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Math Help - [SOLVED] Finding the last two digits of a number raised to a large exponent

  1. #1
    Member diddledabble's Avatar
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    [SOLVED] Finding the last two digits of a number raised to a large exponent

    Okay so if you are given 7^{999} how would you figure out the last two digits of the munber using mod.
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  2. #2
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    Quote Originally Posted by diddledabble View Post
    Okay so if you are given 7^{999} how would you figure out the last two digits of the munber using mod.
    Hi

    You need to find 7^{999} mod 100
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  3. #3
    Member diddledabble's Avatar
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    But how do you work through it. I did it for 7^9999 and got 143 but I am not confident on my steps. I don't want anyone to solve my homework. That is why I altered my question to 7^999. I just need a good detailed step by step explanation. Thanks
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  4. #4
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    Use Euler's theorem: (a,m) = 1 \ \Rightarrow a^{\phi(m)} \equiv 1 \ (\text{mod } m) where m \geq 1

    So we know: \phi (100) = \phi \left(2^2 \cdot 5^2\right) = \cdots = 40

    Thus: 7^{40} \equiv 1 \ (\text{mod } 100)

    Now use the property: a \equiv b \ (\text{mod } m) \ \Rightarrow \ a^n \equiv b^n \ (\text{mod } m)
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    Hello, diddledabble!

    Find the last two digits of: . 7^{999}
    Consider the two-digit endings of consecutive powers of 7 . . .

    . . \begin{array}{c|c}<br />
\text{Power} & \text{Ends in} \\ \hline<br />
7^1 & 07 \\ 7^2 & 49 \\ 7^3 & 43 \\ 7^4 & 01 \\ 7^5 & 07 \\ 7^6 & 49 \\ \vdots & \vdots \end{array}

    The endings form a four-step cycle: . \{07, 49, 43, 01\}


    Since 999 \equiv 3 \text{(mod 4)},\;7^{999} has the third ending: . 43

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  6. #6
    Member diddledabble's Avatar
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    Soroban,
    Your answer was sooo easy to understand. Just what I was looking for.
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