Final answer:
In big endian representation, memory location 100, 101, 102, and 103 will store the binary representation of the integer 2047.
Step-by-step explanation:
In big endian representation, the most significant byte is stored at the lowest memory address. Since x is a 32-bit integer, it will require 4 bytes of memory. The binary representation of 2047 in 32 bits is 00000000 00000000 00000111 11111111.
Therefore, in memory location 100, the binary representation of the first byte (LSB) will be 11111111. In location 101, the second byte will be stored as 00000111. In location 102, the third byte will be stored as 00000000. And in location 103, the fourth byte (MSB) will be stored as 00000000.
If a processor uses big endian representation and x is a 32-bit integer stored at the memory address 100, and if x holds the value 2047, then the memory contents will be as follows: At address 100, we will have the most significant byte (MSB), and at address 103, we will have the least significant byte (LSB). Since 2047 in binary is represented as 0000 0000 0000 0000 0000 0111 1111 1111, in big endian format, the MSB is stored first. Accordingly, memory address 100 will contain 00000000 (0x00 in hexadecimal), address 101 will have 00000000 (0x00), address 102 will have 00000111 (0x07), and address 103 will have 11111111 (0xFF).