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!!
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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
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4472
8378
4175
6449
Stephen Kloder
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4472
8378
4175
6449
Jeff Smith
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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
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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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