For every $n \in \mathbb{N}$, there exist integers $k$ such that $n | k$ and $k$ contains only zeroes and ones in its decimal representation. Let $f(n)$ denote the least possible number of ones in any such $k$. Determine whether there exists a constant $C$ such that $f(n) < C$ for all $n \in \mathbb{N}$. 
1. **Understanding the Problem:**
   We need to determine if there exists a constant \( C \) such that for every natural number \( n \), the function \( f(n) \) (which denotes the least possible number of ones in any integer \( k \) that is divisible by \( n \) and contains only the digits 0 and 1) is less than \( C \).

2. **Classical Result:**
   The classical result mentioned states that any nonzero multiple of \( 10^m - 1 \) has a digit sum of at least \( 9m \). This result can be derived from properties of numbers in base 10.

3. **Digit Sum Analysis:**
   Consider a number \( k \) that is a multiple of \( n \) and contains only the digits 0 and 1. The digit sum of \( k \) is simply the number of ones in \( k \). If \( k \) is a multiple of \( 10^m - 1 \), then the digit sum of \( k \) must be at least \( 9m \).

4. **Implication for \( f(n) \):**
   For \( n = 10^m - 1 \), any multiple of \( n \) must have a digit sum of at least \( 9m \). Therefore, \( f(10^m - 1) \geq 9m \).

5. **Non-existence of Constant \( C \):**
   Since \( f(10^m - 1) \geq 9m \) and \( m \) can be arbitrarily large, \( f(n) \) can grow without bound. Hence, there cannot exist a constant \( C \) such that \( f(n) < C \) for all \( n \in \mathbb{N} \).

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