Answer for John Gowland's puzzle:
A=44
B=72
C=837
D=175
E=64
F=49
a=48
b=777
c=28
d=314
e=46
f=59
--Marcis

4472
8378
4175
6449 
--David Wilson

Hi Ed,

A really nice puzzle from John and one which just requires a wee bit of thought to reduce the possibilities.  Realising that the terminal digit of P_Q_R is the same as the terminal digit of the cube sum helps enormously.  Tabulating these reveals for instance that if P ends in a 6 and Q in a 2 then R must end in 2 or 7.  Using this information along with the fact that B and E are multiples of 4 cracks the puzzle.  I only used my old Casio graphics calculator to do this.  No computer required!!

       The solution is

       4472
       8378
       4175
       6449

       Cheers,

Alastair Cuthbertson in sunny St. Andrews!!
------------------------------------------------------------

Linear Cubes:
4472
8378
4175
6449
Not the best form of presenting these:
(16,50,33)
(22,18,59)
(44,46,64)
(48,72,15)
(98,28,27)
(98,32,21)
Juha Saukkola
----------------------------------------------------------
4472
8378
4175
6449
Stephen Kloder
----------------------------------------------------------
4472
8378
4175
6449
Jeff Smith
--------------------------------------------------------
the only solutions to P^3 + Q^3 + R^3 = P_Q_R for two digit P,Q,R are
P   Q   R
16  50  33
22  18  59
34  10  67
44  46  64
48  72  15
98  28  27
98  32  21

Realizing Q shares its second digit with R's first digit from the 
positioning of e and E and the first row in the table, and that A/2 must 
yield another P from the table, P,Q,R must be 44,46,64
From here, i completed the puzzle through algebraic computation from the 
table and the use of the solution set listed above.
Crossnumber:

4472
8378
4175
6449

Table:

P,  Q,  R

44,46,64
22,18,59
48,72,15
16,50,33
98,28,27
98,32,21

Regards,
-David Perryman
----------------------------------------------------
Ed,

   I've recently discovered the wonders of Perl.  Pretty fun language.  My 
first program in it helped me solve John Gowland's "Linear Cubes" problem.  Here's 
the answer:

   4472
   8378
   4175
   6449


Clint Weaver
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