0 o 1
A 128-bit register can store 2 128th (over 3.40 × 10 38th) different values. The range of integer values that can be stored in 128 bits depends on the integer representation used.
24, or 16 (0 through 15) One binary digit (bit) can have 21 values (0 or 1). Two bits can have 22 values. Three bits can have 23 values. A five-bit number can have 25 values... and so on...
1. A single bit can represent two different values, 0 and 1. Then simply take the largest of those two possible values, 1, and that's your answer.
Since there are 8 bits in between, and they can assume any of the two values (0 or 1), that results in a total of 28 different combinations.
0 and 1
A 128-bit register can store 2 128th (over 3.40 × 10 38th) different values. The range of integer values that can be stored in 128 bits depends on the integer representation used.
Two: '0' or '1'
For signed 32 bit values: 2^31-1 = 0x7FFFFFFF = 2,147,483,647 For unsigned 32 bit values: 2^32-5 = 0xFFFFFFFB = 4,294,967,291
24, or 16 (0 through 15) One binary digit (bit) can have 21 values (0 or 1). Two bits can have 22 values. Three bits can have 23 values. A five-bit number can have 25 values... and so on...
Texas is too big, but you can fit in a Ohio, Maryland, and part of West Virginia. But seriously though...I am assuming that you mean "how many unique combinations of 1's and 0's can be stored in a 16 bit register." The answer is given as 2 (the number of different possibilities per digit) raised to the power of 16. The answer is 65536, made up of 0 through 65535. An 8 bit register can represent 256 different values, 0 through 255.
1. A single bit can represent two different values, 0 and 1. Then simply take the largest of those two possible values, 1, and that's your answer.
8
one bit in two output states true or false
It can have 0 to 1 It can have 0 to 1
An N-bit integer holds 2N different values.For an unsigned integer, the range of values is 0..2N-1 thus.For a signed integer using 2s complement, the range is -2N-1..+2N-1-1.Therefore, the largest positive number that can be stored using 8 bits is 255.
A simple mathematical formula to calculate the maximum address space of a register is 2^N, where N is the bitrate. 2^8 = 256. So 0-255, and 255 is the largest value. A 16-bit register (2^16) is 65536, so 0-65535
No, not any more than the weight of an abacus increases as the number it represents increases. In fact, you could argue that the disk is already full when you originally buy it. Each bit is set... albeit to values to mean that nothing is contained. Whenever you 'write' to disk you are replacing the previous bit values to new ones.