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Letter Cubes 4 (Posted on 2003-10-07) Difficulty: 3 of 5
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 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 today's cubes.

Can you recover the 6 letters on each die?
BECK
COZY
DEWY
FLAW
GAPE
JOVE
LAIR
MASK
PLOT
RASH
SAFE
SULK
TOWN
VOTE

See The Solution Submitted by DJ    
Rating: 4.4545 (11 votes)

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Solution Explanation | Comment 2 of 5 |
Ok, here goes =)

First note that Q and X are the two letters not used.
From SAFE and FLAW you know that L and W must be on the same cubes as E and S. So we could either have E&L on a cube with S&W on another cube, or E&W with S&L. But from DEWY we know that E and W cannot be on the same cube, so it must be E&L with S&W.

By going through all the words that have an E or an L in them, we can eliminate many letter that we know can't be on the same cube with them. Luckily we eliminate so many that we see only H,M,N and Z are left, so they must be on the same cube as E and L. Good, one cube done =)

Now let's look at JOVE and VOTE. These share E,O and V, which means T&J must be on the same cube. From TOWN we see that T and W can't be on the same cube. So, just to capture where we are, here is a table of what we know:

Cube 1: E H L M N Z
Cube 2: S W
Cube 3: T J
Cube 4: ?

From TOWN we know that O must be on Cube 4.
From PLOT we know that P must be on Cube 2.
From VOTE we know that V must be on Cube 2.

From DEWY we know that D and Y must be Cubes 3 and 4. But from COZY we see that Y cannot be on Cube 4. So Y must be on Cube 3 and D must be on Cube 4. And, again from COZY, we see that C must be on Cube 2.

The letters that are unplaced so far are ABFGIKRU. From all the words that SWPVC are in, we can eliminate all of those except for I. Now Cube 2 is done and we have:

Cube 1: E H L M N Z
Cube 2: S W P V C I
Cube 3: T J Y
Cube 4: O D

A cannot be on the same cube as FGKR. Since there are only two cubes with open spaces, this means that FGKR must all be on the same cube. The only one with 4 open spots left is Cube 4.

So the Solution is:

Cube 1: E H L M N Z
Cube 2: S W P V C I
Cube 3: T J Y A B U
Cube 4: O D F G R K

Later!


  Posted by nikki on 2003-10-08 17:34:07
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