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Posers and Puzzles

Posers and Puzzles

  1. 02 Jul '04 01:44
    See how many integers you can form, starting with 1 and going through 10, by using only the digit 4 four times - no more, no less - and the multiplication, division, addition and/or subtraction signs.

    Samples:

    1 = 44/44
    2 = 4/4 + 4/4

    By adding the square-root sign, 11 through 18 are readily obtainable. So start with 3 and see how far you can go.

    -Ray.
  2. 02 Jul '04 12:59
    3= (4+4+4)/4
    4= (4-4)/4 + 4
    5= ((4x4)+4)/4
    6= 4+(4+4)/4
    7= 4+4-4/4
    8= (4+4)x4/4
    9= 4+4+4/4
    10=(44-4)/4
    11=44/sqrt(4x4)
    12=4x4-sqrt(4x4)
    13=44/4+sqrt(4)
    14=4x4-(4/sqrt(4))
    15=44/4+4
    16=4+4+4+4
    17=4x4+4/4
    18=4x4+4/sqrt(4)
  3. 02 Jul '04 13:45
    Good work!

    -Ray.
  4. 02 Jul '04 16:48 / 2 edits
    19 = 4! - (4 + 4/4)
    20 = 4! - 4 + 4 - 4
    21 = 4! -4 + 4/4
    22 = 4! - sqrt(4) + 4 - 4
    23 = 4! - sqrt(4) + 4/4
    24 = 4*4 + 4 + 4
    25 = 4! + sqrt(4) - 4/4
    26 = 4! + sqrt(4) + 4 - 4
    27 = 4! + 4 - 4/4
    28 = 4! +4 - 4 + 4
    29 = 4! + 4 + 4/4
    30 = 4! + sqrt(4) + sqrt(4) + sqrt(4)

    and not sure this is optimal, but...
    31 = 4! + 4/(.4 recurring) - sqrt(4)

    EDIT
    better is:
    31 = 4!+ (4! + 4)/4
  5. 02 Jul '04 18:56
    I think that after passing 30, all mathematical symbols are acceptable. Therefore, your recurrence dot is acceptable.

    Keep going!

    -Ray.
  6. 04 Jul '04 08:44
    Originally posted by rgoudie
    I think that after passing 30, all mathematical symbols are acceptable.
    Are also standard functions acceptable? Is it valid if I define ONE(x) as a function such that ONE(4) = 1, that could be defined with standard functions as:

    floor(cubic root(x)) or
    ceil(log(x)) or
    sign(x)

    This way,

    32 = (4+4) x sqrt(4x4)
    33 = 4! + 4 + 4 + ONE(4)
    34 = 4! + 4 + 4 + sqrt(4)
    35 = 4! + (4! / sqrt (4)) - ONE(4)
    36 = 4^4 - 4! - 4

    -- Germano
  7. 04 Jul '04 14:56
    Is Log acceptable?....
  8. 04 Jul '04 15:42
    If square root is acceptable, then so is log.
    I'm not too sure about that one function.

    -Ray.
  9. 05 Jul '04 10:29
    Originally posted by rgoudie
    I'm not too sure about that [b]one function.

    -Ray.
    [/b]
    it's only a way to say one of the other (valid, as you confirmed) three... just substitute it with ceil (log(4)), this is the sense. Anyways, 36 was wrong, it's quite easily

    36 = (4+4)x4 + 4

    while 37... I really can't figure it! 8-)
  10. 05 Jul '04 21:46
    without ONE, CEIL etc...

    33 = 44 - 4/(.4 recurring) - sqrt(4)
    34 = 4! + 4 + 4 + sqrt(4)
    35 = 44 - 4/(.4 recurring)
    36 = 44 - 4 - 4
    37 = 4! + 4 + 4/(.4 recurring)
    38 = 44 - 4 - sqrt(4)
    39 = 4!*sqrt(4) - 4/(.4 recurring)
    40 = 44 - sqrt(4) - sqrt(4)

  11. 06 Jul '04 23:11
    If log is acceptable then there is a proof that any integer can be expressed with four fours...
  12. 07 Jul '04 00:20
    With a specific base or will any do?

    -Ray.
  13. 08 Jul '04 17:41 / 1 edit
    Any old kind of log will do. By the way, is the % operation allowed? (which would mean divide by 100, e.g. 20% = 0.2)