What is the remainder when 3^1+3^2+3^3…3^2009 is divided by 8.

Question

What is the remainder when 3^1+3^2+3^3…3^2009 is divided by 8.

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Aaliyah 3 days 2021-10-13T03:40:45+00:00 1 Answer 0

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    2021-10-13T03:42:39+00:00

    Answer:

    4

    Step-by-step explanation:

    To find:

    The remainder, when 3^0 + 3^1 + 3^2 + … +3^2009 is divided by 8

    Approach and Working:

    Rem (3^0/8) = 1

    Rem (3^1/8) = 3

    Rem (3^2/8) = 1

    Rem (3^3/8) = 3

    Rem (3^4/8) = 1

    Rem (3^5/8) = 3 and so on…

    We can observe that,

    For even powers of 3, the remainder is 1

    And for odd powers of 3, the remainder is 3.

    Therefore, we can rewrite the given series, in terms of its remainders, as

    = 1 + 3 + 1 + 3 + 1 + 3 + … + 3 = 1005 pairs of (1 + 3) = 1005 x 4 = 4020

    Rem (4020/8) = 4

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