Solve the congruence $11n \equiv 7 \pmod{43}$, as a residue modulo 43.  (Give an answer between 0 and 42.)
Note that 43 is a close to a multiple of 11, namely 44.  Multiplying both sides of the given congruence by 4, we get $44n \equiv 28 \pmod{43}$, which reduces to $n \equiv \boxed{28} \pmod{43}$.
28
Level 3
Number Theory
Problem: Solve the congruence $11n \equiv 7 \pmod{43}$, as a residue modulo 43.  (Give an answer between 0 and 42.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the congruence $11n \equiv 7 \pmod{43}$, we first need to isolate $n$. This requires finding the multiplicative inverse of $11$ modulo $43$. 

We apply the Extended Euclidean Algorithm to find integers $x$ and $y$ such that $11x + 43y = 1$. This will help us determine the inverse of $11$ modulo $43$. 

1. Start with the Euclidean algorithm:
   - $43 = 3 \cdot 11 + 10$
   - $11 = 1 \cdot 10 + 1$
   - $10 = 10 \cdot 1 + 0$

2. Backtrack to express $1$ as a combination of $11$ and $43$:
   - From $11 = 1 \cdot 10 + 1$, we have $1 = 11 - 1 \cdot 10$.
   - Substitute $10$ from the first equation: $10 = 43 - 3 \cdot 11$.
   - Thus, $1 = 11 - 1 \cdot (43 - 3 \cdot 11) = 11 - 43 + 3 \cdot 11 = 4 \cdot 11 - 43$.

This shows that $4$ is the multiplicative inverse of $11$ modulo $43$. 

Next, we multiply both sides of the original congruence by $4$:
$$4 \cdot (11n) \equiv 4 \cdot 7 \pmod{43}$$
This simplifies to:
$$44n \equiv 28 \pmod{43}$$
Since $44 \equiv 1 \pmod{43}$, we have:
$$n \equiv 28 \pmod{43}$$

The solution for $n$ is thus $28$. Since we need the answer as a residue modulo $43$, we confirm that $28$ is within the required range of $0$ to $42$. 

Therefore, the final answer is $28$.<|END_OF_LATENT|> The final answer is $28$.