Project Euler Problem 17: How many letters would be needed to write all the numbers in words from 1 to 1000?

What I learn from this problem:

Yeah, not fourty. I didn't get right answer until I saw this forty was recorded as 5 letters instead of 6 letters.

Anyway.

Project Euler 17

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

There are 3 ways, as far as I know.

1. http://www.mathblog.dk/project-euler-17-letters-in-the-numbers-1-1000/

2. http://blog.functionalfun.net/2008/08/project-euler-problem-17-converting.html

3. This.

First way would be using pattern of occurrence of the words. You know, word one exist in number 1 and 51 and 100 and bla bla bla. I don't really want to deal with this one.

Second way, it is by actually calculating the letters. The Words for specific numbers stored in a dictionary (I don't know what is that). Though this guy made it for more.

My way is similar, but in term of I already calculate it by hand for specific numbers. So I only add numbers according to the noOfLetter I already see first, with some more algorithm to automate the calculate for larger number. Let's see the specific number list.

num, word, noOfLetter

1, "one", 3
2, "two", 3
3, "three", 5
4, "four", 4
5, "five", 4
6, "six", 3
7, "seven", 5
8, "eight", 5
9, "nine", 4
10, "ten", 3
11, "eleven", 6
12, "twelve", 6
13, "thirteen", 8
14, "fourteen", 8
15, "fifteen", 7
16, "sixteen", 7
17, "seventeen", 9
18, "eighteen", 8
19, "nineteen", 8
20, "twenty", 6
30, "thirty, 6"
40, "forty", 5
50, "fifty", 5
60, "sixty", 5
70, "seventy", 7
80, "eighty", 6
90, "ninety", 6
100, "onehundred", 10
200, "twohundred", 10
300, "threehundred",12
400, "fourhundred",11
500, "fivehundred",11
600, "sixhundred",10
700, "Sevenhundred",12
800, "eighthundred",12
900, "ninehundred",11
1000, "onethousand",11

When I see number like 931 = "ninehundred and thirty one", I can see it can be divided by three(3) parts. So, first I calculate the number of letters for "ninehundred", then "thirty", then "one". Bear in mind, there is also "and" which only happened if the hundred don't have tens or unit. It this, I refer hundred to 100-999, tens to 10-99 and unit to 1-9. So, I am planning to make 3 different methods related to each other.

1. Unit[1-9]/noOfLetter(i)
    I only need to return the specific value for each of them.
2. Tens[10-99]/noOfLetterTen(i)
    For 11-19, the term is not divide-able. So we require their own noOfLetter. Then, we can set the value for 20,30,40,50,60,70,80 and 90. Why that only? Because we can re-use Unit part. So, when it is for 21, do recursive of noOfLetterTen(20) + noOfLetter(1). This, as you can see, actually "twentyone" = "twenty"+"one". But there are a need of splitting the number. I'll tell you later.
3. Hundreds[100-999]/noOfLetterHundred(i)
    We need to remember that "and" only happen if there are more words after, e.g., "onehundred". So how we can divide it? I remember our modulus (%) of integer division can split numbers.

Let's test this: 1234

1234/1000=1

1234%1000=234

1234/100 =12

1234%100=34

1234/10 = 123

1234%10 = 4

So, from here, we can reduce our algorithm significantly. Bear in mind, "onehundred" can be divided into "one" and "hundred". "hundred" have noOfLetter=7. "and" have numberOfLetter=2. Something I have done to look at the pattern and create its formula.

i	Name	Return value	i/100	I%100	i/10	i%10
1000	One thousand	11	10
100	One hundred	value(i%100)+7=3+7=10	1	0
200		value(i%100)+7=10	2	0
300		value(i%100)+7=12	3	0
345	Three hundred and forty five	Value(300) + 3[and] + value(40) + value(5)	3	45	4	5
400		value(i%100)+7=11	4	0
500		value(i%100)+7=11	5	0
600		value(i%100)+7=10	6	0
700		value(i%100)+7=5+7=12	7	0
800		value(i%100)+7=12	8	0
900		value(i%100)+7=11	9	0
1	One	3
2	Two	3
3	Three	5
4	Four	4
5	Five	4
6	Six	3
7	Seven	5
8	Eight	5
9	Nine	4
10	ten	3			1	0
11	Eleven	6			1	1
12	Twelve	6			1	2
13	Thirteen	8			1	3
14	Fourteen	8			1	4
15	Fifteen	7			1	5
16	Sixteen	7			1	6
17	Seventeen	9			1	7
18	Eighteen	8			1	8
19	Nineteen	8			1	9
20	Twenty	6			2	0
30	Thirty	6			3	0
40	Forty	5			4	0
45	Fourty-five	Value(40)+value(5)=5+4=10	45		4	5
50	Fifty	5			5	0
60	Sixty	5			6	0
70	Seventy	7			7	0
80	Eighty	6			8	0
90	Ninety	6			9	0
91	Ninety one	Value(90)+value(1)=6+3=9 Value(i - i%10) + value(i%10)	`		9	1

So, for method noOfLetterTen : noOfLetterTen(i-i%10) + noOfLetter(i%10); You see, it split by subtracting i with i%10 for the ten part and modulus i%10 for unit part. While for method noOfLetterHundred, : noOfLetter(i/100) + 7 + and + noOfLetterTen(i%100);

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