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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.)
    Some Thoughts re(3): Missing numbers | Comment 8 of 27 |
    (In reply to re(2): Missing numbers by Nick Reed)

    > Ooooo, I don't like that greatest-integer thing. That's cheating... ;-) I'm going to try to make do without needing that...

    Cheating? What kind of geek are you? :-P> Everything not explicitly laid out in the rules is fair game.

    > I like the 32, 33, 67 and 71 (and implied 31) though - missed those, even though I've used √9 in other places...

    D'oh! I missed 31? I really had a hard time waking up this morning...

    31: 96 / √9 - 1
    47: 9 x (6 - 1) + |_ln 9_|
    49: (9 - |_√6_|)^(√9 - 1)
    89: 91 - √√16
      Posted by friedlinguini on 2002-08-15 05:38:47

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