2^12=4096
To find out how many different values can represented by a certain number of bits, we can use the following formula 2n-1 and that is because the first number is always a zero.Based on that 6 bits = 26- 1= 64-1=637 bits= 27-1= 1278 bits= 28-1=25510 bits= 210-1=1023# of bits1=12=33=74=155=316=637=1278=2559=51110=1023
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...
1000
65,536
2
There are 256 possible values (or characters) in 8 bits.
n2 -1
24 = 16
64 or 123
256 (162)
Binary bits are necessary to represent 748 different numbers in the sense that binary bits are represented in digital wave form. Binary bits also have an exponent of one.
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.