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Math Help - Finding units digit

  1. #1
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    Finding units digit

    Hi all,

    I got one question here.

    Find the units digit in the sum 1^{3}-2^{3}+3^{3}-4^{3}+...+39^{3}

    Thanks!
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  2. #2
    Super Member Bacterius's Avatar
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    Can't you just evaluate it and read the last digit ? Or it is specifically stated you must do it the analytic way ?
    By the way it is not a sum ... well not completely
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  3. #3
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    I am supposed to find the units digit without evaluating the sum.

    There must be hidden pattern, i can't figure it out yet.
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  4. #4
    Super Member Bacterius's Avatar
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    Difference of two cubes, maybe ? You could look at your expression like this :

    (1^3 - 2^3) + (3^3 - 4^3) + ... + (37^3 - 38^3) + 39^3

    Does that make sense ?

    Or maybe by considering modular arithmetic modulo 10, which would conservate only the last digit ...
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  5. #5
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    I tried grouping the terms, still couldn't see any pattern.

    I used calculator to find the sum, which is 30800. So the units digit is zero.
    But how to do this without evaluating the sum?
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  6. #6
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    I'd split it into 1^3 + 3^3 + 5^3 +... 39^3
    and 2^3 + 4^3 +6^3 +... 38^3 and then consider the difference.
    Write down the units digits for cubes of odd numbers and then for even numbers. You'll see a pattern!
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  7. #7
    Super Member Bacterius's Avatar
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    All good then ... is it possible to user-delete newly irrelevant posts of ours ?
    Last edited by Bacterius; November 6th 2009 at 08:54 PM. Reason: Became irrelevant
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  8. #8
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    GREAT! there is a pattern. ThankS
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  9. #9
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    [QUOTE=Bacterius;399101] (1^3 - 2^3) = (1 - 2)(1^2 + 2 \times 1 \times 2 + 2^2)
    = (1 - 2)(1^2 + 4 + 2^2) = (1 - 2)(1 + 4 + 4) = -1 \times 9 = -9

    But 1^{3}-2^{3}=-7, not -9.
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  10. #10
    Super Member Bacterius's Avatar
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    Yes, I made a small (but fatal) error, sorry.
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