An AVR Atmega32 has 0x1FFF as the address of its last byte of its ROM (Program Memory). What is the size of the ROM in bytes?

8191 bytes = 7.99 Kb

Explanation:

As the last address of the ROM is 0x 1FFF. This address present in  Hexa-decimal format. We convert this hexa decimal number to decimal number to find the total number of memory locations.  The conversion is given below:

In decimal

F= 15

so,

1FFF = 1 x 16³ + 15 x 16² + 15 x 16¹ + 15 x 16⁰

        =   1 x 4096 + 15 x 256 + 15 x 16 + 15

        =  4096 + 3840 + 240 + 15

1FFF =  8191

8191 is the total number of memory locations. In question it is mentioned that each location contains 1 byte so, total memory is 8191 bytes.

If we want to convert this memory size into kilo bytes then we divide the answer with 1024 as each kilo in computer system is equal to 1024.

1K byte = 1024

so,

8191 bytes/1024 = 7.99 kB.

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nope i prefer two because it is simpler, i don't like all the powers you get

explanation:

but that's my opinion