Suppose the page size in a computing environment is 1KB. Give the page
number and the offset for the following:

a.	899 (a decimal number) 

b.	23456 (a decimal number) 

c.	0x3F244 (a hexadecimal number)

d.	0x0017C (a hexadecimal number)

Dear siball,


If the page size is 1 KB, then there are 1024 bytes from the start of one
page to the next. Thus, all the addresses from 0 to 1023 are in the first
page; all addresses from 1024 to 2047 are in the second page; and so on.

If we are given a physical address, we can determine its page number
by dividing it by 1024. The remainder of this division then gives us
the offset.


a.

If we divide 899 by 1024, we obtain 0 with a remainder of 899.

  899 / 1024  =  0 rem 899

Therefore, the page number is 0 and the offset is 899.
  

b.

  23456 / 1024  =  22 rem 928
  
The page number is 22 and the offset is 928.


c.

1024 in hexadecimal is 0x400.

  0x3F244 / 0x00400  =  252 rem 580
  
The page number is 252 and the offset is 580.


d.

  0x0017C / 0x00400  =  0 rem 380
  
The page number is 0 and the offset is 380.


Regards,

leapinglizard

Fast, clear, well explained answer. Just what I was looking for. Thank
you very much.