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 ==Table==   ==Table== 
−  # D = 1,2
 +  { border=1 
−  # A = 3,4
 +   1  D  S  I  E 
−  # N = 4
 +   
−  # I = 1
 +   2  P  Y  E  D 
−  # E = 1, (9 or 10)
 +   
−  # S = 1
 +   3  R  A  E  S 
−  # P = 2
 +   
−  # R = 3
 +   4  A  N  B  T 
−  # Y = 1
 +   
 +   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 
 +  } 
   
−  ==Reasoning (Cont'd)==
 
−  # Apply to SPRAYED=16. E=D=1=>Y=1. Problem.
 
   
−  ==Reasoning==  +  Another 70 pointer is 10JQK101010 = '''trouper''' 
−   
−  # 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 4sequence, 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 4sequence.
 
−  ### 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.
 
−  # 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.
 