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

[Joffe]

Have you ever heard of that challenge where you have to make each number from four 4s? This list is the best I have been able to come up with, will you help me fill in the gaps?
Any solution that is simpler than one here can take its place, for example if you come up with a solution to 33 that doesn't use a decimal symbol it is simpler.
EDIT: An underline indicates repetition: (i.e: .4 = .44444 / 4/9).

1 = 4*4/(4*4)
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 = (4*4+4!)/4
11 = (4+4!)/4+4
12 = (4-4/4)*4
13 = (4+4!+4!)/4
14 = 4!/4+4+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 = 4!-(4+4)/4
23 = 4!-4^(4-4)
24 = 4*4+4+4
25 = 4!+(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!)/4
31 = (4+4!)/4+4!
32 = 4^4/(4+4)
33 = (4-.4)/.4+4!
34 = 4!/4+4+4!
35 = (4.4/.4)+4!
36 = (4+4)*4+4
37 = 4/.[U]4[/U]+4+4!
38 = 44-4!/4
39 = (4*4-.4)/.4
40 = (4^4/4)-4!
41 = (4*4+.4)/.4
42 = 4!+4!-4!/4
43 = 44-4/4
44 = 4*4+4+4!
45 = (4!/4)!/(4*4)
46 = (4!-4)/.4 - 4
47 = 4!+4!-4/4
48 = (4*4-4)*4
49 = 4!+4!+4/4
50 = (4*4+4)/.4
51 = 4!/.4-4/.[U]4[/U]
52 = 44+4+4
53 = 44+4/.[U]4[/U]
54 = (4!/4)^4/4!
55 = (4!-.4)/.4-4
56 = 4!+4!+4+4
57 = 4/.[U]4[/U]+4!+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)/.4
63 = (4^4-4)/4
64 = 4^(4-4/4)
65 = 4^4+4/4
66 = (4+4!)/.4-4
67 = (4+4!)/.[U]4[/U]+4
68 = 4*4*4+4
69 = (4+4!-.4)/.4
70 = (4^4+4!)/4
71 = (4!+4.4)/.4
72 = (4-4/4)*4!
73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u]
74 = (4+4!)/.4+4
75 = (4!/4+4!)/.4
76 = (4!-4)*4-4
77 = (4!-.[u]4[/u])/.[u]4[/u]+4!
78 = (4!x.[u]4[/u]+4!)/.[u]4[/u]
79 = ([sup].4[/sup]√4-.4)/.4
80 = (4*4+4)*4
81 = (4/4-4)^4
82 = 4!/.[U]4[/U]+4!+4
83 = (4!-.4)/.4+4!
84 = (4!-4)*4+4
85 = (4/.4+4!)/.4
86 = (4-.4)x4!-.4
87 = 4!x4-4/.[U]4[/U]
88 = 4^4/4+4!
89
90 = (4!/4)!/(4+4)
91
92 = (4!-4/4)*4
93
94 = (4+4!)/.4 + 4!
95 = 4!*4-4/4
96 = 4!*4+4-4
97 = 4!*4+4/4
98 = (4!+.4)*4+.4
99 = (4!+4!-4)/.[U]4[/U]
100 = 4*4/(.4*.4)

Thanks:

• ceptimus: 71
• LarrrSDonald: 51,53,82,87,79

 

[ceptimus]

71 = (4! + 4.4) / .4

...and if we can use square roots:

37 = (sqrt(4) + 4!)/sqrt(4) + 4!
51 = (4! - sqrt(4)) / .4 - 4
53 = sqrt(4) / .4 + 4! + 4!
57 = (4! - .4) / .4 - sqrt(4)
67 = (sqrt(4) + 4!) / .4 + sqrt(4)
78 = 4 * (4! - 4) - sqrt(4)
79 = (4! - sqrt(4)) / .4 + 4!
82 = 4 * (4! - 4) + sqrt(4)
89 = (sqrt(4) + 4!) / .4 + 4!
91 = 4 * 4! - sqrt(4) / .4

 

[Joffe]

Thanks for 71 ceptimus!
The others I am not as interested in because (as I probably should have stated) I am not considering square roots because they require a 2, I would however consider 4th roots so long as the 4 is counted. No triggonometric functions either, basically whatever techniques I have used so far.

 

[ArielGenesis]

if we can use zero as if in decimal, why couldn't we use 2 as if in square root as well as trigonometry function and, most probabaly also included, logarithm. If you are about using arithmetich signs only, then decimal should not be used.

 

[ArielGenesis]

and also exclemation mark !, as 4! should include other number such 1, 2 and 3

 

[LarrrSDonald]

If you're ok with .\bar{4} = \frac{4}{9} then you have:

51 = \frac{4!}{4}-\frac{4}{.\bar{4}}
53 = 44+\frac{4}{.\bar{4}}
82 = \frac{4!}{.\bar{4}}+4!+4
87 = 4!*4-\frac{4}{.\bar{4}}

Possibly others, I haven't had time to seach around that much.

 

[RandallB]

51 = \frac{4!}{4}-\frac{4}{.\bar{4}}

Should be:
51 = \frac{4!}{.4}-\frac{4}{.\bar{4}}

 

[LarrrSDonald]

You're right, lost a decimal point. Sorry 'bout that.
[EDIT] If you're willing to consider 4th root, then perhaps you'd consider \sqrt[.4]{4} = 32 and thus
79 = \frac{(\sqrt[.4]{4} - .4)}{.4}

 

[Joffe]

LarrrSDonald: I like your ideas! Your repetition idea got me a whole heap of answers including 37! And this latest one, 79 is great lateral thinking, Keep em coming.

ArielGenesis: If I were to include square root signs then it might as well be called the four 4 or 2s challenge, it would make it far too easy. Using decimals is still bending a little (hence the reason to try to try to eliminate some of the solutions that use them) but since it does not make a whole number it keeps the game challenging. And if I were to include trig I could make combinations like: sin-1 ( cos ( 4 ) ) = 86, which is clearly too convenient.

EDIT: only 3 remain!

 

[Alkatran]

4 = (4-4)/4+4

4 != 0/4

I'd go for 4 = 4! - 4*4 - 4

 

[Joffe]

4 = (4-4)/4+4

4 != 0/4
I'd go for 4 = 4! - 4*4 - 4

(4-4)/4 + 4
= 0/4 + 4
= 0 + 4
= 4

Perhaps you misinterpreted my grouping symbols.

 

[RandallB]

LarrrSDonald: I like your ideas! Your repetition idea got me a whole heap of answers including 37!

BUT you didn't say what 37 was! Maken us work for it eh?
Did you use:
37 = \frac{{4}*{4}+{.\bar{4}}}{.\bar{4}}

 

[ceptimus]

He's got 37 = 4 / .4 + 4 + 4! in the table above

(he used an underline to show recurring)

 

[Joffe]

Indeed, here are two more:

89 = (4-.4-4%)/4%

91 = (4-.4+4%)/4%

93 is all that remains!

 

[ceptimus]

I can only do it with the forbidden square root.

93 = 4 \times 4! - \sqrt{\frac{4}{.\bar{4}}}

 

[croxbearer]

If logarithm is allowed:

93 = 4 x 4! - log(4/.4%)

 

[Joffe]

There is only one problem with that, there is an implied 10. I would definantly accept log in base 4 and maby the natural logarithm of an expression but that (albeit quite novel) just seems to violate in my opinion.

If I don't ever find a better solution I think I will settle for that though croxbearer.

 

[Markii]

Another way to get 93 using an exponent of 0 is 4! x 4 - 4 - 4°

 

[Fahim555]

Yo, I got 89!

 

[Fang_Lou]

5 = (4x4+4)/4

 

[animefreak1]

i wanted to get 1020 but i don't know how to do it.

 

[Fang_Lou]

How about this:
1020=4^4x4-4

 

[animefreak1]

hey thanks a lot i really mean it too

 

[RichardM]

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]

here are the answers -

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]

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.