Logic Problem Solution:
Letter Cubes Nr. 2
According to the introduction, each of the 24 faces of the Letter
Cubes has a different letter on it. Given the word CAVE, we
arbitrarily assign C to cube 1, A to cube 2, V to cube 3, and E
to cube 4. From the word FARE, F can't be on cube 2 with A or cube
4 with E; and from the word CLEF, F isn't on cube 1 with C. F is on
cube 3 with V. The R in FARE then must be on cube 1 with C, and the
L in CLEF must be on cube 2 with A. From VARY, the Y is on cube 4.
From FLUB, the letter U can't be on cube 2 with AL or on 3 with VF.
From DUPE, the U also can't be on cube 4 with EY, so the U is on cube
1 with CR. Then in FLUB, the B comes from cube 4. The P in DUPE
isn't on cube 1 with U or cube 4 with E, and from RASP the P isn't on
cube 2 with A. The P is on one face of cube 3. In DUPE, then, the D
must be on cube 2; and in RASP, the S must be on cube 4. From POEM,
the M isn't on cube 3 with P or cube 4 with E; nor from the word SMUG
is the M on cube 1 with U. The M is on cube 2. The O in POEM is on
cube 1, and the G in SMUG is on cube 3. The W in GREW must be on a
face of cube 2. The I in LOIN isn't on cube 1 (O) or cube 2 (L),
while the I in TIRE isn't on cube 4 (E). I is a letter on cube 3.
Therefore, the N in LOIN is on cube 4; and the T in TIRE is the sixth
letter on cube 2. From KITH, the H isn't on cube 2 with T or cube 3
with I. From HAZY, H also isn't on cube 4 with Y. H is on cube 1.
In KITH, the K completes cube 4; and in HAZY, the Z is the last letter
on cube 3. By elimination, the letter J is on cube 1. In sum, the
faces on the four cubes are
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