Assuming it is an unsigned int (i.e. no negatives) it would be 11111111111 which is 2047.
Another way to think about it is 11bits can represent 2048 different values, and since it starts at 0 that would be 2048 - 1 which is 2047.
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In 11 bits, the largest number that can be represented is 2^11 - 1, which equals 2047. This is because with 11 bits, you have 2^11 (2048) possible combinations, but since we start counting from 0, the largest number represented is one less, which is 2047.
Oh, dude, the largest number you can represent with 11 bits is 2047. It's like having a party with 2047 guests, but your house can only fit that many - no plus ones allowed! So yeah, 2047 is the big boss in the 11-bit number world.
The largest number you can represent with 11 bits is 2047. You start at 0 and go all the way up to 2^11 - 1, which equals 2047. So, there you have it, the big kahuna of 11-bit numbers.
Well, honey, with 6 bits, you can represent numbers from 0 to 63. So, technically speaking, the largest number you can make with 6 bits is 63. Don't go expecting any bigger miracles with just 6 bits, darling.
Assuming you interpret the bits as an unsigned number, that would be 1111111111 in binary, or 1023 (210 - 1) in decimal.
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)
It is Bin(1111111111111111) = Dec(65535).
Assuming the number holds an unsigned number, the range of numbers for 4 bits is 0-7. 7 is a prime number.