What is the decimal equivalent of the largest binary integer that can be
obtained with (a) 11 bits and (b) 25 bits?
The largest decimal is .25 or 25%.
The largest integer that is not the product of two or more different primes would be the largest prime number. Because there are an infinite number of prime numbers, there is no largest integer that is not the product of two or more different primes.
-3
The set of positive integers is {1, 2, 3, 4, 5, ...}. Because the values approach positive infinity there is no largest positive integer. If you pick any positive integer, you can always create a larger one by simply adding 1 to it. So there can be no largest positive integer.
The smallest 5 digit integer is -99999. The largest 5 digit integer is 99999. The sum is therefore 0.
The largest integer is 211 - 1 which is 2048 - 1 = 2047
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(2^32) - 1
It is 216 - 1 = 65535.
It is Bin(1111111111111111) = Dec(65535).
65,535 in decimal = 1111111111111111 in binary.
The one with the largest integer part (the part before the decimal point).
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.
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.
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 integer equal to -100 is -100. The largest integer less than -100 is -101.
the largest integer of distinct power is 28