Trinomial Cube #2
Design and Copyright : Ernest Treloar (2013).
(a, b, c) = (2, 3+α, 4-α).
☞ “Trinomial Cube” Puzzles
A |
a³ + bc² |
B |
ab² + abc |
C |
abc + b²c |
D |
a²b + abc |
E |
a²b + abc |
F |
ab² + b²c |
G |
ab² + b³ |
H |
a²b + bc² |
I |
abc + ac² |
J |
a²c + b²c |
K |
a²c + a²c |
L |
ac² |
|
Pieces | 16 |
L-M are congruent. |
Selection | random |
Goal | 9×9×9 |
Holes | 0 |
Solutions | 2 |
Trinomial Cube #2a
Arranged : 石野恵一郎 (ISHINO Keiichiro) (2013).
(a, b, c) = (2, 3+α, 4-α).
A |
abc + bc² |
B |
ac² + b²c |
C |
abc + b²c |
D |
a²b + bc² |
E |
a²b + bc² |
F |
abc + ac² |
G |
a²c + b²c |
H |
a²b + ac² |
I |
ab² + abc |
J |
ab² + abc |
K |
a²c + abc |
L |
a²c + ab² |
|
Pieces | 15 |
Selection | random |
Goal | 9×9×9 |
Holes | 0 |
Solutions | 1 |
Trinomial Cube #2b
Arranged : 石野恵一郎 (ISHINO Keiichiro) (2013).
(a, b, c) = (2, 3+α, 4-α).
A |
bc² + bc² |
B |
ab² + c³ |
C |
abc + bc² |
D |
ac² + b²c |
E |
b³ + ac² |
F |
a²c + b²c |
G |
b²c + a²b |
H |
abc + ab² |
I |
a³ + ac² |
J |
a²b + abc |
K |
a²c + a²c |
L |
a²b + ab² |
|
Pieces | 15 |
M-O are congruent. |
Selection | random |
Goal | 9×9×9 |
Holes | 0 |
Solutions | 1 |
Jan 25, 2013 by
k16@chiba.email.ne.jp