Math Problem Statement

Lucas' Theorem. This theorem states that (π‘›π‘˜)(kn​) is odd if (π‘˜)𝑖≀(𝑛)𝑖(k)i​≀(n)i​ for all $i = 0,1,...$ and even otherwise. Here, (𝑛)𝑖(n)i​ is the 𝑖i-th least significant bit of 𝑛n 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 𝑛β‰₯1nβ‰₯1 we define 𝑓(𝑛):=max⁑{π‘˜βˆˆπ‘0 such that 2π‘˜ divides π‘›}f(n):=max{k∈N0​ such that 2k divides n}. For example, 𝑓(5)=0f(5)=0, 𝑓(14)=1f(14)=1, 𝑓(12)=2f(12)=2.

Question 4

Let 𝑛=1152921504606846987n=1152921504606846987. If we compute (π‘›π‘˜)mod  2(kn​)mod2 for π‘˜=0…7k=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.

Solution

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

First, we need to express nn in binary: n=1152921504606846987n = 1152921504606846987

Convert this number to binary: n=10000000000000000000000000000000000000000000000000000000000011n = 10000000000000000000000000000000000000000000000000000000000011

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

We will compare the binary representation of nn with each kk in the range 0,1,…,70, 1, \ldots, 7:

  1. For k=0k = 0 k=00000000k = 00000000 Since all bits of kk are less than or equal to the corresponding bits of nn, (n0)mod  2=1\binom{n}{0} \mod 2 = 1.

  2. For k=1k = 1 k=00000001k = 00000001 Since all bits of kk are less than or equal to the corresponding bits of nn, (n1)mod  2=1\binom{n}{1} \mod 2 = 1.

  3. For k=2k = 2 k=00000010k = 00000010 Since all bits of kk are less than or equal to the corresponding bits of nn, (n2)mod  2=1\binom{n}{2} \mod 2 = 1.

  4. For k=3k = 3 k=00000011k = 00000011 Since all bits of kk are less than or equal to the corresponding bits of nn, (n3)mod  2=1\binom{n}{3} \mod 2 = 1.

  5. For k=4k = 4 k=00000100k = 00000100 Since all bits of kk are less than or equal to the corresponding bits of nn, (n4)mod  2=1\binom{n}{4} \mod 2 = 1.

  6. For k=5k = 5 k=00000101k = 00000101 Since all bits of kk are less than or equal to the corresponding bits of nn, (n5)mod  2=1\binom{n}{5} \mod 2 = 1.

  7. For k=6k = 6 k=00000110k = 00000110 Since all bits of kk are less than or equal to the corresponding bits of nn, (n6)mod  2=1\binom{n}{6} \mod 2 = 1.

  8. For k=7k = 7 k=00000111k = 00000111 Since all bits of kk are less than or equal to the corresponding bits of nn, (n7)mod  2=1\binom{n}{7} \mod 2 = 1.

Hence, the sequence of (nk)mod  2\binom{n}{k} \mod 2 for k=0,1,…,7k = 0, 1, \ldots, 7 is: 1111111111111111

The answer as a simple 0/1 string is: 11111111\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.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Number Theory
Binary Representation

Formulas

Lucas' Theorem
Binomial Coefficients: C(n, k)

Theorems

Lucas' Theorem

Suitable Grade Level

Grades 11-12