If $n$ is is an integer such that $4n+3$ is divisible by $11,$ find the form of $n$ and the remainder of $n^{4}$ upon division by $11$.
1. **Determine the form of \( n \):**
   Given that \( 4n + 3 \) is divisible by \( 11 \), we can write this as:
   \[
   4n + 3 \equiv 0 \pmod{11}
   \]
   Subtracting 3 from both sides, we get:
   \[
   4n \equiv -3 \pmod{11}
   \]
   Since \(-3 \equiv 8 \pmod{11}\), we can rewrite the equation as:
   \[
   4n \equiv 8 \pmod{11}
   \]
   To solve for \( n \), we need to find the multiplicative inverse of 4 modulo 11. The multiplicative inverse of 4 modulo 11 is a number \( x \) such that:
   \[
   4x \equiv 1 \pmod{11}
   \]
   By trial or using the Extended Euclidean Algorithm, we find that \( x = 3 \) because:
   \[
   4 \cdot 3 = 12 \equiv 1 \pmod{11}
   \]
   Multiplying both sides of \( 4n \equiv 8 \pmod{11} \) by 3, we get:
   \[
   n \equiv 8 \cdot 3 \pmod{11}
   \]
   \[
   n \equiv 24 \pmod{11}
   \]
   \[
   n \equiv 2 \pmod{11}
   \]
   Therefore, \( n \) can be written in the form:
   \[
   n = 11k + 2 \quad \text{for some integer } k
   \]

2. **Find the remainder of \( n^4 \) upon division by 11:**
   Given \( n \equiv 2 \pmod{11} \), we need to find \( n^4 \mod 11 \):
   \[
   n \equiv 2 \pmod{11}
   \]
   \[
   n^4 \equiv 2^4 \pmod{11}
   \]
   Calculating \( 2^4 \):
   \[
   2^4 = 16
   \]
   Finding the remainder when 16 is divided by 11:
   \[
   16 \equiv 5 \pmod{11}
   \]
   Therefore:
   \[
   n^4 \equiv 5 \pmod{11}
   \]

The final answer is \( \boxed{ 5 } \)