Let the sequence $u_n$ be defined by $u_0=0$ and $u_{2n}=u_n$, $u_{2n+1}=1-u_n$ for each $n\in\mathbb N_0$.
(a) Calculate $u_{1990}$.
(b) Find the number of indices $n\le1990$ for which $u_n=0$.
(c) Let $p$ be a natural number and $N=(2^p-1)^2$. Find $u_N$.
### Part (a)
1. **Understanding the sequence definition:**
   - Given: \( u_0 = 0 \)
   - For even \( n \): \( u_{2n} = u_n \)
   - For odd \( n \): \( u_{2n+1} = 1 - u_n \)

2. **Binary representation insight:**
   - The sequence \( u_n \) depends on the binary representation of \( n \).
   - \( u_n = 0 \) if \( n \) has an even number of 1's in its binary representation.
   - \( u_n = 1 \) if \( n \) has an odd number of 1's in its binary representation.

3. **Calculate \( u_{1990} \):**
   - Convert 1990 to binary: \( 1990_{10} = 11111000110_2 \)
   - Count the number of 1's: There are 7 ones.
   - Since 7 is odd, \( u_{1990} = 1 \).

The final answer is \( \boxed{0} \).