If we have a device that can be in one of two states, we can store a single digit of a base 2, or **binary**, number on that device. If we have many of these devices, we can store larger numbers, one binary digit per device.

In the picture below, we have a machine with buckets on a conveyor belt. To store a number, the machine fills certain buckets with water. We can consider this machine as representing a base 2 number, where each bucket represents a digit. Filled buckets represent a 1 and empty buckets represent a 0.

What number is stored on the machine? Answer in base 10.

Whether it’s stored in RAM, in a bank of vacuum tubes, or on a conveyor belt of water buckets, digital data in its most basic form is just a series of individual binary digits, also known as **bits**.
One bit represents one digit in base 2, which is either 0 or 1.

In this course, we’ll adopt the convention used in some programming languages to prefix binary numbers with `0b`

. What is the base 10 number 173 in binary? Prefix the binary with `0b`

.

How many bits are required to represent 128 in binary?

**byte**. As we saw in the last problem, the largest number that can be stored in a single byte is 255. Clearly, we require three digits to represent 255 in base 10. What is the smallest base that can represent 255 using only 2 digits?

Base 16 is also referred to as **hexadecimal**, or **hex** for short. In this course, we’ll adopt the convention used in some programming languages to prefix hexadecimal numbers with `0x`

. For example, since 32 in decimal is $2 \times 16^1 + 0 \times 16^0$, it will be represented as `0x20`

.

Since one hexadecimal digit needs to represent up to the value 15 in decimal, we will use letters for some of the digits:

$\begin{array} { | c | c| } \hline \text{dec} & \text{hex} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \vdots & \vdots \\ 9 & 9 \\ 10 & \text{A} \\ 11 & \text{B} \\ 12 & \text{C} \\ 13 & \text{D} \\ 14 & \text{E} \\ 15 & \text{F} \\ \hline \end{array}$

$\begin{array} { | c c r c | } \hline \text{dec} & & & & \text{hex} \\ \hline 0 &= & 0 \times 16^0 & = & 0\text{x}0 \\ 1 &= & 1 \times 16^0 & = & 0\text{x}1 \\ 9 &= & 9 \times 16^0 & = & 0\text{x}9 \\ 10 &= & 10 \times 16^0 & = & 0\text{xA} \\ 11 &= & 11 \times 16^0 & = & 0\text{xB} \\ 12 &= & 12 \times 16^0 & = & 0\text{xC} \\ 15 &= & 15 \times 16^0 & = & 0\text{xF} \\ 16 &= & 1\times 16^1 + 0 \times 16^0 & = & 0\text{x}10 \\ 17 &= & 1\times 16^1 + 1 \times 16^0 & = & 0\text{x}11 \\ 25 &= & 1\times 16^1 + 9 \times 16^0 & = & 0\text{x}19 \\ 26 &= & 1\times 16^1 + 10 \times 16^0 & = & 0\text{x}1\text{A} \\ 31 &= & 1\times 16^1 + 15 \times 16^0 & = & 0\text{x}1\text{F} \\ 32 &= & 2\times 16^1 + 0 \times 16^0 & = & 0\text{x}20 \\ 159 &= & 9\times 16^1 + 15 \times 16^0 & = & 0\text{x}9\text{F} \\ 160 &= & 10\times 16^1 + 0 \times 16^0 & = & 0\text{xA}0 & \\ 255 &= & 15\times 16^1 + 15 \times 16^0 & = & 0\text{xFF} \\ \hline \end{array}$

The decimal number 47872 is represented as `0xBB00`

in hexadecimal. What is the decimal number 48000 in hexadecimal?

As we saw in the previous question, 255 in hexadecimal is `0xFF`

. This happens to be the maximum possible value that can be represented by 2 digits in hexadecimal. Recall that 255 in binary is `0b11111111`

, which is also the maximum possible value that can be represented by 8 digits in binary.

This property makes hexadecimal invaluable in computer science. It allows us to concatenate bytes in either base and get the same result. For example:

$\begin{array} { | c | c| c | } \hline \text{Binary} & \text{Hexadecimal} & \text{Decimal} \\ \hline \text{0b11101111} & \text{0xEF} & 239 \\ \text{0b00111001} & \text{0x39} & 057 \\ \hline \end{array}$

$\text{0b1110111100111001} = 61241$,

$\text{0xEF39} = 61241$.

Same result. But if we concatenate the two decimal values, we get something completely different: 239057.

We normally deal with integers of fixed sizes in multiples of bytes. Recall that a byte is 8 bits. It’s common to see 8-bit, 16-bit, 32-bit, and 64-bit integers in contemporary code. The largest value of a 32-bit integer is $2^{32} - 1$ which is 4294967295 in decimal. If we store a much smaller number, the remaining unused bits must be set to 0, since sizes are fixed.

What is the decimal value 763 stored in a 32-bit integer, padding the unused bits with `0`

s? Answer in hexadecimal.