It depends on the number representation you use and how much resolution you need. If you use a floating point representation, the stored number can be extremely large but have poor resolution.
If you need to have a resolution of one count and include a zero, then the largest is 2^20 - 1
Bits Of Metal are useful for Tinkering and Trading.
Eight bits to the octet. The values are 0-255.
8 Bits is one Byte. Half of a byte (4 bits) is a nibble.
2 bytes=16 bits make a word
2 bytes 8 bits in a byte
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.
65,535 in decimal = 1111111111111111 in binary.
Bits administrator
It isn't. The largest unsigned number that can be stored in one byte (8 bits) is 2^8 - 1 = 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.
0xffffffffffffffff As an unsigned 64-bit integer, this represents the value 18,446,744,073,709,551,615. However, as a signed 64-bit integer, this only represents the value -1. The signed range is -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 making 0x7fffffffffffffff the largest possible positive value, and 0x8000000000000000 the smallest possible negative value.
What is the decimal equivalent of the largest binary integer that can be obtained with (a) 11 bits and (b) 25 bits?
The highest unsigned integer is 255; The highest signed integer is 127.
Assuming the number holds an unsigned number, the range of numbers for 4 bits is 0-7. 7 is a prime number.
From -524287 to 524288
The largest integer is 211 - 1 which is 2048 - 1 = 2047
There are 8 bits in a byte, so a two byte integer would be 16 bits. The largest 16 bit integer possible would be 11111111111111112, which is 65535 in base 10.