A255-gb disk has 65,536 cylinders with 255 sectors per track and 512 bytes per sector. how many platters and heads does this disk have

15 platters 30 heads First, let's calculate how much space a single surface of a single platter can store. 65536 * 255 * 512 = 8,556,380,160 bytes Now one historical artifact of describing storage is the definition of kilo, mega, and giga. For ram, those values refer to powers of 2 with 1 kilobyte = 2^10, 1 megabyte = 2^20, and 1 gigabyte being 2^30. But for disk space, the kilo, mega, giga, and other prefixes refer to the power of 10 notation. So a 255-GB disk would have 255x10^9 bytes of storage and not 255x2^30 or 273,804,165,120 bytes as some might assume. The reason for this difference is mostly unknown, but most suspect that early marketing efforts used the base 10 notation since it made the sizes sound larger and they were relying on selling to people who were not aware of the difference between the base 2 sizes and the base 10 sizes. I personally saw an advertisement in the early 1980's where an advertiser attempted this same trick by advertising an early 8-bit computer as having "65K" of memory. This is technically true since the amount of memory was 65536 bytes. But all programmers would consider that computer as having 64K of memory. Thankfully, that practice of using base 10 didn't catch on for RAM, but unfortunately, it did catch on for disk storage. In any case, let's do the division to see how many surfaces are required to have at least 255 gigabytes using the base 10 notation. 255000000000 / 8556380160 = 29.80232239 Since due to marketing regulations you have to have at least as much as you're advertising, we can round the value of 29.8 up to 30. So the disk has 30 surfaces upon which to store data. Each platter has 2 surfaces, so there's 30/2 = 15 platters. And each surface has to have at least 1 head, so for this problem, there are 30 heads. To illustrate the difference. If the 255-gb disk were specified in base-2 notation, it would have not 255,000,000,000 bytes of storage, but would have 273,804,165,120 bytes of storage instead. And in order to handle that much storage, it would require 32 surfaces or 16 platters and 32 heads. If you look at the technical specifications for manufacturers of disk drives, you're certain to see a footnote somewhere in the specifications indicating that for disk storage, 1 MB = 1,000,000 bytes; 1 GB = 1,000,000,000 bytes, and 1 TB = 1,000,000,000,000 bytes. And that for buffers, 1 MB = 1,048,576 bytes and 1 GB = 1,073,741,824 bytes.

The solution to the system of equations (-2,2)