1

Binary number mod 3. Find:

$1101101001010111010110111011111001_2 \bmod 3$

You may not use a calculator.

I know how to convert a small binary number, but I do not know what approach to take with a larger one like this? Any thoughts on how you would proceed?

J. W. Tanner
  • 60,406
Maxxie
  • 43

2 Answers2

2

Note that $\,2\equiv -1\pmod 3\,$, so it follows that $\,2^n\equiv (-1)^n\pmod 3\,$, and binary numbers are just sums of powers of 2. Can you take it from there?

A.J.
  • 3,892
2

$2^{even}\equiv (-1)^{even} \equiv 1 \pmod 3$

$2^{odd} \equiv (-1)^{odd} \equiv -1 \pmod 3$.

So

$1101101001010111010110111011111001_2 \pmod 3$...

Marking the odd powers in red and the evens in blue

$\color{red}1\color{blue}10\color{blue}1\color{red}10\color{red}100\color{blue}10\color{blue}10\color{blue}1\color{red}1\color{blue}10\color{blue}10\color{blue}1\color{red}10\color{red}1\color{blue}1\color{red}10\color{red}1\color{blue}1\color{red}1\color{blue}1\color{red}100\color{blue}1_2 \pmod 3$

That's $12$ to and even power and $10$ to an odd power.

So $\color{red}1\color{blue}10\color{blue}1\color{red}10\color{red}100\color{blue}10\color{blue}10\color{blue}1\color{red}1\color{blue}10\color{blue}10\color{blue}1\color{red}10\color{red}1\color{blue}1\color{red}10\color{red}1\color{blue}1\color{red}1\color{blue}1\color{red}100\color{blue}1_2 \pmod 3$

$\equiv \color{blue}{12}-\color{red}{10} \equiv 2\pmod 3$

Note that if you have $11$ next to each other that's an even and an odd and they cast each other out.

fleablood
  • 124,253