A 4-byte (32-bit) storage means the integer 165 is stored using 4 bytes (32 bits). Let's break it down in binary and hexadecimal.

Step 1: Convert 165 to Binary

The decimal number 165 can be converted to binary:

$165_{10} = 10100101_2$

Since we are using 4 bytes (32 bits), we need to pad it with leading zeros:

$00000000 \ 00000000 \ 00000000 \ 10100101_2$

Step 2: Convert to Hexadecimal

To convert binary to hexadecimal, we group the 32-bit number into 4-bit sections:

$0000 \ 0000 \ 0000 \ 0000 \ 0000 \ 0000 \ 1010 \ 0101$

Each 4-bit group is converted to hexadecimal:

$0 \ 0 \ 0 \ 0 \ 0 \ 0 \ A \ 5$

Thus, in hexadecimal, the 4-byte storage of 165 is:

$00 00 00 A5_{16}$

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Final Answer:

Binary Representation (4 Bytes):

$00000000 \ 00000000 \ 00000000 \ 10100101_2$

Hexadecimal Representation (4 Bytes):

$00 \ 00 \ 00 \ A5_{16}$

Do you want me to explain how to convert decimal to binary and hex in more detail?

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Related Questions

1. How is the negative number -165 stored in 4 bytes using two’s complement?
2. How does endianness (big-endian vs. little-endian) affect the storage of 165?
3. How would 165 be stored in a system using 2 bytes instead of 4?
4. What is the binary and hexadecimal representation of 165 in an 8-bit system?
5. How does floating-point representation differ from integer storage?

Tip: When converting decimal to binary, always divide by 2 and record the remainders from bottom to top!