# Alice

From CSWiki

## Table

- D = 1,2
- A = 3
- N = 4
- I = 1
- E = 1, (9 or 10)
- S = 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 b=1. Then 4*x+3*b+3 = 13. 4*x=7. Impossible.
- Assume that b=2.
- Assume that the sequence is second. Then I1=2,E2=3,S1=4,D1=1,A1=1,N1=1,D2=1. But when applied to PRAISED, E2=1 a contradiction.
- Assume that the sequence is first. Then D1=2,A1=3,N1=4, D2=1, I1=1, E2=1,S1=1.
- Apply to PRAISED. No matter what, the triplet must be last then. If D=D1, then E2=2 and we have a contradiction. Thus, D=D2, E2=1.

- Assume that we have a 4-sequence.
- Popping since the above is correct. PRAI = 10. If this is a sequence, then P=1, R=2,A=3, I=4. But we already have 4 1's, so this is not the case. Therefore, the sequence is PRA and I is part of a quartet. P+P+1+P+2 = 9 => 3*P = 6 => P=2, R=3,A=4.