Suppose that $x^{10} + x + 1 = 0$ and $x^100 = a_0 + a_1x +... + a_9x^9$. Find $a_5$.
1. Given the equation \( x^{10} + x + 1 = 0 \), we can express \( x^{10} \) in terms of \( x \):
   \[
   x^{10} = -x - 1
   \]

2. We need to find \( x^{100} \). Notice that:
   \[
   x^{100} = (x^{10})^{10}
   \]
   Substituting \( x^{10} = -x - 1 \) into the equation, we get:
   \[
   x^{100} = (-x - 1)^{10}
   \]

3. We expand \( (-x - 1)^{10} \) using the binomial theorem:
   \[
   (-x - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k
   \]
   This expansion gives us:
   \[
   (-x - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k
   \]

4. We need to express \( x^{100} \) in the form \( a_0 + a_1 x + a_2 x^2 + \cdots + a_9 x^9 \). Notice that the highest power of \( x \) in the expansion of \( (-x - 1)^{10} \) is \( x^{10} \). However, since \( x^{10} = -x - 1 \), we can reduce any term with \( x^{10} \) or higher powers back to a polynomial of degree less than 10.

5. Specifically, we are interested in the coefficient \( a_5 \). From the binomial expansion:
   \[
   (-x - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k
   \]
   The coefficient of \( x^5 \) in this expansion is:
   \[
   \binom{10}{5} (-1)^5 = \binom{10}{5} (-1)^5 = \binom{10}{5} (-1) = -\binom{10}{5}
   \]

6. Calculating \( \binom{10}{5} \):
   \[
   \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
   \]

7. Therefore, the coefficient \( a_5 \) is:
   \[
   a_5 = -252
   \]

The final answer is \( \boxed{-252} \)