Difference between revisions of "Alice"
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==Reasoning== | ==Reasoning== | ||
| + | # 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. | ||
# 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. | # 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: | # DANDIES=13. Speculate: | ||
## Assume that we have a 4-sequence. | ## Assume that we have a 4-sequence. | ||
| − | ### Assume that the sequence is first. Then D1=1,A1=2,N1=3,D2=4,I1=1, | + | ### Assume that the sequence is first. Then D1=1,A1=2,N1=3,D2=4,I1=1,E2=2,S1=1. |
### 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. A contradiction (because of E). |
| − | #### | + | #### Then D=D1. Then the last three must be 1 => E2=1. A contradiction. |
| − | # | + | ### Pop. Assume that the sequence is second. Then D1=1,A1=1,N1=1,D2=1,I1=2,E2=3,S1=4. Again, applying to praised yields the same contradiction that E2=1. |
| + | ## Pop. We must have a 3 sequence. | ||
Revision as of 16:29, 22 August 2006
Table
- D1 = 1
- A1 = 2
- N1 = 3
- D2 = 4
- I1 = 1
- E1 = 1
- S1 = 1
Reasoning
- 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.
- 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,E2=2,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. A contradiction (because of E).
- Then D=D1. Then the last three must be 1 => E2=1. A contradiction.
- Pop. Assume that the sequence is second. Then D1=1,A1=1,N1=1,D2=1,I1=2,E2=3,S1=4. Again, applying to praised yields the same contradiction that E2=1.
- Pop. We must have a 3 sequence.
- Assume that we have a 4-sequence.
