Difference between revisions of "Alice"

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### Apply this information to PRAISED=13.   
 
### 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.
 
#### 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.

Revision as of 16:19, 22 August 2006

Puzzle_Boat

[1]

Table

  1. D1 = 1
  2. A1 = 2
  3. N1 = 3
  4. D2 = 4
  5. I1 = 1
  6. E1 = 1
  7. S1 = 1

Reasoning

  1. 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.
  2. DANDIES=13. Speculate:
    1. Assume that we have a 4-sequence.
      1. Assume that the sequence is first. Then D1=1,A1=2,N1=3,D2=4,I1=1,E1=1,S1=1.
      2. Apply this information to PRAISED=13.
        1. 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.
        2. 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.
  3. 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.