
Question 4

Let 𝑛=1152921504606846987*n*=1152921504606846987. If we compute (𝑛𝑘)mod  2(*kn*​)mod2 for 𝑘=0…7*k*=0…7, which sequence do we get? 

To solve the problem of computing \(( \binom{n}{k} ) \mod 2\) for \( k = 0 \) to \( k = 7 \), where \( n = 1152921504606846987 \), we need to use Lucas' theorem.

### Lucas' Theorem
Lucas' theorem provides a way to compute binomial coefficients modulo a prime number. For a prime \( p \) and non-negative integers \( n \) and \( k \), the binomial coefficient \(\binom{n}{k} \mod p\) can be computed using the base-\( p \) representation of \( n \) and \( k \).

If \( n \) and \( k \) are represented in base \( p \) as:
\[ n = n_0 + n_1 p + n_2 p^2 + \cdots + n_m p^m \]
\[ k = k_0 + k_1 p + k_2 p^2 + \cdots + k_m p^m \]
where \( 0 \leq n_i, k_i < p \), then:
\[ \binom{n}{k} \equiv \prod_{i=0}^{m} \binom{n_i}{k_i} \mod p \]

### Application to the Problem
Here, \( p = 2 \), and we need to compute \(( \binom{n}{k} ) \mod 2\) for \( k = 0, 1, 2, \ldots, 7 \).

First, let's convert \( n \) and \( k \) to their binary (base-2) representation. The binary representation of \( n = 1152921504606846987 \) is:
\[ n = 100000000000000000000000000000000000000000000000000011 \]

Now, consider \( k = 0, 1, 2, \ldots, 7 \) in binary:
- \( k = 0 \): \( 0 \)
- \( k = 1 \): \( 1 \)
- \( k = 2 \): \( 10 \)
- \( k = 3 \): \( 11 \)
- \( k = 4 \): \( 100 \)
- \( k = 5 \): \( 101 \)
- \( k = 6 \): \( 110 \)
- \( k = 7 \): \( 111 \)

### Applying Lucas' Theorem
For each \( k \), we use the binary digits of \( n \) and \( k \) to determine \(\binom{n_i}{k_i} \mod 2\).

- If any \( k_i > n_i \), then \(\binom{n_i}{k_i} = 0\) because you cannot choose more elements than available.
- If \( k_i \leq n_i \), then \(\binom{n_i}{k_i} \mod 2 = 1\) if \( k_i = 0\) or \( k_i = 1 \).

Let's analyze each \( k \):

1. **\( k = 0 \):**
   \[
   \binom{n}{0} \mod 2 = 1
   \]

2. **\( k = 1 \):**
   \[
   \binom{n}{1} \mod 2 = 1
   \]

3. **\( k = 2 \):**
   \[
   \binom{n}{2} \mod 2 = 0 \quad \text{(since \( 2 \) has more than one bit set and corresponding bits in \( n \) are 0)}
   \]

4. **\( k = 3 \):**
   \[
   \binom{n}{3} \mod 2 = 0 \quad \text{(same reasoning as above)}
   \]

5. **\( k = 4 \):**
   \[
   \binom{n}{4} \mod 2 = 0 \quad \text{(same reasoning as above)}
   \]

6. **\( k = 5 \):**
   \[
   \binom{n}{5} \mod 2 = 0 \quad \text{(same reasoning as above)}
   \]

7. **\( k = 6 \):**
   \[
   \binom{n}{6} \mod 2 = 0 \quad \text{(same reasoning as above)}
   \]

8. **\( k = 7 \):**
   \[
   \binom{n}{7} \mod 2 = 0 \quad \text{(same reasoning as above)}
   \]

### Conclusion
The sequence \(( \binom{n}{k} \mod 2 )\) for \( k = 0, 1, 2, 3, 4, 5, 6, 7 \) is:
\[ 1, 1, 0, 0, 0, 0, 0, 0 \]

Would you like more details or have any other questions?

Here are some other questions you might consider:
1. How do you derive Lucas' theorem?
2. What are some applications of binomial coefficients in combinatorics?
3. How can you compute large binomial coefficients efficiently?
4. What are other prime modulus properties in binomial coefficients?
5. Can Lucas' theorem be generalized for non-prime moduli?

**Tip:** Always convert large numbers to their binary representation when dealing with modulo \( 2 \) operations, as it simplifies the problem significantly.

Learn how to compute binomial coefficients modulo 2 using Lucas' Theorem with a detailed explanation for each coefficient from 0 to 7. Understand the application of binary representation and modular arithmetic in combinatorics. Suitable for advanced mathematics enthusiasts.

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