# Difference between revisions of "Alice"

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#### 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. | #### 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. | + | # 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. |

## Revision as of 16:24, 22 August 2006

## Table

- D1 = 1
- A1 = 2
- N1 = 3
- D2 = 4
- I1 = 1
- E1 = 1
- S1 = 1

## 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.
- 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.

- Assume that we have a 4-sequence.
- 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.