Hello Everyone, 

i'm trying to make a strong marathon for number theory .. which will be in Pre-Olympiad level

Please if you write any problem don't forget to indicate its number and if you write a solution please indicate for what problem also to prevent the confusion that happens in some marathons. 

it will be preferred to write the source of the problem.
please , show detailed solutions , and please post some challenging Pre-Olympiad problems..
remember ,, different approaches are most welcome :)

now .. let's get our hands dirty :lol: :


let $ f(n)$ denote the sum of the digits of $ n$. Now let $ N \equal{} {4444^{4444}}$.
Find $ f\left( {f\left( {f\left( N \right)} \right)} \right)$.
To solve the problem, we need to find \( f(f(f(N))) \) where \( N = 4444^{4444} \) and \( f(n) \) denotes the sum of the digits of \( n \).

1. **Understanding the Sum of Digits Function \( f(n) \)**:
   The function \( f(n) \) represents the sum of the digits of \( n \). A key property of this function is that \( f(n) \equiv n \pmod{9} \). This is because the sum of the digits of a number is congruent to the number itself modulo 9.

2. **Applying the Property to \( N \)**:
   We need to find \( 4444^{4444} \mod 9 \). First, we find \( 4444 \mod 9 \):
   \[
   4444 = 4 + 4 + 4 + 4 = 16 \implies 16 \mod 9 = 7
   \]
   Therefore, \( 4444 \equiv 7 \pmod{9} \).

3. **Exponentiation Modulo 9**:
   Next, we need to find \( 7^{4444} \mod 9 \). We observe the powers of 7 modulo 9:
   \[
   7^1 \equiv 7 \pmod{9}
   \]
   \[
   7^2 \equiv 49 \equiv 4 \pmod{9}
   \]
   \[
   7^3 \equiv 7 \cdot 4 = 28 \equiv 1 \pmod{9}
   \]
   We see that \( 7^3 \equiv 1 \pmod{9} \). Therefore, \( 7^{4444} \equiv (7^3)^{1481} \cdot 7^1 \equiv 1^{1481} \cdot 7 \equiv 7 \pmod{9} \).

4. **Sum of Digits Function Iteration**:
   Since \( N \equiv 7 \pmod{9} \), we have \( f(N) \equiv 7 \pmod{9} \). The sum of the digits of \( N \) will be a number that is congruent to 7 modulo 9. Let's denote this sum as \( S_1 \).

   Applying the sum of digits function again, \( f(S_1) \equiv 7 \pmod{9} \). Let's denote this sum as \( S_2 \).

   Finally, applying the sum of digits function one more time, \( f(S_2) \equiv 7 \pmod{9} \).

5. **Conclusion**:
   Therefore, \( f(f(f(N))) \equiv 7 \pmod{9} \). Since the sum of the digits of a number that is congruent to 7 modulo 9 is 7, we conclude that:
   \[
   f(f(f(N))) = 7
   \]

The final answer is \( \boxed{7} \).