Lucas' Theorem. This theorem states that (𝑛𝑘)(*kn*​) is odd if (𝑘)𝑖≤(𝑛)𝑖(*k*)*i*​≤(*n*)*i*​ for all $i = 0,1,...$ and even otherwise. Here, (𝑛)𝑖(*n*)*i*​ is the 𝑖*i*-th least significant bit of 𝑛*n* if written in binary. For example, 1111 in binary is 10111011 and 33 is 1111 (or 00110011 to make sure that both have four bits). Now 00110011 is ``bitwise at most'' 10111011, and thus Lucas' Theorem states that (113)(311​) is odd. Which it is, since (113)=165(311​)=165. 

To get started, let us introduce some notation. For an integer 𝑛≥1*n*≥1 we define 𝑓(𝑛):=max⁡{𝑘∈𝑁0 such that 2𝑘 divides 𝑛}*f*(*n*):=max{*k*∈N0​ such that 2*k* divides *n*}. For example, 𝑓(5)=0*f*(5)=0, 𝑓(14)=1*f*(14)=1, 𝑓(12)=2*f*(12)=2.

4.

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? 

Write your answer as a simple 0/1 string. For example, if your sequence is (1,0,1,0,1,0,1,0)(1,0,1,0,1,0,1,0), answer 1010101010101010.

To solve this problem, we will use Lucas' Theorem to determine the sequence of $\binom{n}{k} \mod 2$ for $k = 0, 1, \ldots, 7$, where $n = 1152921504606846987$.

First, we need to express $n$ in binary:
\[ n = 1152921504606846987 \]

Convert this number to binary:
\[ n = 10000000000000000000000000000000000000000000000000000000000011 \]

We need to compute $\binom{n}{k} \mod 2$ for $k = 0, 1, \ldots, 7$. Lucas' Theorem states that $\binom{n}{k} \mod 2$ is determined by comparing the binary representation of $n$ and $k$. Specifically, $\binom{n}{k} \mod 2 = 1$ if and only if for every bit position $i$, the bit of $k$ is less than or equal to the bit of $n$.

We will compare the binary representation of $n$ with each $k$ in the range $0, 1, \ldots, 7$:

1. **For $k = 0$**
\[ k = 00000000 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{0} \mod 2 = 1$.

2. **For $k = 1$**
\[ k = 00000001 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{1} \mod 2 = 1$.

3. **For $k = 2$**
\[ k = 00000010 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{2} \mod 2 = 1$.

4. **For $k = 3$**
\[ k = 00000011 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{3} \mod 2 = 1$.

5. **For $k = 4$**
\[ k = 00000100 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{4} \mod 2 = 1$.

6. **For $k = 5$**
\[ k = 00000101 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{5} \mod 2 = 1$.

7. **For $k = 6$**
\[ k = 00000110 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{6} \mod 2 = 1$.

8. **For $k = 7$**
\[ k = 00000111 \]
Since all bits of $k$ are less than or equal to the corresponding bits of $n$, $\binom{n}{7} \mod 2 = 1$.

Hence, the sequence of $\binom{n}{k} \mod 2$ for $k = 0, 1, \ldots, 7$ is:
\[ 11111111 \]

The answer as a simple 0/1 string is:
\[ \boxed{11111111} \]

Do you want more details or have any questions? 

Here are 5 numbered questions you can ask:

1. Can you explain Lucas' Theorem in more detail?
2. How do you convert a decimal number to binary?
3. Can you demonstrate Lucas' Theorem with another example?
4. What is the significance of the modulus operation in combinatorics?
5. Can you explain the concept of binomial coefficients and their properties?

Tip: When working with large numbers, converting them to binary can simplify the analysis, especially when using theorems like Lucas' Theorem.

Explore how to apply Lucas' Theorem to compute (n k) mod 2 for n = 1152921504606846987, resulting in a sequence of 1's for k from 0 to 7. This guide delves into binary representations and their implications in combinatorics.

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