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Math Help - Induction Proof again :(

  1. #1
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    Induction Proof again :(



    Hi all,
    Once again I'm pretty stumped at this one.
    3^3 = 27
    3^4 = 81
    3^5 =243
    3^6 = 729
    3^7 = 2187

    Now these are the same practically. An even second to last digit and a seven as the last digit. So the pattern will loop. Just proving it seems a right pain! Any help at all would be appreciated.
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  2. #2
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    Quote Originally Posted by Jimbobobo View Post


    Hi all,
    Once again I'm pretty stumped at this one.
    3^3 = 27
    3^4 = 81
    3^5 =243
    3^6 = 729
    3^7 = 2187

    Now these are the same practically. An even second to last digit and a seven as the last digit. So the pattern will loop. Just proving it seems a right pain! Any help at all would be appreciated.
    I can't see your linked image, and I doubt anyone else can either.

    RonL
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  3. #3
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    Prove by induction that in the decimal form 3^n, the second from end digit is even.
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  4. #4
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    Quote Originally Posted by Jimbobobo View Post
    Prove by induction that in the decimal form 3^n, the second from end digit is even.
    Base case n=3 then 3^3=27 and so it is true that the second least significant digit is even.

    Suppose it true for some k, that is the second least significant digit is even.
    Now 3^k is odd, and not divisible by 5, so the least significant digit is one of:

    1,\ 3,\ 7,\ 9.

    Now the carry when these are multiplied by 3 is even (either 0 or 2), and
    3 times the second least significant digit of 3^k is even. Even plus an even is
    even so the second least significant digit of 3^{3+1} is even, which
    completes the induction step.

    So we have proven that for all n>2 the second least significant digit is even.

    RonL
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  5. #5
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    Hello, Jimbobobo!

    I have a slight variation of RonL's solution.


    Prove by induction that in the decimal form of 3^n, the tens-digit is even.
    We note that 3^n ends in: 1, 3, 7, or 9.

    And 3^{n+1} ends in: 03, 09, 21, or 27.
    . . Hence, there is a "carry" of either 0 or 2.


    Verify S(1)\!:\;\;3^1 \:=\:03 . . . true!


    Assume S(k)\!:\;\;3^k has an even tens-digit.

    . . This means that: . 3^k is of the form: . 10(even) + \{1,3,7,9\}

    . . Hence: . 3^k \:=\:10(even) + \{1,3,7,9\}


    Multiply by 3: . 3^{k+1} \;=\;3\left[10(even) + \{1,3,7,9\}\right]

    . . \text{and we have: }\;3^{k+1} \;=\;\underbrace{10\!\cdot\!3(even)}_{\text{even}} + \underbrace{3\!\cdot\!\{1,3,7,9\}}_{\text{carry: 0 or 2}}

    \text{Hence: }\;3^{k+1} \;=\;10\cdot\underbrace{[3(even) + \{0,2\}]}_{\text{even}} + \{1,3,7,9\}


    Therefore, 3^{k+1} has an even tens-digit.
    . . The inductive proof is complete.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I had this wild idea of using binary, but I got nowhere.

    Consider: . 3 \:=\:11_2

    I thought that . \left(11_2\right)^n would lead to a neat solution,
    . . but I didn't find one.

    Anyone care to give it a try?

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