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    04 Nov '14 18:13
    Are calculators allowed?
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    04 Nov '14 21:19
    The post that was quoted here has been removed
    If you really want to rely on your brain, then no scratch work allowed.

    If you write stuff down, then you are also relying on your fingers, paper and pen/pencil.
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    The post that was quoted here has been removed
    I didn't know this was a 'credit' assignment. I thought you were arguing that using a pencil and paper to do calculations was a greater mental challenge than simply using a calculator.

    I then suggested that having to visualize the problem and remember the answers would be an even greater mental challenge and therefore more fun!
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    05 Nov '14 01:46
    The post that was quoted here has been removed
    Why do I not solve the problem?

    I do not find multiplication by hand to be an entertaining endeavor.
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  10. Standard memberDeepThought
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    05 Nov '14 04:41
    Originally posted by Eladar
    Why do I not solve the problem?

    I do not find multiplication by hand to be an entertaining endeavor.
    Lot's of multiplication isn't needed, and isn't that hard anyway as it's just multiplication by 22.

    I've got some ideas about this which I'll post since no one seems to have made any progress. First note that:

    (1978^A)%1000 = (1978%1000 * (1978^A-1))%1000 = (978 * (1978^(A-1))%1000,

    as (ab)%c = (a(b%c))%c, iterating this we get:

    (1978^A)%1000 = (978^A)%1000.

    For N some integer:

    (N*978)%1000 = (N*(1000 - 22))%1000 = (1000*N - 22*N)%1000 = 1000 - (22*N)%1000

    so we never need to multiply by anything bigger than 22.

    If (978^A)%1000 = (978^B)%1000, then (22^A)%1000 = (22^B)%1000 which means we only need to look at powers of 22.

    (22^A)%1000 = (((2^A)%1000) * (11^A)%1000)%1000

    This is a method for generating random numbers on a computer. The sequence repeats after at least 400 steps as 11^A has least significant digit 1 and 2^A has least significant digit one of {2, 4, 6, 8}, so there are only 4 possibilities for the least significant digit which gives B = 1, A = 401 and C = 402 at most. I need to think about it to get any further.
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    05 Nov '14 19:503 edits
    The post that was quoted here has been removed
    Yes, I'm much too stupid to know how to multiply.

    Gotcha.

    And much like any puzzle, once you know the proper approach the result is much easier. I'm too normal to enjoy thinking about useless problems for long periods of time. If being normal makes be dumb, so be it.

    Btw, Isn't there an entire forum on this site devoted to puzzles?
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    06 Nov '14 00:11
    Originally posted by DeepThought


    (1978^A)%1000 = (978^A)%1000.

    It doesn't take high levels of math to understand that the last three digits of a multiplication problem has nothing to do with place values greater than the hundreds place.
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    06 Nov '14 00:121 edit
    The post that was quoted here has been removed
    Yeah, I can see where you'd have problems making the distinction between thinking and thinking about useless things.

    Btw, the belief that race has no effect on intelligence is just as racists as the opposite belief. Any assumption based on race is by definition racist.

    In other words, everyone is a racist. You simply believe your form of racism doesn't stink.
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