Determine the largest integer $n$ such that $2^n$ divides the decimal representation given by some permutation of the digits $2$, $0$, $1$, and $5$. (For example, $2^1$ divides $2150$. It may start with $0$.)
1. **Identify the permutations of the digits 2, 0, 1, and 5:**
   The possible permutations of the digits 2, 0, 1, and 5 are:
   \[
   \{2015, 2051, 2105, 2150, 2501, 2510, 1025, 1052, 1205, 1250, 1502, 1520, 5012, 5021, 5102, 5120, 5201, 5210, 0125, 0152, 0215, 0251, 0512, 0521\}
   \]

2. **Check divisibility by powers of 2:**
   We need to determine the highest power of 2 that divides any of these permutations. We will check each permutation to see how many trailing zeros (which indicate factors of 2) they have.

3. **Evaluate each permutation:**
   - **2015:** Ends in 5, not divisible by 2.
   - **2051:** Ends in 1, not divisible by 2.
   - **2105:** Ends in 5, not divisible by 2.
   - **2150:** Ends in 0, divisible by \(2^1\).
   - **2501:** Ends in 1, not divisible by 2.
   - **2510:** Ends in 0, divisible by \(2^1\).
   - **1025:** Ends in 5, not divisible by 2.
   - **1052:** Ends in 2, divisible by \(2^1\).
   - **1205:** Ends in 5, not divisible by 2.
   - **1250:** Ends in 0, divisible by \(2^1\).
   - **1502:** Ends in 2, divisible by \(2^1\).
   - **1520:** Ends in 0, divisible by \(2^1\).
   - **5012:** Ends in 2, divisible by \(2^1\).
   - **5021:** Ends in 1, not divisible by 2.
   - **5102:** Ends in 2, divisible by \(2^1\).
   - **5120:** Ends in 0, divisible by \(2^4\) (since \(5120 = 2^4 \times 320\)).
   - **5201:** Ends in 1, not divisible by 2.
   - **5210:** Ends in 0, divisible by \(2^1\).
   - **0125:** Ends in 5, not divisible by 2.
   - **0152:** Ends in 2, divisible by \(2^1\).
   - **0215:** Ends in 5, not divisible by 2.
   - **0251:** Ends in 1, not divisible by 2.
   - **0512:** Ends in 2, divisible by \(2^1\).
   - **0521:** Ends in 1, not divisible by 2.

4. **Determine the highest power of 2:**
   From the above evaluations, the permutation **5120** is divisible by \(2^4\), which is the highest power of 2 among all permutations.

5. **Conclusion:**
   The largest integer \(n\) such that \(2^n\) divides the decimal representation given by some permutation of the digits 2, 0, 1, and 5 is \(n = 4\).

The final answer is \(\boxed{4}\)