21 bits.
8
Using n bits, you can count to 2n - 1. This is for unsigned integers. So 10 bits = 210 - 1 = 1023 14 bits = 214 - 1 = 16383 To count to 511 you need log2(511+1) = log2(512) = 9 bits. To count to 63 you need log2(63+1) = log2(64) = 6 bits.
What is the decimal equivalent of the largest binary integer that can be obtained with (a) 11 bits and (b) 25 bits?
0.0000001
65,535
Count them: 643(10)=1010000011(2)
9 bits
103
Assuming you start from 0, you need at least 4 bits. 15 in binary: 15 = 8 + 4 + 2 + 1 = 1111₂
8
17 bits would allow a value up to 131071.
how many bits are needed to represent decimal values ranging from 0 to 12,500?
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.
There are 16 decimal numbers that can be represented by 4-bits.
Using n bits, you can count to 2n - 1. This is for unsigned integers. So 10 bits = 210 - 1 = 1023 14 bits = 214 - 1 = 16383 To count to 511 you need log2(511+1) = log2(512) = 9 bits. To count to 63 you need log2(63+1) = log2(64) = 6 bits.
8 bits if unsigned, 9 bits if signed
41 in decimal is 0100 0001 in BCD (this is 8 bits not 6 bits)41 in decimal is 101001 in binary (this is 6 bits, but binary not BCD)There is no 6 bit BCD representation of the decimal number 41!