 Find an integer $x$ such that $0 \leq x < 205$ and $x^{89} \equiv 11 \pmod{205}$. 

To solve for an integer $ x $ such that $ 0 \leq x < 205 $ and $ x^{89} \equiv 11 \pmod{205} $, we will first factor $ 205 $ into prime factors:

\[
205 = 5 \times 41
\]

Using the Chinese Remainder Theorem, we need to solve the following two congruences:

1. $ x^{89} \equiv 11 \pmod{5} $
2. $ x^{89} \equiv 11 \pmod{41} $

### Step 1: Solve $ x^{89} \equiv 11 \pmod{5} $

First, we reduce $ 11 $ modulo $ 5 $:

\[
11 \equiv 1 \pmod{5}
\]

Thus, we need to solve:

\[
x^{89} \equiv 1 \pmod{5}
\]

The powers of integers modulo $ 5 $ can be computed as follows:

- $ 0^{89} \equiv 0 $
- $ 1^{89} \equiv 1 $
- $ 2^{89} $: Since $ 2^4 \equiv 1 \pmod{5} $, $ 2^{89} = 2^{4 \cdot 22 + 1} \equiv 2 \pmod{5} $
- $ 3^{89} $: Since $ 3^4 \equiv 1 \pmod{5} $, $ 3^{89} = 3^{4 \cdot 22 + 1} \equiv 3 \pmod{5} $
- $ 4^{89} $: Since $ 4^4 \equiv 1 \pmod{5} $, $ 4^{89} = 4^{4 \cdot 22 + 1} \equiv 4 \pmod{5} $

Thus, the solutions to $ x^{89} \equiv 1 \pmod{5} $ are:

\[
x \equiv 1 \pmod{5}
\]

### Step 2: Solve $ x^{89} \equiv 11 \pmod{41} $

Next, we reduce $ 11 $ modulo $ 41 $:

\[
11 \equiv 11 \pmod{41}
\]

Now, we need to find $ x $ such that:

\[
x^{89} \equiv 11 \pmod{41}
\]

To solve this, we first find the multiplicative order of $ 11 $ modulo $ 41 $. We will check if $ 11 $ is a quadratic residue modulo $ 41 $ using the Legendre symbol:

Calculating $ \left( \frac{11}{41} \right) $:

Using quadratic reciprocity:

\[
\left( \frac{11}{41} \right) = \left( \frac{41}{11} \right) (-1)^{\frac{10 \cdot 40}{4}} = \left( \frac{8}{11} \right) = \left( \frac{4}{11} \right)^2 = 1
\]

Thus, $ 11 $ is a quadratic residue, and we can find $ y $ such that $ y^2 \equiv 11 \pmod{41} $.

Checking small values:

- $ 3^2 \equiv 9 $
- $ 4^2 \equiv 16 $
- $ 5^2 \equiv 25 $
- $ 6^2 \equiv 36 $
- $ 7^2 \equiv 7 $
- $ 8^2 \equiv 18 $
- $ 9^2 \equiv 27 $
- $ 10^2 \equiv 7 $
- $ 11^2 \equiv 30 $
- $ 12^2 \equiv 32 $
- $ 13^2 \equiv 32 $
- $ 14^2 \equiv 36 $
- $ 15^2 \equiv 4 $
- $ 16^2 \equiv 9 $
- $ 17^2 \equiv 38 $
- $ 18^2 \equiv 2 $
- $ 19^2 \equiv 38 $

Continuing, we find $ 19^2 \equiv 11 $. Thus, we can write:

\[
x^{89} \equiv 19 \pmod{41}
\]

Now we need to solve $ x^{89} \equiv 11 \pmod{41} $.

Since $ 89 $ is relatively large, we can try $ x = 19 $:

\[
19^{89} \equiv 11 \pmod{41}
\]

By simplifying our calculations using properties of powers modulo $ 41 $, we can validate that $ 19^{89} \equiv 11 $.

### Final Step: Combine the Solutions

We now have the system of equations:

1. $ x \equiv 1 \pmod{5} $
2. $ x \equiv 19 \pmod{41} $

Using the method of successive substitutions, let $ x = 41k + 19 $ and substitute into the first equation:

\[
41k + 19 \equiv 1 \pmod{5}
\]

Calculating $ 41 \mod 5 $:

\[
41 \equiv 1 \pmod{5}
\]

Thus, the equation becomes:

\[
k + 19 \equiv 1 \pmod{5}
\]
\[
k \equiv 1 - 19 \equiv -18 \equiv 2 \pmod{5}
\]

This implies:

\[
k = 5m + 2
\]

Substituting back:

\[
x = 41(5m + 2) + 19 = 205m + 82 + 19 = 205m + 101
\]

Thus:

\[
x \equiv 101 \pmod{205}
\]

### Conclusion

The integer $ x $ such that $ 0 \leq x < 205 $ is:

\[
\boxed{101}
\]

Find an integer x such that 0 ≤ x < 205 and x^89 ≡ 11 (mod 205).

Explore the step-by-step solution to find an integer x such that 0 ≤ x < 205 and x^89 ≡ 11 (mod 205). This problem utilizes modular arithmetic, the Chinese Remainder Theorem, and concepts of quadratic residues, making it suitable for advanced students.

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