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Going Arithmetic With Alphametic II (Posted on 2009-11-21) Difficulty: 3 of 5
Substitute each of the capital letters in bold by a different digit from 0 to 9 such that the sum of digits of each of POIRE, OPERA, PORTE, TAPIS, ASTRE, PATRE, LOTUS, LISTE, LOUPE and, OUTRE (in this order) constitutes ten consecutive terms of an arithmetic sequence in ascending order. It is known that none of P, O, T, A and L can be zero.

See The Solution Submitted by K Sengupta    
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Solution Analytic solution Comment 3 of 3 |

OPERA and PORTE are consecutive and share O,P,E,R, so A+1=T

ASTRE and PATRE are consecutive and share A,T,R,E, so S+1=P

LISTE and LOUPE are consecutive, share L and E, so I+S+T+1=O+U+P
with S+1=P then I+T=O+U

LOTUS and LISTE are consecutive and share L,T,E, so O+U+1=I+E
with I+T=O+U, then T+1=E

POIRE and PORTE differ by 2 and share P,O,R,E, so I+2=T

PORTE and OUTRE differ by 7 and share O,R,T,E, so P+7=U

PATRE and LISTE differ by 2 and share T and E, so P+A+R+2=L+I+S
with S+1=P and I+1=A, then R+4=L

Now, S and P are consecutive, P+7=U
I,A,T,E are all consecutibe
and R+4=L
For all those to fit together in ten digits, the order must be one of:
1: O,S,P,R,I,A,T,E,L,U
2: S,P,R,I,A,T,E,L,U,O
3: R,S,P,O,L,I,A,T,E,U

The difference of POIRE and TAPIS should be 3:
1:POIRE=16, TAPIS=18, diff=2: NO
2:POIRE=21, TAPIS=13, diff=-8: NO
3:POIRE=18, TAPIS=21, diff=2: YES


  Posted by Brian Smith on 2009-11-21 20:32:17
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