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Math Help - divide by 3?

  1. #1
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    Cool divide by 3?

    hi folks,

    I hope this is the right place to post this! I have no idea how to proceeed with the following question:

    Show that for all positive integral values of n

    [HTML]7 <sup>n</sup> + 2 <sup>2n+1</sup>[/HTML]

    is divisible by 3.

    I tried a few terms as follows:

    n = 1. 7 ^ 1 + 2 ^ 3 = 7 + 8 = 15
    n = 2. 7
    ^ 2 + 2 ^ 5 = 49 + 32 = 81
    n = 3. 7
    ^ 3 + 2 ^ 7 = 343 + 128 = 471

    and these are all divisible by 3 but how do I handle the general case?

    I thought of expanding the expression i.e.

    2.2 ^ n. 2 ^ n - (1 - 8) ^ n

    and using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?

    regards and thanks
    Simon

    p.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!
    Last edited by s_ingram; June 20th 2009 at 04:50 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by s_ingram View Post

    I thought of expanding the expression i.e.

    2.2 ^ n. 2 ^ n - (1 - 8) ^ n

    and using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?
    I suggest writing 7^n+2^{2n+1}=(2^2+3)^n+2^{2n+1} and then using the binomial theorem to expand (2^2+3)^n.

    Quote Originally Posted by s_ingram View Post
    p.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!
    To enter math. equations you can use LaTeX. (see http://www.mathhelpforum.com/math-help/latex-help/)
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  3. #3
    Senior Member Twig's Avatar
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    (2^{2}+3)^{n} = 2^{2n} + \binom{n}{1}2^{2(n-1)}3^{1} + ... + \binom{n}{k}2^{2(n-k)}3^{k} + ... + 3^{n} =

     = 2^{2n} + 3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}]

    So we can write:

     3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(1+2) =  3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(3)

    Now factor out a 3 from both terms.

     3[(\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}) + 2^{2n} ]

    And because we have a number times 3, the result is divisible by 3.

    Note: If there aint a careless mistake somewhere in this mess, I am surprised.
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  4. #4
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    thanks to Twig and flyingsquirrel. You guys make it look so easy!
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  5. #5
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    If you do it by induction here is the last step.
    \begin{array}{rcl}<br />
   {7^{n + 1}  + 2^{2n + 3} } &  =  & {7^{n + 1}  + 7 \cdot 2^{2n + 1}  - 7 \cdot 2^{2n + 1}  + 2^{2n + 3} }  \\<br />
   {} & {} & {7\left( {7^n  + 2^{2n + 1} } \right) - 2^{2n + 1} \left( {7 - 2^2 } \right)}  \\<br />
   {} & {} & {7\left( {7^n  + 2^{2n + 1} } \right) - 2^{2n + 1} \left( 3 \right)}  \\ \end{array}
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  6. #6
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    Hello, s_ingram!

    How about an inductive proof?


    Show that for any positive integer n \!:\;\;7^n + 2^{2n+1} is divisible by 3.

    Verify S(1)\!:\;\;7^1 + 2^3 \:=\:7+8 \:=\:15 ...
    divisible by 3.


    Assume S(k)\!:\;\;7^k + 2^{2k+1} \:=\:3a\;\text{ for some integer }a.


    Add 6\!\cdot\!7^k + 3\!\cdot\!2^{2k+1} to both sides:

    . . 7^k + {\color{blue}6\!\cdot\!7^k} + 2^{2k+1} + {\color{blue}3\!\cdot\!2^{2k+1}} \;=\;3a + {\color{blue}6\!\cdot\!7^k + 3\!\cdot\!2^{2k+1}}

    . . (1 + 6)\!\cdot\!7^k + (1 + 3)\!\cdot\!2^{2k+1} \;=\;3\left(2a + 2\!\cdot\!7^k + 2^{2k+1}\right) .
    a multiple of 3

    . . . . . . . . 7\!\cdot\!7^k + 2^2\!\cdot2^{2k+1} \;=\;3b\;\;\text{ for some integer }b


    Therefore: . . 7^{k+1} + 2^{2k+3} \;=\;3b


    We have proved S(k+1).
    . . The inductive proof is complete.

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  7. #7
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    Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Or you can just note that

    7^n\equiv 1^n \equiv 1 \mod 3,

    2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \mod 3,

    so that 7^n+2^{2n+1} \equiv 0 \mod 3.
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  9. #9
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    Quote Originally Posted by s_ingram View Post
    Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!
    Hi

    7^{n+1} + 2^{2n+3} = 7(7^{n} + 2^{2n+1})-7 \cdot 2^{2n+1} + 2^{2n+3} = 7 \cdot 3a + 2^{2n+1}(-7+2^2) = 7 \cdot 3a -3 \cdot 2^{2n+1}

    Therefore
    7^{n+1} + 2^{2n+3} = 3(7a-2^{2n+1})
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