To store the hexadecimal number FF, we need to convert it to binary first. FF in hexadecimal is equivalent to 1111 1111 in binary, which requires 8 bits to represent. Each hexadecimal digit corresponds to 4 bits in binary, so two hexadecimal digits (FF) require 8 bits to store.
Eight.
9 bits
FA2B = 2 bytes = 4 nibbles = 16 bits 1 1 1 1 . 1 0 1 0 . . . 0 0 1 0 . 1 0 1 1
8
103
17 bits would allow a value up to 131071.
4 bits equal to half byte.8 bits is one byte.when converting hexadecimal digits to binary, each hexadecimal digits will take 4 binary digits, which means 4 bits.Because one binary digit means one bit having two values [true/false] or [on/off] like that.. [0/1]we can represent one hexadecimal digit as 4 bits like..for [7] as hexadecimal, we can say [0111] in bits.
Convert each group of 4 bits into one hexadecimal digit. 1010 is "A" in hexadecimal, so this particular number is "AA".
Count them: 643(10)=1010000011(2)
Binary: 1 bit Octal: 3 bits Hexadecimal: 4 bits Decimal: somewhere between 3 and 4 bits. In theory, about 3.32 bits.
Yes, a byte is 8 bits, and a one hexadecimal digit takes up four bits, so two hexadecimal digits can be stored in a byte. The largest hexadecimal digit is F (which is 15 in base ten.) In base two, this converts to 1111, which takes up four bits, which is why it only takes four bits to store a hexadecimal digit. With 8 bits, two hexadecimal digits can be stored (FF would be 11111111, which is 8 bits), and 8 bits make up a byte. Generally, 4 bits are always used to store a hexadecimal digit, using leading zeros where necessary. For example, the hexadecimal digit 5 would be stored as 0101, and the hexadecimal digits 5A would be stored as 01011010.