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| | ==Table== | | ==Table== |
| − | # D1 = 1
| + | {| border=1 |
| − | # A1 = 2
| + | | 1 || D || S || I || E |
| − | # N1 = 3
| + | |- |
| − | # D2 = 4
| + | | 2 || P || Y || E || D |
| − | # I1 = 1
| + | |- |
| − | # E1 = 1
| + | | 3 || R || A || E || S |
| − | # S1 = 1
| + | |- |
| | + | | 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 |
| | + | |} |
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| − | ==Reasoning==
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| − | # 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.
| + | Another 70 pointer is 10JQK101010 = '''trouper''' |
| − | # 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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| − | # DANDIES=13. Speculate:
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| − | ## Assume that we have a 4-sequence.
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| − | ### Assume that the sequence is first. Then D1=1,A1=2,N1=3,D2=4,I1=1,E2=2,S1=1.
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| − | ### Apply this information to PRAISED=13.
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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. A contradiction (because of E).
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| − | #### Then D=D1. Then the last three must be 1 => E2=1. A contradiction.
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| − | ### 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.
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| − | ## Pop. We must have a 3 sequence.
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| − | ### Assume that b=2.
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| − | #### 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.
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| − | #### Assume that the sequence is first. Then D1=2,A1=3,N1=4, D2=1, I1=1, E2=1,S1=1.
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| − | ##### 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=D1, E2=1, S1=1.
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