65,535
The smallest number is zero. The largest number is 63.
31 - it's binary equivalent is 11111
What is the decimal equivalent of the largest binary integer that can be obtained with (a) 11 bits and (b) 25 bits?
If using the compressed format, where a byte holds two decimal digits (because only 4 bits are needed to make nine), so two bytes would be four decimal digits, the largest which is 9999.
65,535
There are 16 decimal numbers that can be represented by 4-bits.
If the 8 bits represent a signed number, the range is usually -128 to +127. This is -27 to 27-1.
Largest 8 bit unsigned number is 11111111 binary which is the number 255 in decimal. In hexadecimal 255 is represented as FF In octal 255 is represented as 377. The related link below will help.
The smallest number is zero. The largest number is 63.
It is 127 = 2^7 - 1
674
To achieve the answer to what the decimal equivalent of the largest binary number with five places (or bits) is, work this equation: The formula is 2_ -1 where n is the number of bits. That will get you where you need to be.
The largest number is 11111111111111 which is 215 - 1. In decimal, that is 32767.
11b which is 1*2 + 1*1 = 3 would be for two bits. But a byte is 8 bits, so 2 bytes is 16 bits. The largest binary number is [2^16 - 1], which is 65535 (base ten)
255 is the largest decimal number. 1 byte (8 bits) can store 256 *values* (0-255).
The largest unsigned integer is 26 - 1 = 63, giving the range 0 to 63; The largest signed integer is 25 - 1 = 31, giving the range -32 to 31.