose!

PaulT
Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

Postby PaulT » Wed May 02, 2007 2:40 pm UTC

poizan42 wrote:
PaulT wrote:Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?


sqrt(4) = 2
4*4 = 4^(sqrt(4)) = 4^2 = 16
44 + 16 = 60
where's the problem?


Well, there's no problem per se. But where's the elegance?

EvanED
Posts: 4331
Joined: Mon Aug 07, 2006 6:28 am UTC
Location: Madison, WI
Contact:

Postby EvanED » Wed May 02, 2007 2:49 pm UTC

poizan42 wrote:
mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

2. you aren't allowed to do 4^(5/2)


4^(5/2) is, in LaTeX notation, \sqrt[.4]{4]; in other words, the 0.4th root of 4.

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Wed May 02, 2007 3:11 pm UTC

poizan42 wrote:hotaru: what does that ~ mean? and ** is that exponentiation?

~
and yes, ** is exponentiation.

User avatar
mattmacf
Posts: 90
Joined: Sat Sep 02, 2006 11:19 pm UTC
Contact:

Postby mattmacf » Wed May 02, 2007 3:25 pm UTC

poizan42 wrote:
mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

1. too slow!
2. you aren't allowed to do 4^(5/2)
3. you lose!


Haha, my bad. I explained the .4th root thing a couple posts ago (I think it's on the first page). And yes, I was very late posting that. I tend to multitask when posting and get sidetracked terribly easily :P

I hope I can make it up to you with
83 = (4/.4...)^sqrt(4) + sqrt(4)

...where the .4... = 0.44444444... repeating infinitely. Usually you'd use the bar on top of it to denote that, but I don't believe you can do that with ASCII :|

User avatar
evilbeanfiend
Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

Postby evilbeanfiend » Wed May 02, 2007 4:02 pm UTC

hotaru wrote:
poizan42 wrote:hotaru: what does that ~ mean? and ** is that exponentiation?

~
and yes, ** is exponentiation.


a FORTRANer?

edit: nevermind spotted the link now
in ur beanz makin u eveel

EvanED
Posts: 4331
Joined: Mon Aug 07, 2006 6:28 am UTC
Location: Madison, WI
Contact:

Postby EvanED » Wed May 02, 2007 5:27 pm UTC

84 = 44 * sqrt(4) - 4

(I'm pretty sure I got that one right this time ;-))

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 5:52 pm UTC

mattmacf wrote:Haha, my bad. I explained the .4th root thing a couple posts ago (I think it's on the first page).

oic. By why not use 4^(1/.4) ? I think it's more clear that you are only using fours that way.

85 = floor((4^4*√(4))/Γ(4))
86 = Γ(4)!/(4+4) - 4
87 = √(4)^Γ(4) + 4! - φ(φ(4))
88 = Γ(4)!/(4+4) - √(4)
EDIT: got beaten on that one.
Last edited by poizan42 on Wed May 02, 2007 6:23 pm UTC, edited 3 times in total.

PaulT
Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

Postby PaulT » Wed May 02, 2007 6:14 pm UTC

88=44+44

Hooray!

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Wed May 02, 2007 6:17 pm UTC

89 = Γ(4)!/(4+4) - φ(φ(4))
90 = √(4)^Γ(4) + 4! + √(4)
91 = Γ(4)!/(4+4) + φ(φ(4))
92 = Γ(4)!/(4+4) + √(4)
93 = 4!*4 - √(4) - φ(φ(4))
94 = 4!*4 - 4 + √(4)
95 = 4!*4 - √(4) + φ(φ(4))
96 = 4!*4 + 4 - 4
97 = 4!*4 + √(4) - φ(φ(4))
98 = 4!*4 + 4 - √(4)
99 = 4!*4 + 4 - φ(φ(4))
100 = 4!*4 + Γ(4) - √(4)
101 = 4!*4 + 4 + φ(φ(4))
102 = 4!*4 + 4 + √(4)
103 = 4!*4 + Γ(4) + φ(φ(4))
104 = 4!*4 + Γ(4) + √(4)
105 = 4!*4 - φ(φ(4)) + 4 + Γ(4)
106 = 4!*4 + 4 + Γ(4)

Okey... think I should stop now

EDIT: unicode ftw!
Last edited by poizan42 on Thu May 03, 2007 8:59 am UTC, edited 1 time in total.

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Getting small numbers

Postby poizan42 » Wed May 02, 2007 8:14 pm UTC

I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem ;))

1 with 1 four: φ(φ(4))
1 with 2 fours: 4/4
1 with 3 fours: 4 - √(4) - φ(φ(4)) or just φ(φ(4)) + 4 - 4
1 with 4 fours: 4/4 * 4/4

2 with 1 four: √(4)
2 with 2 fours: 4 - √(4)
2 with 3 fours: √(4) + 4 - 4
2 with 4 fours: 4 - 4 + 4 - √(4) or 4/4 + 4/4

3 with 1 four: π(Γ(4)) where π is the prime counting function
was: floor(ln(4!))
could also be: ~-4 (but I don't like it - it's like the predecessor function - you can get anything by applying it repeatedly to some big value)
3 with 2 fours: 4 - φ(φ(4))
3 with 3 fours: √(4) + 4/4
3 with 4 fours: 4 - √(4) + 4/4

4 with 1 four: 4
4 with 2 fours: √(4) + √(4)
4 with 3 fours: 4 + 4 - 4
4 with 4 fours: 4 + (4-4)*4

5 with 1 four: π(π(φ(Γ(Γ(4)))))
was: round(log((4!)!))
5 with 2 fours: 4 + φ(φ(4))
5 with 3 fours: 4 + 4/4
5 with 4 fours: 4 + √(4) - 4/4

6 with 1 four: Γ(4)
6 with 2 fours: 4 + √(4)
6 with 3 fours: 4 + 4 - √(4)
6 with 4 fours: 4 - 4 + 4 + √(4)

7 with 1 four: floor(exp(2))
or: ~-φ(4!)
7 with 2 fours: Γ(4) + φ(φ(4))
7 with 3 fours: Γ(4) + 4/4
7 with 4 fours: Γ(4) + √(4) - 4/4

8 with 1 four: φ(4!)
was: floor(Γ(log(√(√(√(√(exp(4))))))))
8 with 2 fours: 4 + 4
8 with 3 fours: 4 + √(4) + √(4)
8 with 4 fours: 4 - 4 + 4 + 4

9 with 1 four: π(4!)
was: cototient(round(√(Γ(4)!)))
and: floor(√(exp(exp(log(√(√(√(√((4!)!)))))))))
9 with 2 fours: φ(4!) + φ(φ(4))
9 with 3 fours: φ(4!) + 4/4
9 with 4 fours: φ(4!) + √(4) - 4/4

10 with 1 four: π(π(Γ(Γ(4))))
was: floor(log(exp(4!)))
10 with 2 fours: Γ(4) + 4
10 with 3 fours: Γ(4) + √(4) + √(4)
10 with 4 fours: Γ(4) + 4 + 4 - 4

Someone who can find something better for the ones using floor and round?

EDIT: found something better for 8 and 9 - 9 is still ugly though
EDIT2: better ones for 3, 5, 9 and 10 using the prime counting function
Last edited by poizan42 on Fri May 04, 2007 1:13 pm UTC, edited 4 times in total.

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Thu May 03, 2007 8:29 am UTC

Because i can... and i don't have anything better to do :)

107 = Γ(Γ(4)) - 4!/√(4) - φ(φ(4))
108 = 4!*4 + 4!/sqrt(4)
109 = Γ(Γ(4)) - 4!/√(4) + φ(φ(4))
110 = Γ(Γ(4)) - Γ(4) - √(4) - √(4)
111 = Γ(Γ(4)) - 4 - 4 - φ(φ(4))
112 = 4!*4 + 4*4
113 = Γ(Γ(4)) - 4 - √(4) - φ(φ(4))
114 = Γ(Γ(4)) - √(4) - √(4) - √(4)
115 = Γ(Γ(4)) - √(4) - √(4) - φ(φ(4))
116 = Γ(Γ(4)) - 4 + 4 - 4
117 = Γ(Γ(4)) - 4 + 4/4
118 = Γ(Γ(4)) - √(4) + 4 - 4
119 = Γ(Γ(4)) - φ(φ(4)) + 4 - 4
120 = Γ(Γ(4)) + (4 - 4)/4
121 = Γ(Γ(4)) + φ(φ(4)) + 4 - 4
122 = Γ(Γ(4)) + √(4) + 4 - 4
123 = Γ(Γ(4)) + √(4) + 4/4
124 = Γ(Γ(4)) + 4 + 4 - 4
125 = Γ(Γ(4)) + 4 + 4/4
126 = Γ(Γ(4)) + √(4) + √(4) + √(4)
127 = Γ(Γ(4)) + 4 + √(4) + φ(φ(4))
128 = Γ(Γ(4)) + 4 + √(4) + √(4)
129 = Γ(Γ(4)) + 4 + 4 + φ(φ(4))
130 = Γ(Γ(4)) + 4 + 4 + √(4)
131 = Γ(Γ(4)) + Γ(4) + 4 + φ(φ(4))
132 = Γ(Γ(4)) + Γ(4) + 4 + √(4)
133 = Γ(Γ(4)) + Γ(4) + Γ(4) + φ(φ(4))
134 = Γ(Γ(4)) + Γ(4) + Γ(4) + √(4)
135 = Γ(Γ(4)) + 4*4 - φ(φ(4))
136 = Γ(Γ(4)) + Γ(4) + Γ(4) + 4
137 = Γ(Γ(4)) + 4*4 + φ(φ(4))
138 = Γ(Γ(4)) + 4*4 + √(4)
139 = Γ(Γ(4)) + 4! - 4 - φ(φ(4))
140 = Γ(Γ(4)) + 4*4 + 4
141 = Γ(Γ(4)) + 4! - 4 + φ(φ(4))
142 = Γ(Γ(4)) + 4! - 4 + √(4)
143 = Γ(Γ(4)) + 4! - 4/4
144 = Γ(Γ(4)) + 4! * 4/4
145 = Γ(Γ(4)) + 4! + 4/4
146 = Γ(Γ(4)) + 4! + Γ(4) - 4
147 = Γ(Γ(4)) + 4! + √(4) + φ(φ(4))
148 = Γ(Γ(4)) + 4! + √(4) + √(4)
149 = Γ(Γ(4)) + 4! + 4 + φ(φ(4))
150 = Γ(Γ(4)) + 4! + 4 + √(4)
151 = Γ(Γ(4)) + 4! + Γ(4) + φ(φ(4))
152 = Γ(Γ(4)) + 4! + 4 + 4
153 = Γ(Γ(4)) + 4! + φ(4!) + φ(φ(4))
154 = Γ(Γ(4)) + 4! + φ(4!) + √(4)
155 = Γ(Γ(4)) + φ(Γ(Γ(4))) + 2 + φ(φ(4))
156 = Γ(Γ(4)) + 4! + φ(4!) + 4
157 = Γ(Γ(4)) + φ(Γ(Γ(4))) + 4 + φ(φ(4))
158 = Γ(Γ(4)) + 4! + φ(4!) + Γ(4)
159 = Γ(Γ(4)) + φ(Γ(Γ(4))) + Γ(4) + φ(φ(4))
160 = Γ(Γ(4)) + 4! + 4*4

If someone doesn't have something that can calculate φ(Γ(Γ(4))) it's = φ(120) = 32

User avatar
Cosmologicon
Posts: 1806
Joined: Sat Nov 25, 2006 9:47 am UTC
Location: Cambridge MA USA
Contact:

Postby Cosmologicon » Thu May 03, 2007 3:38 pm UTC

poizan42 wrote:I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem ;))

1 with 1 four: φ(φ(4))
2 with 1 four: √(4)
3 with 1 four: floor(ln(4!))
4 with 1 four: 4
5 with 1 four: round(log((4!)!))
6 with 1 four: Γ(4)
7 with 1 four: floor(exp(√(4)))
8 with 1 four: floor(Γ(log(√(√(√(√(exp(4))))))))
9 with 1 four: cototient(round(√(Γ(4)!)))

Man, this could make the greatest Sudoku ever!

Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

User avatar
umbrae
Posts: 336
Joined: Sat Sep 09, 2006 2:44 pm UTC
Contact:

Postby umbrae » Thu May 03, 2007 3:57 pm UTC

Nifty.

Then you can just do it such that you receive N+4, and then subtract 4 for your solution, using four fours.

(Math hax)

User avatar
Torn Apart By Dingos
Posts: 817
Joined: Thu Aug 03, 2006 2:27 am UTC

Postby Torn Apart By Dingos » Thu May 03, 2007 3:59 pm UTC

Or, if we allow Peano's successor function S, we can write every N as N=S(S(S(...S(4/4)...))). Or for more breaking-the-rules-and-ruining-the-fun smart-assery, we can write every N as {{},{{}},{{{}}},...}, requiring no fours at all!

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Thu May 03, 2007 4:05 pm UTC

you can also do -~-~-~...-~(4/4)

User avatar
evilbeanfiend
Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

Postby evilbeanfiend » Thu May 03, 2007 4:29 pm UTC

yes clearly this is a puzzle which calls for elegant solutions rather than scalable solutions.
in ur beanz makin u eveel

User avatar
cmacis
Posts: 754
Joined: Wed Dec 13, 2006 5:22 pm UTC
Location: Leeds or Bradford, Thessex
Contact:

Postby cmacis » Thu May 03, 2007 5:20 pm UTC

This was far more fun and useful than doing particle dynamics.
li te'o te'a vei pai pi'i ka'o ve'o su'i pa du li no
Mathematician is a function mapping tea onto theorems. Sadly this function is irreversible.
QED is Latin for small empty box.
Ceci n’est pas une [s]pipe[/s] signature.

User avatar
kyrmse
Posts: 1
Joined: Thu May 10, 2007 2:46 pm UTC
Location: Brasil

Postby kyrmse » Thu May 10, 2007 2:48 pm UTC

If I'm not mistaken, the simplest (IMHO) solution for 3 has not yet been posted:

3 = (4+4+4)/4

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Thu May 10, 2007 3:04 pm UTC

i: (-4/4)^(sqrt(4)/4)

User avatar
blob
Posts: 350
Joined: Thu Apr 05, 2007 8:19 pm UTC

Postby blob » Thu May 10, 2007 5:02 pm UTC

There's a similar game where you use the digits of the current year...

1 = 2-0!+0*7
2 = 2+0*0*7
3 = -2-0!-0!+7

(they don't have to be in order ... that's just a bonus)
Avatar yoinked from Inverloch.

"Unless ... unless they kill us, then animate our corpses as dead zombies to fight for them. Then I suppose they've taken our lives, AND our freedom."
- Elan, OOTS 421.

User avatar
fortyseventeen
Ask for a lame title, receive a lame title
Posts: 88
Joined: Fri Mar 02, 2007 3:41 am UTC
Location: SLC, UT, USA
Contact:

Postby fortyseventeen » Thu May 10, 2007 8:43 pm UTC

hotaru wrote:you can also do -~-~-~...-~(4/4)


Um, yeah...that's pretty much a major hack. Bits don't exist in the analytical world of integers, nor to binary representations of negative numbers.

EDIT: As such, I'm gonna try a redo on 51:

51 = 4! + 4! + phi(phi(4+4)), or 4/phi(4), or phi(sqrt(4)+sqrt(4))...

hmm...is phi() too much of a cheat, too?
Last edited by fortyseventeen on Thu May 10, 2007 8:54 pm UTC, edited 1 time in total.
Quick, what's schfifty-five minus schfourteen-teen?

ZeroSum
Cooler than Jeff
Posts: 2903
Joined: Tue May 08, 2007 10:10 pm UTC

Postby ZeroSum » Thu May 10, 2007 8:53 pm UTC

Cosmologicon wrote:
poizan42 wrote:I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem ;))

1 with 1 four: φ(φ(4))
2 with 1 four: √(4)
3 with 1 four: floor(ln(4!))
4 with 1 four: 4
5 with 1 four: round(log((4!)!))
6 with 1 four: Γ(4)
7 with 1 four: floor(exp(√(4)))
8 with 1 four: floor(Γ(log(√(√(√(√(exp(4))))))))
9 with 1 four: cototient(round(√(Γ(4)!)))

Man, this could make the greatest Sudoku ever!

Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

You can make that a four 4s solution (excepting 1, I guess) by replacing with one of the √√ pairs with a 4th root.

User avatar
blob
Posts: 350
Joined: Thu Apr 05, 2007 8:19 pm UTC

Postby blob » Thu May 10, 2007 9:20 pm UTC

ZeroSum wrote:
Cosmologicon wrote:Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

You can make that a four 4s solution (excepting 1, I guess) by replacing with one of the √√ pairs with a 4th root.

Or just replace one of the 4s with √(4*4)
Avatar yoinked from Inverloch.

"Unless ... unless they kill us, then animate our corpses as dead zombies to fight for them. Then I suppose they've taken our lives, AND our freedom."
- Elan, OOTS 421.

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Thu May 10, 2007 9:54 pm UTC

fortyseventeen wrote:EDIT: As such, I'm gonna try a redo on 51:

51 = 4! + 4! + phi(phi(4+4)), or 4/phi(4), or phi(sqrt(4)+sqrt(4))...

hmm...is phi() too much of a cheat, too?


i think (4!-4+.4)/.4 is better...

User avatar
Patashu
Answerful Bignitude
Posts: 378
Joined: Mon Mar 12, 2007 8:54 am UTC
Contact:

Postby Patashu » Fri May 11, 2007 5:28 am UTC

cmacis wrote:This was far more fun and useful than doing particle dynamics.


Useful?

Maybe if a terrorist ever points a gun to your head and demands you solve the four fours problem...

User avatar
gmalivuk
GNU Terry Pratchett
Posts: 26726
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There
Contact:

Postby gmalivuk » Fri May 11, 2007 3:36 pm UTC

hotaru wrote:i think (4!-4+.4)/.4 is better...


That feels more like cheating than phi(4), definitely.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
---
If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

(he/him/his)

User avatar
Alpha Omicron
Posts: 2765
Joined: Thu May 10, 2007 1:07 pm UTC

Postby Alpha Omicron » Sat May 12, 2007 1:57 am UTC

I spent much of this school day playing this game. It gets rather easy when you allow yourself enough obscure functions.
Here is a link to a page which leverages aggregation of my tweetbook social blogomedia.

User avatar
hotaru
Posts: 1045
Joined: Fri Apr 13, 2007 6:54 pm UTC

Postby hotaru » Sat May 12, 2007 3:01 am UTC

gmalivuk wrote:
hotaru wrote:i think (4!-4+.4)/.4 is better...


That feels more like cheating than phi(4), definitely.

which part of it feels more like cheating than phi(4)?

poizan42
Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark
Contact:

Postby poizan42 » Sat May 12, 2007 7:29 pm UTC

hotaru wrote:
gmalivuk wrote:
hotaru wrote:i think (4!-4+.4)/.4 is better...


That feels more like cheating than phi(4), definitely.

which part of it feels more like cheating than phi(4)?

The part about the hidden zero? The part about doing things relying on base 10? phi(4) is not relying on any specific representation of the number - nor is it relying on hiding something.


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 12 guests