In the game of Letter Cubes, a different letter of the alphabet is on each face of each of the 4 cubes so that 24 of the 26 letters of the alphabet, including J, occur. Words are formed by rearranging and turning the cubes so that the top letters spell a common 4-letter word. The 14 words below have been made using the cubes.

CAVE
CLEF
DUPE
FARE
FLUB
GREW
HAZY
KITH
LOIN
POEM
RASP
SMUG
TIRE
VARY

Can you recover the 6 letters on each?

(puzzle originally from www.allstarpuzzles.com)

Just wanted to bring up something someone mentioned in the queue:

Often, puzzles like this are solved when a certain face of a cube can be oriented different ways to represent more than one character (see the <a href=http://perplexus.info/show.php?pid=196>Calendar Cubes series, and others), like using the same face for a 6 and a 9.

Although it is not necessary to solve this problem, as there are exactly 23 letters in the words, and a J is stipulated in the introduction, as is the fact that each face bears a different letter.
We can still try to see if it is possible.

Start with the solution that several people have found:

A D L M T W


B E K N S Y


C H J O R U


F G I P V Z

Two letters that can be represented by the same face are N and Z.
In the solution, they appear on different cubes. As this solution is unique, there are certain direct or indirect reasons that those two letters cannot be on the face of the same cube.
For example, if you replaced the word 'HAZY' with the 'HANY' and tried to solve the problem that way, thinking it will yield a solution in which the same face is used, you will find that it is impossible.

Similarly, I and H (which could be rotations of each other) appear on different cubes. Moreover, they are in the same word (KITH), and certainly could not be represented by the same face.

Lastly, M and W are 180° rotations of each other. These letters do appear on the same cube, so if you removed the W, for instance, each word with a W in it could still be read with an upside-down M.
All the faces must still bear a different letter, so the now-extra face must contain a Q or an X, which were omitted from the original problem.

Thus, an alternate solution might be:

A D L M T _


B E K N S Y


C H O R U _


F G I P V Z

in which one of the blanks contains a J, and the other blank is filled with either a Q or an X (four possible arrangements this way).

There are no other letters that could be represented with the same face.

Therefore, there are perhaps five ways to solve this problem.

Posted by DJ on 2003-06-18 18:33:37