Math Problem Statement

Let 𝑛

1152921504606846987 n=1152921504606846987. If we compute ( 𝑛 π‘˜ ) m o d

2 ( k n ​ )mod2 for π‘˜

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 10101010 10101010.

Solution

To determine the sequence of (nk)mod  2\binom{n}{k} \mod 2 for k=0…7k = 0 \ldots 7 with n=1152921504606846987n = 1152921504606846987, we need to use Lucas' theorem. Lucas' theorem gives a way to compute binomial coefficients modulo a prime number (in this case, 2) by looking at the binary representations of nn and kk.

First, we convert nn to its binary representation: n=1152921504606846987n = 1152921504606846987 n in binary=100000000000000000000000000000000000000000000000000011n \text{ in binary} = 100000000000000000000000000000000000000000000000000011

We then compute the binomial coefficients (nk)mod  2\binom{n}{k} \mod 2 for k=0k = 0 to k=7k = 7 by examining the bits of nn and kk.

Here are the binary representations:

  • k=0k = 0: 000000
  • k=1k = 1: 001001
  • k=2k = 2: 010010
  • k=3k = 3: 011011
  • k=4k = 4: 100100
  • k=5k = 5: 101101
  • k=6k = 6: 110110
  • k=7k = 7: 111111

Using Lucas' theorem, (nk)mod  2\binom{n}{k} \mod 2 is determined by the individual bits of nn and kk. For each bit position where kk has a 1, nn must also have a 1 for (nk)\binom{n}{k} to be odd (1 mod 2). If any bit position where kk has a 1 but nn has a 0, (nk)\binom{n}{k} is even (0 mod 2).

Let's evaluate:

  • For k=0k = 0: always 1 (since (n0)=1\binom{n}{0} = 1).
  • For k=1k = 1: The rightmost bit of nn is 1.
  • For k=2k = 2: The second bit from the right of nn is 1.
  • For k=3k = 3: Both the first and second bits from the right of nn are 1.
  • For k=4k = 4: The third bit from the right of nn is 0.
  • For k=5k = 5: The third bit from the right of nn is 0.
  • For k=6k = 6: The third bit from the right of nn is 0.
  • For k=7k = 7: The third bit from the right of nn is 0.

Thus, we get the sequence: (nk)mod  2\binom{n}{k} \mod 2 for k=0k = 0 to k=7k = 7 as: 1,1,1,1,0,0,0,01, 1, 1, 1, 0, 0, 0, 0

This can be written as the binary string: 1111000011110000

If you have any questions or need further details, feel free to ask!

Here are some related questions you might be interested in:

  1. How does Lucas' theorem work for prime moduli other than 2?
  2. Can you show a detailed example of Lucas' theorem for a smaller number?
  3. How can we compute (nk)mod  2\binom{n}{k} \mod 2 for larger ranges of kk?
  4. What are the properties of binomial coefficients modulo a prime number?
  5. How can Lucas' theorem be extended to composite moduli?

Tip: Lucas' theorem is a powerful tool for computing binomial coefficients modulo prime numbers, especially useful in combinatorial number theory and computer science for solving problems efficiently.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Number Theory
Binary Arithmetic

Formulas

Lucas' Theorem for binomial coefficients modulo a prime

Theorems

Lucas' Theorem

Suitable Grade Level

Grades 11-12