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Zero to a Hundred in 1996 (Posted on 2002-08-15) Difficulty: 3 of 5
Using the digits in 1996 and any operations (but not mathematical constants), try to write equations that have the numbers from 0 to 100 as the answer.

For example with 1995:

  • 0 = 1*(9-9)*5
  • 2 = (19-9)/5
    etc.

    Provide as many as you can. Digits 1,9,9 and 6 do not have to appear in order. (But each digit has to be used - 1 and 6 once, 9 twice.)

    This is more of a game than a puzzle

  • See The Solution Submitted by levik    
    Rating: 3.4000 (10 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Solution re: Not quite full solution | Comment 7 of 27 |
    (In reply to Not quite full solution by Nick Reed)

    51 = ((9+1)*6)-9
    52 = (6*9)+1-sqrt(9)
    53 = (6*9)-(1^9)
    54 = ((6-1)*9)+9
    55 = ((1/9)+6)*9
    56 = (6*9)-1+sqrt(9)
    57 = ((9-1)*6)+9
    58 = (6*9)+1+sqrt(9)
    59 = 69-9-1
    60 = 61-(9/9)
    61 = 61+9-9
    62 = (6*9)+9-1
    63 = (9*1)+(6*9)
    64 = (6*9)+1+9
    65 = (9*9)-16
    66 = ((9-1)*9)-6
    67 = 69+1-sqrt(9)
    68 = 69-(1^9)
    69 = 69*(1^9)
    70 = 69+(1^9)
    71 = 69-1+sqrt(9)
    72 = ((6+1)*9)+9
    73 = (6*9)+19
    74 = (9*9)-1-6
    75 = (9*9*1)-6
    76 = (9*9)+1-6
    77 = 96-19
    78 = ((9-1)*9)+6
    79 = 69+9+1
    80 = (9*9)-(1^6)
    81 = (9*9)*(1^6)
    82 = (9*9)+(1^6)
    83 = 99-16
    84 = ((9+1)*9)-6
    85 = (9*9)+sqrt(16)
    86 = (9*9)+6-1
    87 = (9*9)+(6*1)
    88 = (9*9)+6+1
    89 = ((6!)/9)+(9*1)
    90 = 9*(9+(1^6))
    91 = ???
    92 = 99-1-6
    93 = 99-(6*1)
    94 = 99+1-6
    95 = 96-(1^9)
    96 = 96*(1^9)
    97 = 96+(1^9)
    98 = 99-(1^6)
    99 = 99*(1^6)
    100 = 99+(1^6)

    (Note: "^" is "to the power of" and "!" is factorial...)
      Posted by Nick Reed on 2002-08-15 05:20:26

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