Let \( A = { 0, 1, 2, \ldots, 15 } \). We need to find the number of subsets \( S \) of \( A \) such that if \( n \in S \) and \( m \equiv n + 1 \pmod{16} \), then \( m \in S \).
This condition indicates that if an element \( n \) is chosen to be in the subset, the next element in the sequence, \( n+1 \) (taken modulo 16), must also be included.
We can simplify our analysis by breaking down the elements into equivalence classes based on the modulo operation. More specifically, we can think in terms of cycles, since adding 1 wraps around from 15 back to 0.
An alternate way to view this problem is to note that every element \( n \) that is included in set \( S \) mandates certain other elements to be included based on the condition provided. Consider \( S \) based upon which of the groups of consecutive elements we allow into \( S \).
Step 1: Partitioning into cycles
We create a cycle of elements as follows:
- The first cycle is \( {0, 1, 2, 3, \ldots, 15} \).
This cycle can be viewed as a circular arrangement of choices, where choosing any number in the sequence forces us to decide about all elements in that circular arrangement.
Step 2: Defining choices
We notice that choosing whether to include 0 or not leads directly to whether to include other elements:
- If we include \( 0 \), we must include \( 1 \) (since if \( n = 0 \), then \( m = 1 \)).
- Continuing this logic around the cycle, if we include any element \( n \), we must include all elements \( n, n+1, n+2, \ldots \) until we hit back at \( n \).
Step 3: Setting up free choices
However, we can choose to start including elements from any point in the cycle or choose to skip an entire cycle. Each cycle allows either:
- To include the entire cycle (which forces all members of that sequence),
- To skip the entire cycle altogether (which gives an empty set).
Therefore, we can treat the cycle as having \( 0 \) (not included) or \( 1 \) (included) decisions, meaning for each element there are essentially two choices.
Conclusion of Choices
Each cycle can either be included completely (which fixes all of those positions) or excluded entirely. This gives us:
\[ \text{Total subsets} = 2^k \]
where \( k \) is the number of independent (unselected) cycles, which translates to the total groups of elements such that if \( n \) is in the set, \( n+1 \mod{16} \) is in there too.
Since our choices are only between including or not including an entire cycle of numbers, \( k = 16 \) (the total size of \( A \)). Thus, the total number of subsets \( S \) that satisfy the condition is given by:
\[ \boxed{16} \]
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