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Math Help - tens digit

  1. #1
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    tens digit

    Determine the tens digit of 3^1999. Could you find a second solution?
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  2. #2
    o_O
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    Finding the remainder once divided by 100 will give you the tens and ones digits. For example, 539 \equiv 39 \ (\text{mod 100}).

    So the problem becomes: 3^{1999} \equiv a \ (\text{mod 100})

    By Fermat's theorem: 3^{\varphi (100)} \equiv 1 \ (\text{mod } 100)

    But: \varphi (100) = \varphi (2^2 \cdot 5^2) = \varphi (2^2) \varphi (5^2) = 2^1(2-1) \cdot 5^1(5-1)=40

    So: 3^{40} \equiv 1 \ (\text{mod } 100)

    Raise both sides by 50 (the closest we'll get the power to 1999):
    \begin{aligned} \left(3^{40}\right)^{50} & \equiv 1^8 \ (\text{mod } 100) \\ 3^{2000} & \equiv 1 \ (\text{mod } 100) \\ & \ \ \vdots \end{aligned}
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  3. #3
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    thank you so much
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