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:

$\boxed{101}$