Is the virtual address 0x1FE0C0 below the "line" or above the line?

I'm studying z/OS and I was asked if the virtual address 0x1FE0C0 is:

A. above the line.

B. above the bar and below the line.

C. above the line and below the bar.

D. below the bar.

I choose D as my answer, but my teacher wrote that it's wrong.

My logic for choosing D was that to represent 0x1FE0C0 in binary, you will need 21 bits. The line marks the area that can be addressed by 24 bits, so that's why I believe the address is below the line. From that I could eliminate answer A and C. Answer B was eliminated because the bar is above the line, so if the address is above the bar it wi开发者_如何学编程ll also be above the line. So we are left with answer D which seems correct to me, because if the address is below the line it is also below the bar (31 bit address space).

So what is the correct answer?

This is driving me crazy and I would appreciate your help.

1FE0C0 is 6 hex characters so it is a 24bit address, based on this page it is below the line which is also below the bar so D is the correct answer.

Converting 0x1FE0C0 to decimal, 2089152, is less than 2GB (2 147 483 648 bytes), so the only possible answer is D. Below the bar.

B. is impossible, and the others state above the line, more than 2GB which it is not.