### 1. Divide the number repeatedly by 2:

#### Keep track of each remainder.

#### We stop when we get a quotient that is equal to zero.

- division = quotient +
**remainder**; - 23 042 019 ÷ 2 = 11 521 009 +
**1**; - 11 521 009 ÷ 2 = 5 760 504 +
**1**; - 5 760 504 ÷ 2 = 2 880 252 +
**0**; - 2 880 252 ÷ 2 = 1 440 126 +
**0**; - 1 440 126 ÷ 2 = 720 063 +
**0**; - 720 063 ÷ 2 = 360 031 +
**1**; - 360 031 ÷ 2 = 180 015 +
**1**; - 180 015 ÷ 2 = 90 007 +
**1**; - 90 007 ÷ 2 = 45 003 +
**1**; - 45 003 ÷ 2 = 22 501 +
**1**; - 22 501 ÷ 2 = 11 250 +
**1**; - 11 250 ÷ 2 = 5 625 +
**0**; - 5 625 ÷ 2 = 2 812 +
**1**; - 2 812 ÷ 2 = 1 406 +
**0**; - 1 406 ÷ 2 = 703 +
**0**; - 703 ÷ 2 = 351 +
**1**; - 351 ÷ 2 = 175 +
**1**; - 175 ÷ 2 = 87 +
**1**; - 87 ÷ 2 = 43 +
**1**; - 43 ÷ 2 = 21 +
**1**; - 21 ÷ 2 = 10 +
**1**; - 10 ÷ 2 = 5 +
**0**; - 5 ÷ 2 = 2 +
**1**; - 2 ÷ 2 = 1 +
**0**; - 1 ÷ 2 = 0 +
**1**;

### 2. Construct the base 2 representation of the positive number:

#### Take all the remainders starting from the bottom of the list constructed above.

#### 23 042 019_{(10)} = 1 0101 1111 1001 0111 1110 0011_{(2)}

### 3. Determine the signed binary number bit length:

#### The base 2 number's actual length, in bits: 25.

#### A signed binary's bit length must be equal to a power of 2, as of:

#### 2^{1} = 2; 2^{2} = 4; 2^{3} = 8; 2^{4} = 16; 2^{5} = 32; 2^{6} = 64; ...

#### The first bit (the leftmost) is reserved for the sign:

#### 0 = positive integer number, 1 = negative integer number

#### The least number that is:

#### 1) a power of 2

#### 2) and is larger than the actual length, 25,

#### 3) so that the first bit (leftmost) could be zero

(we deal with a positive number at this moment)

#### === is: 32.

### 4. Get the positive binary computer representation on 32 bits (4 Bytes):

#### If needed, add extra 0s in front (to the left) of the base 2 number, up to the required length, 32:

## Number 23 042 019_{(10)}, a signed integer number (with sign),

converted from decimal system (from base 10)

and written as a signed binary (in base 2):

## 23 042 019_{(10)} = 0000 0001 0101 1111 1001 0111 1110 0011

Spaces were used to group digits: for binary, by 4, for decimal, by 3.