no because it can be divided by 2
(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768
Integers are the whole numbers between -32768 and +32767. 62500 is single precision number.
7 is a prime number!
109 is a prime number.
167 is a prime number.
32,768 has only one prime factor: 2
(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768
No and Yes... No because as such the two numbers are different and convention will define what the bit pattern should represent. However, with 16 bits, the UNSIGNED bit pattern for 32768 is the same as the SIGNED pattern for -32768 BUT the two numbers are being represented in different ways (ie in different number systems). If the binary representation is a SIGNED number then the top bit will be set if it is negative and with 16 bits the possible range of numbers is -32768 to +32767 and so it is IMPOSSIBLE to represent +32768. With an UNSIGNED number of 16 bits the top bit has no such special meaning and the range possible is 0 to +65535, so it is IMPOSSIBLE to represent -32768. ie it is IMPOSSIBLE with 16 bits to represent -32768 and +32768 in the same number system as they have the same bit pattern, BUT in different number systems the bit pattern can be used to represent the two numbers as the different number systems cannot represent BOTH -32678 and +32768.
32,768 has a prime factorisation of 215- or 25*25* 25, or (23)5.
a KB is 1024 bytes 32KB = 32*1024 = 32,768 bytes
32 cubed equals to 32768.
Sorry, I really don't understand what values does the compiler give you...-32768..32767 is the value range of a 16-bit-long signed binary number (short in C)
1,000,000,000,000,000,
It is: 32
8x8x8x8x8 equals 32,768. This can be calculated by multiplying 8 by itself five times, which is equivalent to 8 raised to the power of 5. In mathematical terms, this can be represented as 8^5 = 32,768.
Integers are the whole numbers between -32768 and +32767. 62500 is single precision number.
A signed 16 bit number can represent the decimal numbers -32768 to 32767.