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N 4's - Redux

  1. Jan 9, 2006 #1
    So, you've got the "standard" four 4's problem-- Using exactly four 4's (and no other numbers), and any mathematical functions, come up with combinations to get the integers 0-100 (or what-have-you). But a bunch of functions are potentially "cheats", like sqrt(4) which implies the number 2, since you're taking the 2nd root of 4, while you could instead take the (4+4)/4 root of 4, and that would be a totally legal way of getting the same result (you've just used up your full allotment of 4's).

    Ex: using four 4's:
    0 = 4-4+4-4
    1 = (4/4)*(4/4)
    2 = 4-(4+4)/4
    3 = 4+(4^(4-4))
    4 = 4+(4-4)/4

    Hence, the question is: how many 4's are necessary to complete all the integers from 0 to 100? Could you do it with exactly five 4's? Would it take as many as ten 4's? What's the absolute minimum?

    "Cheats" include anything with an implied other number, such as:

    4! - implying 4*3*2*1
    sqrt(4) - implying the "2nd" root of 4
    .4 - implying 4/10
    .4 - implying 4/9
    44 - implying 4*10 + 4
    4C4 - implied factorials buried in there
    log 4 - implied 10
    ln 4 - implied e.

    Other things to stay away from: AND's and OR's (as in bitwise operators), implied bases (such as saying "4*4=10 in base 16", etc).

    Known functions that would be ok:
    +, -, *, /, roots (with specified base), log (with specified base), exponents, modulus.

    The jury's still out on trig functions, since you're potentially implying 360 degrees or 2 pi radians.

    I "solved" this using a computer program to bash away at the possible combinations, although it didn't use trig functions, and there may be other functions out there, so perhaps there's a lower number than the one I came up with. To any attempting this by hand, I'd advise starting at 100 and working your way backwards, rather than starting at 1 and working up...

    What did I get? I found that I needed 4+(4-4)/4+4*4^(4-4) fours in order to get all numbers between 0 and 100.

  2. jcsd
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