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?
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 further 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
Edited on April 10, 2024, 11:59 pm
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