The House of Four Fours
By Jamie on Tuesday, January 29, 2002 - 03:43 pm:

I have a problem to figure out. Here it is. There once was a house in which all numbers were representable with exactly four fours! No matter what number one wanted to represent, it could be done using the four fours in an appropriate way. The job is to figure out how to represent each of the different numbers using four fours. Begin by representing each of the numbers from 0-10.

Thanks for your help,
Jamie

By Graeme Mcrae on Tuesday, January 29, 2002 - 05:18 pm:

Here are some ways to represent some of the numbers between 0 and 100. I left gaps where I can't think of a way to represent that number.

If I found different representations for a number, I opted for the more whimsical. For example, 23=4!-(Ö4+Ö4)/4 was not chosen because it was too straightforward.

0 = (4+4-4-4)

1 = (4+4/4-4)

2 = 4/4+4/4

3 = (4+4+4)/4

4 = 4+4-Ö4-Ö4

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-Ö4)

11 = 4!/Ö4-4/4

12 = (4!+4-4×4)

13 = 4!/Ö4+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-Ö4)

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

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

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

22 = (4!/4+4×4)

23 = 4!!/(4!-Ö4)!/4!

24 = (4+4+4×4)

25 = 4!+4/(Ö4+Ö4)

26 = 4!+(4+4)/4

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

28 = (4!+4+4-4)

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

30 = (4×4×Ö4-Ö4)

31 = 4!+(4!+4)/4

32 = (4×4+4×4)


33 = (   æ
 ú
Ö
  æ
Ö

Ö
44!
 
 
 +Ö4)/Ö4

34 = (4×4×Ö4+Ö4)

35 = 4!+(4!-Ö4)/Ö4

36 = (4!+4+4+4)

37 = 4!+(4!+Ö4)/Ö4

38 = 4!+4×4-Ö4

39=

40 = (4×4×4-4!)

41=

42 = 4!+4×4+Ö4

43 = 44-4/4

44 = (4!+4+4×4)

45 = (Ö4+4)!/4/4

46 = (4!-4/4)×Ö4

47 = 4!+4!-4/4

48 = 4!+4!+4-4


49 =   _________
Ö((4!+4)/4)4
 

50 = 4!×Ö4+4-Ö4

51=


52 =   æ
Ö
  ___
Ö4!4
 
 ×4
 
 +4

53=

54 = 4!×Ö4+4+Ö4

55=


56 =   æ
Ö
  ______
Ö(4!+4)4
 
 ×4
 

57=

58 = (4!+4)×Ö4+Ö4

59=

60 = (4×4×4-4)

61=

62 = (4×4×4-Ö4)

63 = (44-4)/4

64 = (4×4×Ö4×Ö4)

65 = (44+4)/4

66 = (4×4×4+Ö4)

67=

68 = (4×4×4+4)

69=

70 = (44+4!)/4

71=


72 =
Ö
 
((Ö4+4)4×4)
 

73=

74 = 4!×4-4!+Ö4

75=

76 = (4!-4)×4-4

77=

78 = (4!-4)×4-Ö4

79=

80 = (4!×4-4×4)

81 = (4-4/4)4

82 = (4!-4)×4+Ö4

83=

84 = (4!-4)×4+4

85=

86 = (4!-Ö4)×4-Ö4

87=

88 = (4!+4×4×4)

89=

90 = 4!×4-4-Ö4

91=

92 = 4!×4-Ö4-Ö4

93=

94 = 4!×4-4+Ö4

95 = (4!×4-4/4)

96 = (4!×4+4-4)

97 = (4!×4+4/4)

98 = 4!×4+4-Ö4

99=

100 = 4!×4+Ö4+Ö4

By Gavin Adams on Tuesday, January 29, 2002 - 10:33 pm:

Define (a1,a2,a3,a4,a5,a6,a7,a8,a9...,aN) be a binary string with each number aN corresponding to the number N if aN = (4/4)/(4/4)=1
So:
1 = ((4/4)/(4/4),,,,,,....,,,,)
10 = (,,,,,,,,,(4/4)/(4/4),,,,....,,,,)
And so forth
So tee hee