It differs slightly depending on what platform your computer uses or what language you are using.
for the Java programming language, which is platform independent, it is 64-bits.
17 bits would allow a value up to 131071.
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.
Count them: 643(10)=1010000011(2)
It requires 6 bits to address 64 words. It does not matter what the word size is.
In ASCII code, each letter, number or punctuation mark takes one byte, or 8 bits. That gives you 256 discrete combinations. Two letters take 2 bytes, or 16 bits.
9 bits
You need 20 bits of address bus to address 1 Mb of memory.
15 bits
4
The minimum number of address lines required to address 4k of memory is 12.To reach this number, remember that each line has two possibilities and keep doubling as you count off. So one line can be used for two possibilities. Two lines represent four possibilities. Three represent eight. When you get to ten, you have 1024 possibilities. So double to 2048 at 11 and again to 4096 at 12. Or for the shortcut, if you take two to the 12th power, you get 4096.
103
8
In an 8085 system, the memory word size required is 8 bits. This means that each memory location can store 8 bits or one byte of data. The 8085 processor accesses memory locations using these 8-bit memory addresses to read or write data during program execution. The memory word size of 8 bits allows the 8085 system to handle data in small, manageable chunks efficiently.
17 bits would allow a value up to 131071.
how many bits are needed to represent decimal values ranging from 0 to 12,500?
integer data type consumes memory of 4 bytes or 32 bits
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.