Let $m>1$ be a fixed positive integer. For a nonempty string of base-ten digits $S$, let $c(S)$ be the number of ways to split $S$ into contiguous nonempty strings of digits such that the base-ten number represented by each string is divisible by $m$. These strings are allowed to have leading zeroes.

In terms of $m$, what are the possible values that $c(S)$ can take?

For example, if $m=2$, then $c(1234)=2$ as the splits $1234$ and $12|34$ are valid, while the other six splits are invalid.
1. **Initial Claim**: If \( S \) is not divisible by \( m \), then \( c(S) = 0 \). This is because if \( S \) cannot be split into parts where each part is divisible by \( m \), then there are no valid ways to split \( S \).

2. **Divisibility and Splitting**: Consider splitting \( S \) into contiguous nonempty substrings such that each substring represents a number divisible by \( m \). If \( S \) can be split in such a way, then each split point must ensure that the numbers before and after the split are divisible by \( m \).

3. **Binary Choice at Each Split**: For each potential split point in \( S \), there are two choices: either to split or not to split. This binary choice suggests that the number of valid ways to split \( S \) is a power of 2.

4. **Example for \( m = 2 \)**: Consider \( S = 1234 \). The valid splits are:
   - \( 1234 \) (no split, \( 1234 \) is divisible by 2)
   - \( 12|34 \) (both \( 12 \) and \( 34 \) are divisible by 2)
   The other possible splits (e.g., \( 1|234 \), \( 123|4 \), etc.) do not result in all parts being divisible by 2. Thus, \( c(1234) = 2 \), which is \( 2^1 \).

5. **General Case**: For any \( m \), if \( S \) can be split into parts where each part is divisible by \( m \), the number of valid splits will always be a power of 2. This is because each split point offers a binary choice (split or not split), leading to \( 2^k \) possible splits for some integer \( k \).

6. **Construction for Powers of 2**: To illustrate that \( c(S) \) can take any power of 2, consider repeating the number \( m \) multiple times. For example, if \( S = m \) repeated \( k \) times, there are \( 2^k \) ways to split \( S \) such that each part is divisible by \( m \).

7. **Construction for \( c(S) = 1 \)**: If \( S = m-1 \), then \( S \) itself is not divisible by \( m \), and thus \( c(S) = 0 \).

Conclusion:
The possible values that \( c(S) \) can take are always powers of 2.

The final answer is \( \boxed{ c(S) \text{ is always a power of 2} } \).