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Math Help - find the last two digits

  1. #1
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    find the last two digits

    Define the squences a_1, a_2, ... and b_1, b_2, ... by a_1 = b_1 = 7 and a_n+1 = (a_n)^7, b_n+1 = 7^(b_n)

    Find the last digit of a_2009 and of b_2009

    What about the last two digits? or more?
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by ahm0605 View Post
    Define the squences a_1, a_2, ... and b_1, b_2, ... by a_1 = b_1 = 7 and a_n+1 = (a_n)^7, b_n+1 = 7^(b_n)

    Find the last digit of a_2009 and of b_2009

    What about the last two digits? or more?
    Solution for finding out the last digits:
    its clear that a_2=7^7, \, a_3=7^{7^2}, \,a_{2009}=7^{7^{2008}}

    since 7^2 \equiv -1(mod 10),
    we have 7^6 \equiv -1(mod 10)
    so 7^7 \equiv -7(mod 10)
    so (7^7)^7 \equiv 7^{7^2} \equiv 7(mod 10
    so finally 7^{7^{2008}} \equiv 7(mod 10)
    so a_{2009} \equiv 7(mod 10).
    The last digit of a_{2009} is 7.

    for b_{2009} observe that since 7 \equiv 1 (mod 6), b_i \equiv 1(mod 6) for each i.
    now b_{2009} = 7^{b_{2008}}.
    write b_{2008}=6k+1 for some integer k.
    so b_{2009} = 7^{b_{2008}} \equiv 7^{6k+1} \equiv (7^6)^k \cdot 7 (mod 10)
    compute that 7^6 \equiv -1 (mod 10),
    so that b_{2009} \equiv (-1)^k \cdot 7(mod 10).
    convince yourself that b_i \equiv -1(mod4) for each i and hence prove that k is odd.
    this will lead to the result that The last digit of b_{2009} is 3.
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