Happy New Year 2554

December 31, 2010

Happy New Year 2554 (Buddhist calendar equivalent to 2011).

I wish this year will be a great one for everybody.

Last year, I wrote a similar HNY card. First Sakulbuth suggested that we do this again this year. So, here it is. (What do you do when you are too lazy to do mathematics? – Arithmetic.)

Pick your favorite lucky number. If it’s an integer within -100 to 100, we can represent it with digits in “2554” sequentially, using only +, -, *, /, !, √, ., ^. (With exceptions for 69, 75, 83, 87).

PS. Consequently, if you favorite lucky number is an integer out of this range, you can represent it with finite multiple repeats of digits in “2554” sequentially. If you favorite lucky number is a rational number, you can also represent it with finite multiple repeats of digits in “2554.” If you favorite lucky number is irrational, at worst you then can approximate it with sequences of digits in “2554.” If you allow “i,” this goes for those whose favorite lucky number isn’t real too.

Again, happy new year – whichever your favorite lucky numbers are.

0 = 2*(5-5)*4
1 = 2*5 – (5+4)
2 = -(2+5) + (5+4)
3 = -(2^(5-5)) + 4
4 = -2+5+5-4
5 = 2^(5-5) + 4
6 = 2^(5/5) + 4
7 = (2+5)*(5-4)
8 = 2+5+5-4
9 = 2*5 – 5 + 4
10 = 2*5*(5-4)
11 = (-2+5)*5-4
12 = -2+5+5+4
13 = -2-5+5*4
14 = √25 + 5 + 4
15 = -√25 + 5*4
16 = 2+5+5+4
17 = 2-5+5*4
18 = 2*(5+5) – √4
19 = -2+5*5-4
20 = 25/5*4
21 = √25 * 5 – 4
22 = -2^5 + 54
23 = 2^5 – 5 – 4
24 = 25-5+4
25 = 25*(5-4)
26 = 25+(5-4)
27 = -2+5*5+4
28 = 2*(5+5+4)
29 = -25 + 54
30 = 2*5+5*4
31 = 2+5*5+4
32 = (2^5)*(5-4)
33 = 2^5 + (5-4)
34 = 25+5+4
35 = 2*5.5 + 4!
36 = 2^(√5*5) + 4
37 = (2+5)*5*√4
38 = -2+(5+5)*4
39 = (2+5)*5+4
40 = 2*(5+5)*√4
41 = 2^5 + 5 + 4
42 = 2+(5+5)*4
43 = -2 + 5*(5+4)
44 = -(2*5)+54
45 = 25 + 5*4
46 = 2*5*5-4
47 = -2-5+54
48 = 2*5*5 – √4
49 = -2+55-4
50 = 2.5*5*4
51 = 2-5+54
52 = 2^5+5*4
53 = 2+55-4
54 = 2*5*5+4
55 = -2+55+√4
56 = 2^√(5*5) + 4!
57 = -2+55+4
58 = 2*5+5!*.4
59 = 2+55+√4
60 = (-2+5)*5*4
61 = 2+55+4
62 = 2^5+5!/4
63 = (2+5)*(5+4)
64 = 2^(5+5-4)
65 = √25 + 5!/√4
66 = 2+5!*.5+4
67 = 2+5+5!/√4
68 = 2+5!-54
69 = σ(2)+5!-54
70 = (2+5)*5*√4
71 = -25+5!-4!
72 = .2*5!+5!*.4
73 = 25+5!*.4
74 = 2+5!-5!*.4
75 = σ(2)*5*5!/4!
76 = (.2*5!-5)*4
77 = -2+55+4!
78 = .2*5!+54
79 = 25+54
80 = (25-5)*4
81 = 2+55+4!
82 = -2+5!*.5+4!
83 = σ(2^5)+5*4
84 = -2^5+5!-4
85 = 25+5!/√4
86 = 2^5+54
87 = σ(2^√(5*5))+4!
88 = (-2+5!/5)*4
89 = -2-5+5!-4!
90 = 2*5*(5+4)
91 = -25+5!-4
92 = (-2+5*5)*4
93 = 2-5+5!-4!
94 = -2+(√5*5)!-4!
95 = -25 + (√(5^√4))!
96 = 2*5!/5*√4
97 = -25+5!+√4
98 = 2*(-5+54)
99 = -2+5+5!-4!
100 = (√25)*5*4

Note : σ is a sigma function. σ(n) = sum of positive divisor of n.
(If you can find an alternative for 69, 75, 83, 87 without using σ, please leave a comment.)

Happy New Year 2553!

January 2, 2010

<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

The Interest

February 21, 2008

Last week, I got an email from my dad. As usual, he made a weird loan system for the cooperative – well, not so weird actually.

Suppose that you borrow A $ from a bank with k% interest. Generally the bank take k% of A $ then divided it into N times of payment. For example, you borrow 10000 $ with 12% per year interest. You will pay back every month in two years. Therefore, you have to pay the bank 933.34 $ each month.

The cooperative has the main purpose to help members in finances. Hence, the cooperative needs less money than banks do.

Here is his loan system.

The same system, but this time, the interest won’t be calculated first. Each time you pay, the interest is decreased due to how much the capital is left. You will pay back in the same amount of money each time, however. Therefore, each time you pay, the capital part is getting larger and the interest part is getting smaller.

Confuse? Well, I’m confuse. Let’s see the example.

Suppose that you borrow 399000 $ from me. (I don’t have that much money actually. haha) We agree that you will pay me back in 84 months with 5% interest per year (or 0.4167 % per month) with my dad’s system. From the calculation, you have to pay me 5639.43 $ per month.

First month, I take 0.4167 % of 399000 $ which is 1662.5 $. You pay me 5639.43 $ which is 1662.5$ interest and (5639.43-1662.5) = 3976.93 $ capital.

Next month, I take 0.4167% of (399000 – 3976.93) = 395023.07 $ which is 1645.93 $. You pay me 5639.43$ which is 1645.93$ and 3993.50 $ capital.

and so on

in last month, you will pay 5639.43$ as usual, but it will be about 23.4$ interest and 5639.40$ capital.

I think we get the system now.

The problem is how can we calculate ‘how much you have to pay me each month in order to pay me back all the money included interest by the last payment – exactly.’

So my dad asked me this problem. He asked for a formula to calculate ‘the amount of money the borrower needs to pay each month’ when the amount of capital money, the interest rate, and the number of payments are given.

It took me such a long time for this problem because I tried to figure out how he came up with 5639.43$ at first. (Later on I found out that he adjusted it by hand.) First, I tried to find a recursive relation between each payment. I, however, could not see an easy pattern from its recursion.

I then assumed the amount of each payment to be X. Then I wrote out everything that would happen each payment in variables. Bang! I saw a pattern. I then summed up and solved the equation. I came out with a formula in the end.

I think this might me an interesting problem for somebody who is finding something to think in his/her leisure.

Here is the answer I got.

Let;

borrowed money = A
interest per payment = k
the number of payments = N
an amount of money would be paid each time = X

Define

untitled-1.gif

then, X=AG
the sum of interest (I) ; I=(NG-1)A

The result comes out pretty cute, isn’t it?

My First Attempt on LaTeX

February 20, 2008

LaTeX is a high-quality typesetting system; it includes features designed for the production of technical and scientific documentation.

I started to want to learn how to use LaTeX three years ago, maybe, but I had never done it. After installing a full TeX package on my Mac three months ago, I started reading The Not So Short Introduction to LaTeX 2e last week.

I know that LaTeX will be significantly important to my education and career in the not-so-far future. Better learn it now!

My first trial is the abstract of my childish research, the Relation of Multinomial Coefficients on the Forming of Multidimensional Simplex. By the way, this is the first publication of the research on the Internet.

The Relation of Multinomial Coefficients on the Forming of Multidimensional Simplex : Abstract

Hello World!

February 20, 2008

Hey World,

I think that I should write a blog. I have spent too much time which one should I use. First, I thought that I would choose the one that I can modify its theme as I want. Finally, I chose WordPress because its simple but beautiful original theme. For me, it looks so Mac. Now, the header is Aqua.


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