## The four fours problem

**Moderators:** gmalivuk, Moderators General, Prelates

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

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.

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

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...>

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

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

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.

- Gelsamel
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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).

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

- > 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

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)

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.

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

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

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

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

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.

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!

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.

### Getting small numbers

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

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.

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

- Cosmologicon
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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.

- Torn Apart By Dingos
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- evilbeanfiend
**Posts:**2650**Joined:**Tue Mar 13, 2007 7:05 am UTC**Location:**the old world

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)

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.

"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.

- fortyseventeen
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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?

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.

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.

"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.

- Alpha Omicron
**Posts:**2765**Joined:**Thu May 10, 2007 1:07 pm 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.

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.

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