Happy New Year 2553!

<h1>Happy New Year 2553!</h1> (Note : 2553 in Thai year-counting system is equivalent to 2010) Best wish to you all… I feel like this year will be a great one. Why? We can represent all integers from 0 to 100 with digits in “2553″ sequentially!

0 = 2*(5-5)*3

1 = -2+5-5+3

2 = 25÷5-3

3 = √25 – 5 + 3

4 = (2+5+5)÷3

5 = 2-5+5+3

6 = 2+(5÷5)+3

7 = √25 + 5 -3

8 = 2*5-5+3

9 = 2+5+5-3

10 = 25-5*3

11 = -2+5+5+3

12 = 2*5+5-3

13 = √25 + 5+ 3

14 = (2+5)*(5-3)

15 = 2+5+5+3

16 = (2^5)÷(5-3)

17 = 25-5-3

18 = 2*5+5+3

19 = -2+(5!÷5)-3

20 = 2*5*(5-3)

21 = -(2^5)+53

22 = 2+5+5*3

23 = 25-5+3

24 = (-2+5)*(5+3)

25 = 2*5+5*3

26 = -2+5*5+3

27 = 25+5-3

28 = -25+53

29 = 2^(√(5*5)) – 3

30 = 2+5*5+3

31 = (√25)*5+3!

32 = (2+5)*5-3

33 = 25+5+3

34 = (2^5)+5-3

35 = 2^(√(5*5)) + 3

36 = (2+5+5)*3

37 = 2-5+(5!÷3)

38 = -2+5*(5+3)

39 = (2-5+5!)÷3

40 = (2^5)+5+3

41 = (.2*5)+(5!÷3)

42 = 2+5*(5+3)

43 = -(2*5)+53

44 = 2*(5*5-3)

45 = (-2+5)*5*3

46 = -2-5+53

47 = 2*5*5-3

48 = -(√25)+53

49 = (2+5)^(5-3)

50 = -2+55-3

51 = 2*(5!÷5)+3

52 = -2+5÷(.5*)*3!

53 = 2*5*5+3

54 = .2*5+53

55 = (√25)*(5+3!)

56 = (2+5)*(5+3)

57 = 2+5*(5+3!)

58 = √25 + 53

59 = -2+55+3!

60 = 2+5+53

61 = -2+.5*5!+3

62 = 2+(5!÷(5-3))

63 = 2+55+3!

64 = (2^5)*(5-3)

65 = 25+(5!÷3)

66 = 2*5.5*3!

67 = (√25)!-53

68 = (2^5)+(5!*.3)

69 = 2+5!-53

70 = -2+(5!÷5*.3)

<i>71</i> = 2*p(5)*5 + ɸ(ɸ(3))

72 = (2^5) + (5!÷3)

73 = -2+5*5*3

74 = 2+(5!÷5*3)

75 = (√25)*5*3

<i>76</i> = -2*p(5)+ɸ(5^3)

77 = 2+5*5*3

78 = 25+53

<i>79</i> = 2*ɸ(55)-ɸ(ɸ(3))

80 = 2*5*(5+3)

81 = [(-2+5)^5]÷3

82 = 2+(5!+5!)÷3

<i>83</i> = 2*ɸ(55)+3

84 = (√25)!-(5!*.3)

85 = 255÷3

86 = 2+5!-5!*.3

87 = (.2+.5)*5!+3

88 = -2+5!-5*3!

89 = -25+5!-3!

90 = (25+5)*3

91 = -(2^5)+5!+3

92 = 2+5!-5*3!

93 = -(2^5)+(5^3)

94 = -(2^5)+5!+3!

<i>95</i> = p(2*5)+53

96 = 2*(-5+53)

<i>97</i> = ɸ(25)*5-3

98 = -25+5!+3

99 = -(.2*5!)+5!+3

100 = (2*5)^(5-3)

Notation : .5* means repeated decimal, so .5* = .55555555…

p(n) = number of partition of n

ɸ(n) = Euler Phi Function

I tried to use only arithmetic functions. However, I don’t have that much free time. Thus, there must be a compromise on 71, 76, 79, 83, 95, 97 by using number theory functions. If anybody can solve those, please leave me a comment and I will update the list. Also, if anyone find a mistake or a prettier calculation, please tell me too.

Happy New Year!

Nics

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