Difference between revisions of "Alice"

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==Table==
 
==Table==
# D1 = 1
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{| border=1
# A1 = 2
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| 1 || D || S || I || E
# N1 = 3
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|-
# D2 = 4
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| 2 || P || Y || E || D
# I1 = 1
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|-
# E1 = 1
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| 3 || R || A || E || S
# S1 = 1
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|-
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| 4 || A || N || B || T
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|-
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| 5 || D || I || R ||
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|-
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| 6 || R || O || W || N
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|-
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| 7 || A || M || E || G
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|-
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| 8 || A
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|-
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| 9 || S
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|-
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| 10 || P || R || T || E
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|-
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| J || I || R || L || E
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|-
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| Q || N || O || T || V
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|-
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| K || C || U || R || Y
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|}
  
==Reasoning==
 
  
# If a sequence of 3, then the smallest the base of the sequence, b, must be b+(b+1)+(b+2) +1+1+1+1 <= 13.  Thus, 3*b+7 <= 13.  b <= 2. If it's a 4-sequence, then we have b+(b+1)+(b+2)+(b+3)+1+1+1 <= 13.  4*b+9 <= 13.  b <= 1. 
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Another 70 pointer is 10JQK101010 = '''trouper'''
# DANDIES=13.  Speculate:
 
## Assume that we have a 4-sequence.
 
### Assume that the sequence is first.  Then D1=1,A1=2,N1=3,D2=4,I1=1,E1=1,S1=1.
 
### Apply this information to PRAISED=13. 
 
#### If D=D2, then the sequence must end there, and we must have 1+1+1+1+2+3+4.  Giving us  P1=1,R1=1,A2=1,S2=2,E2=3.
 
#### So S<=2, E<=3, A<=2, D<=4.  Apply to ANDRIES=21.  N+R+I >= 10.  So at least N,R, or I must be one we haven't seen yet.
 
# TROUBLE=67.  For a 4 sequence, b,b+1,b+2,b+3,x,x,x; b=10 => x = 9; b=9 impossible; b=8 => x = 10.  For a 3 sequence, b,b+1,b+2,x,x,x,x;  b=10 impossible; b = 9 impossible; b = 8 => x=10.  E1 = 9 or 10.
 

Latest revision as of 08:01, 23 August 2006

Puzzle_Boat

[1]

Table

1 D S I E
2 P Y E D
3 R A E S
4 A N B T
5 D I R
6 R O W N
7 A M E G
8 A
9 S
10 P R T E
J I R L E
Q N O T V
K C U R Y


Another 70 pointer is 10JQK101010 = trouper