# Decimal to binary question [closed]

This is an easy question, but I can't explain it lol, how would you guys explain it?

There is a procedure for converting a decimal number to binary in the following way: repeatedly divide the number by 2, writing the quotient and remainder at each step. Stop when 0 is obtained as a quotient. The sequence of remainders, written backwards (from the way they were obtained) is the binary representation. Here is an example, ﬁnding the binary representation of 49:

|49 24 1 |

|24 12 0 |

|12 6 0 |

|6 3 0 |

|3 1 1 |

|1 0 1 |

So 49 = 110001 (base 2) - Explain why this method works.

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Do you want to know why it works or the details of how it works? –  Thomas Jun 1 '12 at 4:04
either way is good :) –  hihello4 Jun 1 '12 at 4:16
Did my answer help? –  Thomas Jun 1 '12 at 4:49
Sorry, I can't find how this relates to cryptography, the topic of this site. –  Paŭlo Ebermann Jun 1 '12 at 7:26

## closed as off topic by fgrieu, Paŭlo Ebermann♦Jun 1 '12 at 7:25

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In base 10 we write for example $133$ when we mean $$133 = 1 * 10^2 + 3*10^1 + 3*10^0.$$ If we want to write $49$ in base $2$ then note first that: $$49 = 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0.$$ Because of this $49$ is $110001$.
So you want to find the digits in base you, you could start by dividing $49$ by $2$. You get a remainder of $1$ because $49$ is odd. Hence you know that the coefficient in from of $2^0$ is one. You subtract that reminder and divide by $2$. Dividing by $2$ decreases all the exponents by one, so $$24 = 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0.$$ Now you just continue like this dividing by $2$ to find the coefficient infront of $2^0$ (which then is the coefficient in front of $2^1$ in $49$).