Logic Puzzle # 149 Logic Problems Help

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

• A-D-L-M-T-W
• B-E-K-N-S-Y
• C-H-J-O-R-U
• F-G-I-P-V-Z