Kevin is trying to solve an economics question which has six steps. At each step, he has a probability $p$ of making a sign error. Let $q$ be the probability that Kevin makes an even number of sign errors (thus answering the question correctly!). For how many values of $0 \le p \le 1$ is it true that $p+q=1$?

[i]Proposed by Evan Chen[/i]
1. We start by defining the probability \( q \) that Kevin makes an even number of sign errors. Since there are six steps, the possible even numbers of errors are 0, 2, 4, and 6. The probability of making exactly \( k \) errors out of 6 steps, where \( k \) is even, is given by the binomial distribution:
   \[
   q = \sum_{k \text{ even}} \binom{6}{k} p^k (1-p)^{6-k}
   \]
   Therefore, we have:
   \[
   q = \binom{6}{0} p^0 (1-p)^6 + \binom{6}{2} p^2 (1-p)^4 + \binom{6}{4} p^4 (1-p)^2 + \binom{6}{6} p^6 (1-p)^0
   \]

2. Simplifying the binomial coefficients and terms, we get:
   \[
   q = (1-p)^6 + 15 p^2 (1-p)^4 + 15 p^4 (1-p)^2 + p^6
   \]

3. We can use the binomial theorem and symmetry properties to simplify this expression. Notice that:
   \[
   q = \frac{1}{2} \left( (p + (1-p))^6 + (p - (1-p))^6 \right)
   \]
   Since \( p + (1-p) = 1 \) and \( p - (1-p) = 2p - 1 \), we have:
   \[
   q = \frac{1}{2} \left( 1^6 + (2p-1)^6 \right) = \frac{1}{2} \left( 1 + (2p-1)^6 \right)
   \]

4. Given that \( p + q = 1 \), we substitute \( q \) into the equation:
   \[
   p + \frac{1}{2} \left( 1 + (2p-1)^6 \right) = 1
   \]

5. Simplifying the equation, we get:
   \[
   p + \frac{1}{2} + \frac{1}{2} (2p-1)^6 = 1
   \]
   \[
   p + \frac{1}{2} (2p-1)^6 = \frac{1}{2}
   \]
   \[
   2p + (2p-1)^6 = 1
   \]

6. We now solve the equation \( 2p + (2p-1)^6 = 1 \). Consider the possible values of \( 2p-1 \):
   - If \( 2p-1 = 0 \), then \( p = \frac{1}{2} \).
   - If \( (2p-1)^6 = 1 - 2p \), then \( (2p-1)^6 = 1 - 2p \).

7. For \( (2p-1)^6 = 1 - 2p \):
   - If \( 2p-1 = 1 \), then \( p = 1 \), but this does not satisfy \( 1 + 0 = 1 \).
   - If \( 2p-1 = -1 \), then \( p = 0 \), and this satisfies \( 0 + 1 = 1 \).

8. Therefore, the possible values of \( p \) are \( 0 \) and \( \frac{1}{2} \).

The final answer is \( \boxed{2} \).