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Greetings in the day of Mary ... Mother Mary ...
And Evil always to prove God do not exist ... Evil is stronger and the Universe is a small town, or an evil country. But the Universe is Huge.
Now, The Additive Number Theory is equivalent to ... solve a problem up to a certain point or number ... and then solve the rest.
While ... in this Planet (I excuse with you for this statement) ... The Distribution of Prime Number is still an unsolved problem this do not means will be forever unsolved.
There are many problems in "Additive Number Theory" still unsolved.
I reveal today some secrets about my first Math books when I was a teenager.
Let me reveal some other ... between the book I purchase were (without any previous consistent knowledge in German) but simply because included in the Bibliography of G. H. Hardy  An Introduction to Number Theory (5th Ed) ... I do not pay that book, was a gift from my family that lived and live in England ...
... Well ... I translate one of the books ... in a Summer (may be Summer 1979) from German to Spanish including a great number of details on Perfect Numbers and Amicable Numbers.
But let us solve this puzzle ...
The Question is: "In how many ways you can change 1 dollar (100 ¢) ... using coins of 1¢, 5¢, 10¢ (dime), 25¢ (quarter) and 50¢ ? ...
Let us begin ... and consider (translating or simplifying the problem) ...
... that you have a White Juke box where you can insert only and exclusively coins of 1¢ ...
A
Consider this Juke box have only Spanish Songs ... 100 songs ... equivalent to the modes to change 1$ ... and you want to listen to begin ...
Formula V  Eva Maria for ... 25¢.
In how many ways you can insert these coins in the machine to listen that song? ...
You can think about ... but there are Exactly ... ONE MODE ... because you have just 1¢.
Therefore to listen your song ... You need to insert 25 x 1¢ ... You have no other coins.
If you label "A" this Juke box ... or the number of possible coins ... you will have (or begin to have) a Table ... like ...
We will built ... Please understand that A_n = 1 ... for all n.
Now ... we will continue to add ... and consider NOT a Juke box only with coins of 5¢ ... but a Juke box where you can enter coins of 1¢ and 5¢.
B
Call this Juke Box ... B ... and consider it have 100 Italians songs.
Now you have only coins of 1 and 5 cents ... Suppose you want to listen the Italian Song ... "I Pooh ... Portami via" ... that costs 30¢.
In how many ways you can insert coins of 1¢ and 5¢? ...
You can divide the problem in two ... subproblems ... that are:
 In how many ways you can insert ONLY coins of 5¢ ... inside the Juke box ... to listen a song that cost 30¢ and
 In how many ways you can insert (the rest of options) ...
To insert or use only coins of 5¢ ... You will ...
 Use 6 coins of 5¢ ... and listen the song. This is one solution.
 Use 5 coins of 5¢ ... and the rest of 1¢ ... This is a second solution.
 Use 4 coins of 5¢ ... and the rest of 1¢ ... This is a third solution.
 Use 3 coins of 5¢ ... and the rest of 1¢ ... This is a fourth solution.
 Use 2 coins of 5¢ ... and the rest of 1¢ ... This is a fifth solution.
 Use 1 coins of 5¢ ... and the rest of 1¢ ... This is a sixth solution.
 ... the rest ... Only coins of 1¢ ...
Then, The Solution is 7 ... or B_30 = 7.
You can complete the row related to B ... (1 and 5 cents) applying the recurring formula
You can deduct ... for now we have explained only the latest formula ...
Now ... consider a Blue Juke Box ... what range of coins will be possible to include? ... Evidently 1, 5 and 10.
This is the Juke Box ... Cyan (or Celeste) ... and have 100 songs only in French ...
C
In how many ways you can insert the coins ... if you want to listen a French song Lolita ... for 40¢? ...
Again ... we will split the problem ... using previous solution.
We first use the coin with the highest value ... which is 10¢.
Then ... the two solutions we will consider are:
With 10¢ and without ...
 We use 4 coins of 10¢ ... This is a solution.
 We use 3 coins of 10¢ + 2 x 5¢ ... This is a Solution #2.
 + 1 x 5¢ + 5 x 1¢ ... This is a Solution #3.
 + 10 x 1¢ ... This is a Solution #4.
 We use 2 coins of 10¢ + 3 x 5¢ + 5 x 1¢ ... This is a Solution #5.
 + 2 x 5¢ + 10 x 1¢ ... This is a Solution #6.
 + 1 x 5¢ + 15 x 1¢ ... This is a Solution #7.
 + 20 x 1¢ ... This is a Solution #8.
 We use 1 coins of 10¢ + 6 x 5¢ ... This is a Solution #9.
 + 5 x 5¢ + 5 x 1¢ ... This is a Solution #10.
 + 4 x 5¢ + 10 x 1¢ ... This is a Solution #11.
 + 3 x 5¢ + 15 x 1¢ ... This is a Solution #12.
 + 2 x 5¢ + 20 x 1¢ ... This is a Solution #13.
 + 1 x 5¢ + 25 x 1¢ ... This is a Solution #14.
 + 30 x 1¢ ... This is a Solution #15.
 + 40 x 1¢ ... This is a Solution #16.
... with no more coins of 10¢ ... we add to the number of solutions ... the solution of B_40 = 9.
Therefore ... we apply the recurring formula relating C_40 = B_40 + C_(4010) = 9+16= 25.
In the same way we can complete the row relating C ...
Let us move to D ...
D
We will consider 100 songs in Portuguese ... and add a coin for choice: 25¢
Consider you want to listen the Brazilian song "Mas Que Nada" ... for 50¢
Is "evident" that ...
D_50 = C_50 + D_25 ... and this give us ... 49 = 13+36.
We now conclude ... adding 50¢ ...
Our Juke box ... now include 100 English songs ... and E (the labeled Juke Box) will include in the term ... E_100 ... the expected answer for the Problem.
E
This is extremely simple ... if we insert 50¢ ... we reduce the problem to previous one.
Suppose you want to listen ... "Michael Jackson ... You are not alone" ... for 50¢
E_50 = 1 + D_50 = 50.
The Solution is 292 ...
The complete Table is:
n 
0 
5 
10 
15 
20 
25 
30 
35 
40 
45 
50 
55 
60 
65 
70 
75 
80 
85 
90 
95 
100 
An 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
Bn 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
Cn 
1 
2 
4 
6 
9 
12

16 
20 
25 
30 
36 
42 
49 
56 
64 
72 
81 
90 
100 
110 
121 
Dn 
1 
2 
4 
6 
9 
13 
18 
24 
31 
39 
49 
60 
73 
87 
103 
121 
141 
163 
187 
213 
242 
En 
1 
2 
4 
6 
9 
13 
18 
24 
31 
39 
50 
62 
77 
93 
112 
134 
159 
187 
218 
252 
292 
You can also read the Classical Book ... "Basic Programming" by Kemeny (another Martian) and Kurtz.
They solve the Problem using the Computer and a Program in BASIC language.
The Solution for Euro ... with 20¢ is consonant to the Problem that appear in the German Edition related to the Swiss Franc ... that is equivalent to solve the Euro ... The number is 4562. 