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Perplexus Cryptarithm Puzzle (Posted on 2007-03-31) Difficulty: 3 of 5
The cryptarithm FLOOBLE + PUZZLE = PERPLEX has no solutions. But if the word 'PUZZLE' is included twice, then there is a solution.
How many more times can the word 'PUZZLE' appear in the addition and the cryptarithm still have a solution?

See The Solution Submitted by Brian Smith    
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Some Thoughts answer | Comment 2 of 4 |

Let us denote the relationship as:

FLOOBLE + M*(PUZZLE) = PERPLEX, where M denotes the number of times the word PUZZLE is repeated in the alphametic relationship.

Then for 1< M < 10, it can be verified that, the following solutions are possible:

Case I: M= 2

B=0 E=8 F=6 L=2 O=5 P=7 R=3 U=9 X=4 Z=1, so that:

6255028 + 791128 + 791128 = 7837284


Case II: M= 3

B=7 E=2 F=4 L=3 O=9 P=6 R=1 U=0 X=8 Z=5, so that:

4399732 + 605532 + 605532 + 605532 = 6216328


Case III: M= 4

B=3 E=9 F=4 L=1 O=7 P=6 R=0 U=8 X=5 Z=2, so that:

4177319 + 682219 + 682219 + 682219 + 682219 = 6906195


Case IV: M = 7

B=2 E=6 F=1 L=4 O=9 P=3 R=5 U=0 X=8 Z=7; so that:

1499246 + 307746 + 307746 + 307746 + 307746 + 307746
+ 307746 + 307746 = 3653468

It can also be verified that no solution is possible in the
range 11<= M <=30


 

Edited on April 1, 2007, 5:52 am

Edited on April 2, 2007, 11:07 am
  Posted by K Sengupta on 2007-04-01 05:49:53

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