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Math Help - 1983 AIME question

  1. #1
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    Question 1983 AIME question

    Let a_n = 6^n +8^n.
    Determine the remainder on dividing a_83 by 49.

    Thank you very much.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    Let a_n = 6^n +8^n.
    Determine the remainder on dividing a_83 by 49.

    Thank you very much.
    a_83=6^83+8^83

    .......=(7-1)^83 + (7+1)^83

    Now expand the powers using the binomial theorem:

    a_83=[(-1)^83 + 83 (-1)^82 7 + K_1 49] + [(1)^83 + 83 (1)^82 7 + K_2 49]

    for some integers K_1 and K_2. So for some integer K_3

    a_83=2 83 7 + K_3 49

    .......= 1162 + K_3 49 = (23 49 + 35) + K_3 49

    So the required remainder is 35.

    RonL

    (What does AIME stand for?)
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  3. #3
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    Hi Captainblack,

    Could you please solve this question by using the modulo 49? Thank you very much.

    for example:

    6^83 + 8^83 is congruent to x ( mod 49)

    I don't know how to carry on the next working step after this....
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    Hi Captainblack,

    Could you please solve this question by using the modulo 49? Thank you very much.

    for example:

    6^83 + 8^83 is congruent to x ( mod 49)

    I don't know how to carry on the next working step after this....

    a_83=6^83+8^83

    .......=(7-1)^83 + (7+1)^83

    Now expand the powers using the binomial theorem:

    a_83=[(-1)^83 + 83 (-1)^82 7 + K_1 49] + [(1)^83 + 83 (1)^82 7 + K_2 49]

    So

    a_83=2 83 7 mod 49

    .......= 1162 mod 49 = 35 mod 49

    So the required remainder is 35.

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    Hi Captainblack,

    Could you please solve this question by using the modulo 49? Thank you very much.

    for example:

    6^83 + 8^83 is congruent to x ( mod 49)

    I don't know how to carry on the next working step after this....
    Do exaxtly what I did but modulo 49, so:

    6^83 + 8^83 = (7-1)^83 + (7+1)^83 ( mod 49)

    and the rest works almost exactly as before

    RonL

    RonL
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  6. #6
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    a_83=2 83 7 mod 49

    .......= 1162 mod 49


    =======================
    Hi Captainblack,
    I don't get the above step = > 2837 mod 49 equal to 1162 mod 49.

    In fact, I have completely forgotten the binomial thoerem. So feel confused with that part of your work.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    a_83=2 83 7 mod 49

    .......= 1162 mod 49


    =======================
    Hi Captainblack,
    I don't get the above step = > 2837 mod 49 equal to 1162 mod 49.

    In fact, I have completely forgotten the binomial thoerem. So feel confused with that part of your work.
    Its a product: 2*83*7=1162

    and any numbers which are equal are congrunent to the each other modulo anything

    RonL

    RonL
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  8. #8
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    ic Thank you very much.
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