# Challenge Challenge where you have to make each number from four 4s

#### animefreak1

i wanted to get 1020 but i dont know how to do it.

1020=4^4x4-4

#### animefreak1

hey thanks alot i really mean it too

1 = 44/44
7 = 44//4-4
12=(44+4)/4
15=44/4+4
16=4+4+4+4
36=44-4-4
88=44+44

#### RichardM

1000 = 4^4*4-4!
10000 = (4!/4+4)^4
100000000 = (4!*4+4)^4

#### cmb

I remember doing this at school!

I do not recall being permitted to use .4 or %, though.

But I seem to recall we were permitted to use '√' .

93 can also be got by this unintuitive term;

93=4![4-√√√[{√4}^-{4!}]

Without %, 89 and 91 need both '√' and '.4' (though I like Joffe's, with the %) .

#### amits

in 1 post in the puzzles community on (nowadays inactive)orkut, i remember doing it. it involved the usage of double factorials too. ex 7!! = 1*3*5*7 = 105, 8!! = 2*4*6*8 = 384, etc

i was able to find the solutions of all numbers from 1 to about 150 using the double factorial thing alongwith the usual operators like +,-,*,/,sqrt, and using the decimal point. if i find the solutions, i will post them here

#### amits

Code:
1 = 44/44

2 = (4/4) + (4/4)

3 = [(4*4) -4]/4

4 = (4/.4) - (4!/4)

5 = (4!/4) - (4/4)

6 = (4!/4)*(4/4)

7 = 4 + 4 - (4/4)

8 = 4 + 4 + 4 - 4

9 = 4 + 4 + (4/4)

10 = (44 - 4)/4

11 = (4/.4) + (4/4)

12 = (44 + 4)/4

13 = 4! - (44/4)

14 = (4/.4) + sqrt(4) + sqrt(4)

15 = (4*4) - (4/4)

16 = 4 + 4 + 4 + 4

17 = (4*4) + (4/4)

18 = (4/.4) + 4 + 4

19 = 4! - 4 - (4/4)

20 = (4! - 4) + 4 - 4

21 = 4! - 4 + (4/4)

22 = (44/4)*(sqrt(4))

23 = 4! - sqrt(4) + (4/4)

24 = 44 + 4 - 4!

25 = 4! + sqrt(4) - (4/4)

26 = (4*4) + (4/.4)

27 = 4! + 4 - (4/4)

28 = 4! + 4 + 4 - 4

29 = 4! + 4 + (4/4)

30 = 4! + 4 + 4 - sqrt(4)

31 = 4! + 4 + 4 - sqrt(sqrt(sqrt(sqrt...sqrt(4)...)))

32 = (4^4)/(4+4) = 4^(4/(.4*4))

33 = 4! + 4 + 4 + sqrt(sqrt(sqrt(sqrt...sqrt(4)...)))

34 = 4! + 4 + 4 + sqrt(4)

35 = 4! + (44/4)

36 = 44 - 4 - 4

37 = (4!/.4) - 4! + sqrt(sqrt(sqrt(sqrt...sqrt(4)...)))

38 = 44 - 4 - sqrt(4)

39 = 44 - 4 - sqrt(sqrt(sqrt(sqrt...sqrt(4)...)))

40 = (4!/.4) - 4! + 4

41 = 44 - 4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

42 = 44 - 4 + sqrt(4)

43 = 44 - (4/4)

44 = 44 + 4 - 4

45 = 44 + (4/4)

46 = 44 + 4 - sqrt(4)

47 = 4! + 4! - (4/4)

48 = 4! + 4! + 4 - 4

49 = 4! + 4! + (4/4)

50 = 44 + 4 + sqrt(4)

51 = (4! - 4 + .4)/.4

52 = 44 + 4 + 4

53 = 4! + 4! + (sqrt(4)/.4)

54 = 4! + 4! + 4 + sqrt(4)

55 = (4!/.4) - (sqrt(4)/.4)

56 = 4! + 4! + 4 + 4

57 = (4! - sqrt(4))/.4 + sqrt(4)

58 = 4! + 4! + (4/.4)

59 = (4!/.4) - (4/4)

60 = (4!/.4) + 4 - 4

61 = (4!/.4) + (4/4)

62 = (4!/.4) + 4 - sqrt(4)

63 = ((4^4) - 4)/4

64 = 4^(4 - (4/4))

65 = (4!/.4) + (sqrt(4)/.4)

66 = 44 + 4! - sqrt(4)

67 = (4! + sqrt(4))/.4 + sqrt(4)

68 = (4*4*4) + 4

69 = (4! + 4 - .4)/.4

70 = 44 + 4! + sqrt(4)

71 = (4! + 4.4)/.4

72 = 44 + 4! + 4

73 = 4! + 4! + 4! + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

74 = 4! + 4! + 4! + 4

75 = (4! + 4 + sqrt(4))/.4

76 = 4! + 4! + 4! + 4

77 = ( 4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) )^4 - 4

78 = (4! - 4)*4 - sqrt(4)

79 = (4! - sqrt(4))/.4 + 4!

80 = ((4*4) + 4)*4

81 = (4 - (4/4))^4

82 = (4! - 4)*4 + sqrt(4)

83 = (4! - .4)/.4 + 4!

84 = 44*(sqrt(4)) - 4

85 = ( 4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) )^4 + 4

86 = 44*(sqrt(4)) - sqrt(4)

87 = 44*(sqrt(4)) - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

88 = 44 + 44

89 = 44*(sqrt(4)) + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

90 = 44*(sqrt(4)) + sqrt(4)

91 = (4!*4) - (sqrt(4)/.4)

92 = 44*(sqrt(4)) + 4

93 = (4!*4) - 4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

94 = (4!*4) - 4 + sqrt(4)

95 = (4!*4) - (4/4)

96 = 4! + 4! + 4! + 4!

97 = (4!*4) - (4/4)

98 = (4!*4) + 4 - sqrt(4)

99 = (4!*4) + 4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

100 = (4!*4) + sqrt(4) + sqrt(4)

#### amits

Code:
101 = (4!*4) + [sqrt(4)/.4]

102 = (4!*4) + 4 + sqrt(4)

103 = (4! + sqrt 4)*4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

104 = (4!*4) + 4 + 4

105 = (4! + sqrt 4)*4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

106 = (4! + sqrt 4)*4 + sqrt(4)

107 = (4 + (4!/4!!))!! + sqrt(4) [B]double factorials introduced here[/B]

108 = (44/.4) - sqrt(4)

109 = (4 + (4!/4!!))!! + 4

110 = ((4!!)!! + 4!)/4 + 4!!

111 = (4! + 4)*4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

112 = (44/.4) + sqrt(4)

113 = (4! + 4)*4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

114 = (44/.4) + 4

115 = (sqrt(4)/.4)! - (sqrt(4)/.4)

116 = (4 + (4/4))! - 4

117 = (sqrt(4)/.4)! - 4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

118 = (4 + (4/4))! - sqrt(4)

119 = (4 + (4/4))! - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

120 = (4!/.4) + (4!/.4) = 120

121 = (4 + (4/4))! + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

122 = (4 + (4/4))! + sqrt(4)

123 = (sqrt(4)/.4)! + 4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

124 = (4 + (4/4))! + 4

125 = sqrt(sqrt(sqrt {[4 + (4/4)]^(4!)} ))

126 = [(4^4)/sqrt(4)] - sqrt(4)

127 = [(4^4)/sqrt(4)] - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

128 = 4*4*(4+4)

129 = [(4^4)/sqrt(4)] + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...)))

130 = [(4^4)/sqrt(4)] + sqrt(4)

131 = {[(4!!)!]/[(4!!)!!]} + 4! + sqrt(4)

132 = [(4^4)/sqrt(4)] + 4

133 = {[(4!!)!]/[(4!!)!!]} + 4! + 4

134 = (44/.4) + 4!

135 = (sqrt(4)/.4)! + (sqrt(4)/.4)!!

136 = [4! + (4/.4)]*4

#### Joffan

If both (specified base) logarithms and normal roots are permitted, then any n can be represented (actually with only 3 4's):

$$n = -log_{4}(log_{4}(\overbrace{√√√...√√√}^{\text{2n of these}}(4)))$$

Or, to use the extra 4:

$$n = -log_{4}(log_{4}(\overbrace{√√√...√√√}^{\text{2n+1 of these}}(4*4)))$$

Thanks for playing.

#### Greg Bernhardt

Opening this back up because we still need 89, 91,93!

#### cmb

Opening this back up because we still need 89, 91,93!
What are the rules? What is wrong with the 93 above using sqrt and ! .

Is % allowed?

#### cmb

I am thinking about the 'rules' and I think 0.4 is a cheat.

It implies 10. The decimalised column is 'Base 10' and infers a number which is only the value '0.4' by virtue of the assumption the base is 10 and thus the decimal column is worth "10^-1".

If you work in the bizarre base of 'Base 1/22.5' (i.e. the number "10" = 2/45ths (denary)) rather than 'Base 10' you can ALSO include a reference to the required base value using a '4'.

$$.4 + (4/4) (Base { 0.0\bar{4}})$$
and
$$.4 - (4/4) (Base { 0.0\bar{4}})$$

I am not using zeros any more than '.4' implies base 10, so the solution above should be no more nor less 'allowed' than .4 .

Thoughts?

#### dextercioby

Homework Helper
93 can be done.

$93 = 4 \cdot 4! - \frac{4!}{4!!}$

"Challenge where you have to make each number from four 4s"

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