Let $n$ be a positive integer. Find the number of polynomials $P(x)$ with coefficients in $\{0, 1, 2, 3\}$ for which $P(2) = n$.
To find the number of polynomials \( P(x) \) with coefficients in \(\{0, 1, 2, 3\}\) such that \( P(2) = n \), we need to express \( n \) in base 4. This is because the coefficients of the polynomial \( P(x) \) are restricted to the set \(\{0, 1, 2, 3\}\), which corresponds to the digits in base 4.

1. **Express \( n \) in base 4:**
   Any positive integer \( n \) can be uniquely expressed in base 4 as:
   \[
   n = a_k \cdot 4^k + a_{k-1} \cdot 4^{k-1} + \cdots + a_1 \cdot 4 + a_0
   \]
   where \( a_i \in \{0, 1, 2, 3\} \) for all \( i \).

2. **Relate the base 4 representation to the polynomial:**
   The polynomial \( P(x) \) can be written as:
   \[
   P(x) = a_k x^k + a_{k-1} x^{k-1} + \cdots + a_1 x + a_0
   \]
   Given \( P(2) = n \), substituting \( x = 2 \) into the polynomial gives:
   \[
   P(2) = a_k \cdot 2^k + a_{k-1} \cdot 2^{k-1} + \cdots + a_1 \cdot 2 + a_0
   \]
   This is exactly the base 4 representation of \( n \) evaluated at \( x = 2 \).

3. **Count the number of polynomials:**
   Each unique base 4 representation of \( n \) corresponds to a unique polynomial \( P(x) \) with coefficients in \(\{0, 1, 2, 3\}\). Therefore, the number of such polynomials is equal to the number of ways \( n \) can be expressed in base 4.

4. **Conclusion:**
   Since every positive integer \( n \) has a unique base 4 representation, there is exactly one polynomial \( P(x) \) with coefficients in \(\{0, 1, 2, 3\}\) for which \( P(2) = n \).

The final answer is \( \boxed{1} \)