If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.
The largest decimal number with 25 digits is 9,999,999,999,999,999,999,999,999.
The smallest decimal number in the form 2n-1 which is greater than or equal to that is 19,342,813,113,834,066,795,298,815. That corresponds to 284-1.
So, the minimum number of binary bits required to represent the decimal number 25 nines in a row is 84. This is 84 ones in a row. If you want to support negative as well as positive numbers, you will need 85.
Since the largest integer in most compilers is 64 bits, this will require a special library supporting 128 bits, or an arbitrary length decimal library, if you want to manipulate such large numbers in a computer and still retain the precision of an integer.
All I know is that when a number is negative, you convert the decimal into binary and if it is negative you put 1111 before the binary digits.
Computers do not understand decimal notation. All information (both instructions and data) must be converted to a binary representation before the machine can understand it. We use the symbols 0 and 1 (binary notation) but the machine has a variety of physical representations it can use to encode binary data, including transistors, flux transitions, on/off switches and so on.
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.
0xc = 1100 Hexadecimal digits use exactly 4 binary digits (bits). The 0x0 to 0xf of hexadecimal map to 0000 to 1111 of binary. Thinking of the hexadecimal digits as decimal numbers, ie 0x0 to 0x9 are 0 to 9 and 0xa to 0xf are 10 to 15, helps with the conversion to binary: 0xc is 12 decimal which is 8 + 4 → 1100 in [4 bit] binary.
A binary number system has two states '0' '1' for a long word in bits it can be as follows 101010101010101010101011 intimidating RIGHT? it can be represented in groups of 3 bits in octal 10/010/101/010/101/010/101/011= 22525253 digital or in group of 4 bits as 10/1010/1010/1010/1010/1010 = 2AAAAA 111 =7 octal 1111=f F in hexadecimal numbers 1000 =8 1010 =10 or A
10 digits.
The decimal representation of numbers is shorter. Binary number require approx 3.3 times as many digits.
7 digits
The largest decimal number is binary 11111, which is decimal 31.
56 in binary is 111000. Unlike the decimal number system where we use the digits.
In binary, the number 1000 is represented as 1111101000. This is calculated by converting the decimal number 1000 into binary, which involves dividing the number by 2 and recording the remainders. The binary representation uses only the digits 0 and 1, where each digit represents a power of 2.
It is a decimal representation where, after a finite number of digits, all subsequent digits are 0 [or of them all 9].
It takes 7 digits.
Assuming you start from 0, you need at least 4 bits. 15 in binary: 15 = 8 + 4 + 2 + 1 = 1111₂
All I know is that when a number is negative, you convert the decimal into binary and if it is negative you put 1111 before the binary digits.
Octet
Advantage of binary over decimal: information can be recorded and stored in any dichotomous variable: magnetised or not magnetised (most electronic media), pit or no pit (optoelectronic media CDs/DVDs). For decimal it would be necessary to store as 10 different levels of magnetisation or depths of pits. Not so easy to make such a system error-free. Advantage of decimal over binary: fewer "digits" required. Every ten binary digits (1024 values) can be replaced by just a shade more than three decimal digits (1000 values). So the number of digits to be stored is less than a third.