Find $2 \cdot 5^{-1} + 8 \cdot 11^{-1} \pmod{56}$.

Express your answer as an integer from $0$ to $55$, inclusive.
If we first get a "common denominator" as if 2, 5, 8, and 11 represent real numbers rather than residues, we get $$\frac 25 + \frac{8}{11} \equiv \frac{2 \cdot 11 + 8 \cdot 5}{55} \equiv \frac{62}{-1} \equiv -62 \equiv \boxed{50} \pmod{56}.$$Indeed, we can justify this manipulation as follows. Suppose that $n \equiv 2 \cdot 5^{-1} + 8 \cdot 11^{-1} \pmod{56}$; then multiplying both sides of the congruence by $55$ (which is relatively prime with $56$) yields that $-n \equiv 55n \equiv 22 + 40 \equiv 62 \pmod{56}$.
50
Level 5
Number Theory
Problem: Find $2 \cdot 5^{-1} + 8 \cdot 11^{-1} \pmod{56}$.

Express your answer as an integer from $0$ to $55$, inclusive.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to compute the modular inverses of 5 and 11 modulo 56. 

First, we find the modular inverse of 5 modulo 56. The modular inverse \( x \) satisfies the equation \( 5x \equiv 1 \pmod{56} \). We can use the Extended Euclidean Algorithm to find this inverse. 

1. Apply the Euclidean algorithm to find \( \gcd(5, 56) \):
   - \( 56 = 11 \cdot 5 + 1 \)
   - \( 5 = 5 \cdot 1 + 0 \)
   - Thus, \( \gcd(5, 56) = 1 \).

2. Now, backtrack to express 1 as a linear combination of 5 and 56:
   - From \( 1 = 56 - 11 \cdot 5 \), we see that \( -11 \) is a solution for \( 5x \equiv 1 \pmod{56} \).
   - To convert \( -11 \) to a positive equivalent, we compute \( -11 + 56 = 45 \).
   - Therefore, \( 5^{-1} \equiv 45 \pmod{56} \).

Next, we find the modular inverse of 11 modulo 56. We need \( y \) such that \( 11y \equiv 1 \pmod{56} \).

1. Again, apply the Euclidean algorithm:
   - \( 56 = 5 \cdot 11 + 1 \)
   - \( 11 = 11 \cdot 1 + 0 \)
   - Thus, \( \gcd(11, 56) = 1 \).

2. Backtrack to express 1 as a linear combination:
   - From \( 1 = 56 - 5 \cdot 11 \), we find \( -5 \) is a solution for \( 11y \equiv 1 \pmod{56} \).
   - To convert \( -5 \) to a positive equivalent, we compute \( -5 + 56 = 51 \).
   - Therefore, \( 11^{-1} \equiv 51 \pmod{56} \).

Now we substitute the inverses back into the original expression:
- Calculate \( 2 \cdot 5^{-1} + 8 \cdot 11^{-1} \):
  - \( 2 \cdot 45 = 90 \)
  - \( 8 \cdot 51 = 408 \)

Next, we sum these results:
- \( 90 + 408 = 498 \).

Finally, we reduce \( 498 \) modulo \( 56 \):
- Compute \( 498 \div 56 \) which gives approximately \( 8.893 \), so the integer part is \( 8 \).
- Calculate \( 8 \cdot 56 = 448 \).
- Subtract to find the remainder: \( 498 - 448 = 50 \).

Thus, the final answer is \( 50 \).<|END_OF_LATENT|> The final answer is $50$.