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### Problem Summary:
This is about exploring **Pascal's Triangle modulo 2**, focusing on the pattern of even and odd binomial coefficients. The problem involves two tasks:

### (a) Write out the first 9 rows of Pascal's triangle modulo 2.

Pascal's triangle is formed from binomial coefficients \(\binom{n}{k}\), and you are asked to calculate these binomial coefficients modulo 2 for the first 9 rows.

### (b) Prove two properties about binomial coefficients mod 2 for powers of 2:
1. For \(n = 2^m - 1\), 
   \[
   \binom{2^n - 1}{k} \equiv 1 \mod 2 \quad \text{for} \quad 0 \leq k \leq 2^n - 1
   \]
   
2. For \(n = 2^m\), 
   \[
   \binom{2^n}{k} \equiv \begin{cases}
   1 \mod 2 & \text{for } k = 0 \text{ or } k = 2^n, \\
   0 \mod 2 & \text{otherwise}.
   \end{cases}
   \]

---

### Let's begin with part (a):

I'll now generate the first 9 rows of Pascal's triangle modulo 2.Here are the first 9 rows of Pascal's triangle modulo 2:

\[
\begin{matrix}
1 \\
1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{matrix}
\]

These rows exhibit an interesting pattern of zeros and ones, where:
- The binomial coefficients mod 2 appear symmetrically.
- Odd entries are replaced by 1 and even entries by 0.

### Part (b) - Proof Outline:

#### 1. Proving \( \binom{2^n - 1}{k} \equiv 1 \mod 2 \) for \( 0 \leq k \leq 2^n - 1 \):

This result can be explained by Lucas' Theorem, which relates binomial coefficients to their values modulo a prime (in this case, \( p = 2 \)).

- Since all integers from \( 0 \) to \( 2^n - 1 \) can be written in binary as \( n \)-bit numbers, Lucas' Theorem tells us that when calculating \( \binom{2^n - 1}{k} \mod 2 \), the result is always 1 for \( k \leq 2^n - 1 \).

#### 2. Proving \( \binom{2^n}{k} \equiv 1 \mod 2 \) for \( k = 0, 2^n \) and \( 0 \mod 2 \) otherwise:

- Again, using Lucas' Theorem, the binomial coefficient \( \binom{2^n}{k} \mod 2 \) behaves differently. For \( k = 0 \) and \( k = 2^n \), the binomial coefficient is 1. For other values of \( k \), it is 0.

This can also be verified by direct calculation using the definition of binomial coefficients.

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Would you like further details or explanations on any of these points? 

Here are some related questions:
1. How does Lucas' Theorem help in modulo calculations of binomial coefficients?
2. Why do binomial coefficients exhibit a pattern when calculated modulo 2?
3. How does the binary representation of numbers play a role in Lucas' Theorem?
4. What happens to Pascal's triangle modulo a number other than 2, such as 3?
5. How can the pattern in Pascal's triangle modulo 2 help in number theory problems?

**Tip:** In modulo 2 operations, the symmetry and simplicity of the results often reflect deeper properties of number systems, such as binary arithmetic.

Pascal's triangle modulo 2. Write the first 9 rows of Pascal's triangle modulo 2. Prove properties about binomial coefficients modulo 2 for powers of 2.

Learn about Pascal's triangle modulo 2 by exploring the first 9 rows and proving properties of binomial coefficients for powers of 2. This problem involves key concepts like binomial coefficients, modulo arithmetic, and Lucas' Theorem.

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