Let the price of a book be ABCD.EF.

Then 100 books would cost ABCDEF.00

So we subtract the price of 1 book from ABCDEF.00 and get G38254.1H,
the price of 99 books.

So we have this subtraction

     A B C D E F.0 0
       - A B C D.E F 
     ---------------
     G 3 8 2 5 4.1 H

So G=A, 10-F=H and 9-E=1, so E=8

     A B C D 8 F.0 0
       - A B C D.8 F 
     ---------------
     A 3 8 2 5 4.1 H

F cannot be 0, since that would make H be 10.

Depending on whether we will have to borrow from the 8 above the C,
C must be either 2 or 3, so those two cases are:

    A B 2 D 8 F.0 0             A B 3 D 8 F.0 0
      - A B 2 D.8 F               - A B 3 D.8 F 
    ---------------             ---------------
    A 3 8 2 5 4.1 H             A 3 8 2 5 4.1 H

Either way, we know we'll have to borrow from the B to make the 2 a 12.
So 12-A=8 so A is 4.  And since we borrow from B, B=4

So these two cases become

    4 4 2 D 8 F.0 0             4 4 3 D 8 F.0 0
      - 4 4 2 D.8 F               - 4 4 3 D.8 F 
    ---------------             ---------------
    4 3 8 2 5 4.1 H             4 3 8 2 5 4.1 H

Now we see that D can only be 6:

    4 4 2 6 8 F.0 0             4 4 3 6 8 F.0 0
      - 4 4 2 6.8 F               - 4 4 3 6.8 F 
    ---------------             ---------------
    4 3 8 2 5 4.1 H             4 3 8 2 5 4.1 H

That tells us that we did have to borrow from the 8,
so that eliminates the second case and makes F having to
have been scratched with a zero over it, which makes F=1

    4 4 2 6 8 1.0 0            
      - 4 4 2 6.8 1                
    ---------------             
    4 3 8 2 5 4.1 H

So we finally see that H=9.

    4 4 2 6 8 1.0 0            
      - 4 4 2 6.8 1                
    ---------------             
    4 3 8 2 5 4.1 9

So the missing digits are 4 and 9.

[Can you believe that a textbook would cost $4426.81?]  

Edwin