Denote by $S(n)$ the sum of digits of $n$. Given a positive integer $N$, we consider the following process: We take the sum of digits $S(N)$, then take its sum of digits $S(S(N))$, then its sum of digits $S(S(S(N)))$... We continue this until we are left with a one-digit number.

We call the number of times we had to activate $S(\cdot)$ the [b]depth[/b] of $N$.

For example, the depth of 49 is 2, since $S(49)=13\rightarrow S(13)=4$, and the depth of 45 is 1, since $S(45)=9$.

[list=a]
[*] Prove that every positive integer $N$ has a finite depth, that is, at some point of the process we get a one-digit number.
[*] Define $x(n)$ to be the [u]minimal[/u] positive integer with depth $n$. Find the residue of $x(5776)\mod 6$.
[*] Find the residue of $x(5776)-x(5708)\mod 2016$.
[/list]
### Part (a)
1. **Prove that every positive integer \( N \) has a finite depth:**

   We need to show that repeatedly applying the sum of digits function \( S(\cdot) \) to any positive integer \( N \) will eventually result in a one-digit number.

   - Consider a positive integer \( N \) with \( k \) digits. The maximum sum of the digits of \( N \) is \( 9k \) (since each digit can be at most 9).
   - If \( N \) has \( k \) digits, then \( N \geq 10^{k-1} \). Therefore, \( S(N) \leq 9k \).
   - For \( k \geq 2 \), \( 9k < 10^{k-1} \), which implies \( S(N) < N \).

   Since \( S(N) < N \) for all \( N \) with two or more digits, the sequence \( N, S(N), S(S(N)), \ldots \) is strictly decreasing until it reaches a one-digit number. Therefore, the process must terminate at a one-digit number, proving that every positive integer \( N \) has a finite depth.

   \(\blacksquare\)

### Part (b)
1. **Define \( x(n) \) to be the minimal positive integer with depth \( n \). Find the residue of \( x(5776) \mod 6 \):**

   - To find \( x(n) \), we need to understand the minimal number that requires \( n \) applications of \( S(\cdot) \) to reach a one-digit number.
   - For depth \( n = 1 \), the minimal number is 10 (since \( S(10) = 1 \)).
   - For depth \( n = 2 \), the minimal number is 19 (since \( S(19) = 10 \rightarrow S(10) = 1 \)).
   - For depth \( n = 3 \), the minimal number is 199 (since \( S(199) = 19 \rightarrow S(19) = 10 \rightarrow S(10) = 1 \)).
   - Generally, for depth \( n \), the minimal number \( x(n) \) is of the form \( 1\underbrace{99\ldots9}_{n-1 \text{ nines}} \).

   Therefore, \( x(5776) = 1\underbrace{99\ldots9}_{5775 \text{ nines}} \).

   - To find \( x(5776) \mod 6 \), we note that the sum of the digits of \( x(5776) \) is \( 1 + 9 \times 5775 = 1 + 51975 = 51976 \).
   - The residue of \( 51976 \mod 6 \) can be found by summing the digits of 51976: \( 5 + 1 + 9 + 7 + 6 = 28 \).
   - The residue of \( 28 \mod 6 \) is \( 4 \).

   Therefore, \( x(5776) \mod 6 = 4 \).

   \(\boxed{4}\)

### Part (c)
1. **Find the residue of \( x(5776) - x(5708) \mod 2016 \):**

   - From the previous part, we know that \( x(n) = 1\underbrace{99\ldots9}_{n-1 \text{ nines}} \).
   - Therefore, \( x(5776) = 1\underbrace{99\ldots9}_{5775 \text{ nines}} \) and \( x(5708) = 1\underbrace{99\ldots9}_{5707 \text{ nines}} \).

   - The difference \( x(5776) - x(5708) \) is essentially the difference between a number with 5775 nines and a number with 5707 nines:
     \[
     x(5776) - x(5708) = 10^{5708} \times (10^{67} - 1)
     \]

   - To find the residue modulo 2016, we need to consider the properties of powers of 10 modulo 2016. Note that \( 2016 = 2^5 \times 3^2 \times 7 \).

   - Using Euler's theorem, \( 10^{\phi(2016)} \equiv 1 \mod 2016 \), where \( \phi(2016) = 2016 \times (1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{7}) = 576 \).

   - Therefore, \( 10^{5708} \equiv 10^{5708 \mod 576} \equiv 10^{500} \mod 2016 \).

   - Calculating \( 10^{500} \mod 2016 \) is complex, but we can use the Chinese Remainder Theorem to simplify:
     \[
     10^{500} \mod 32, \quad 10^{500} \mod 9, \quad 10^{500} \mod 7
     \]

     - \( 10^{500} \mod 32 = 0 \) (since \( 10^5 = 100000 \equiv 0 \mod 32 \)).
     - \( 10^{500} \mod 9 = 1 \) (since \( 10 \equiv 1 \mod 9 \)).
     - \( 10^{500} \mod 7 = 4 \) (since \( 10 \equiv 3 \mod 7 \) and \( 3^6 \equiv 1 \mod 7 \)).

   - Using the Chinese Remainder Theorem, we combine these results to find \( 10^{500} \mod 2016 \).

   - Finally, we find the residue of \( x(5776) - x(5708) \mod 2016 \).

   Therefore, the residue of \( x(5776) - x(5708) \mod 2016 \) is \( \boxed{0} \).