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  1. #1
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    multiple of 5

    Prove that, for any positive integer n, 3^3n+1 + 2^n+1 is always divisible by 5.
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  2. #2
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    Quote Originally Posted by bearej50 View Post
    Prove that, for any positive integer n, 3^3n+1 + 2^n+1 is always divisible by 5.
    For n= 1, that says 3^{1}+ 2^{1}= 5 is divisible my 5 which is true.

    Assume that, for some positive integer k, 3^{3k+1}+ 2^{k+1} is a multiple of 5. That is, that 3^{3k+1}+ 2^{k+1}= 5i for some integer i.

    3^{3(k+1)+1}+ 2^{(k+1)+1}= (3^{3k+ 4}+ 2^{(k+1)+ 1}= 3^{3}3^{3k+1}+ 2(2^{k+1}
    "Add and subtract" 3^3(2^{k+1} to get
    = 3^3(3^{k+1}+ 2^{k+1})- 3^3(2^{k+1})+ 2(2^{k+1}
    = 27(3^{k+1}+ 2^{k+1})- (27- 2)2^{k+1}
    = 27(5i)- (25)2^{k+1}= 5(27i- (5)2^{k+1}
    which is 5 times an integer.
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  3. #3
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    thank you so much
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  4. #4
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    Hello, bearej50!

    Prove that, for any positive integer n\!:\;\;3^{3n+1} +2^{n+1} is always divisible by 5.

    We have: . N \;=\;3^{3n}\cdot3^1 + 2^{n+1} \;=\;(3^3)^n\cdot3 + 2^{n+1} \;=\;3(27^n) + 2^{n+1} \;= \;3(25 + 2)^n + 2^{n+1}

    . . = \;3\bigg[25^n + {n\choose1}25^{n-1}\cdot2 + {n\choose2}25^{n-2}\cdot 2^2 + \hdots + {n\choose 2}25^2\cdot2^{n-2} + {n\choose1}25\cdot2^{n-1} + 2^n\bigg] + 2^{n+1}

    . . = \;3\underbrace{\bigg[25^n + {n\choose1}25^{n-1}\cdot 2 + \hdots + {n\choose1}25\cdot2^{n-1}\bigg]}_{\text{This is a multile of 5, }5k} \;+\; 3\cdot2^n \;+\; 2^{n+1} \;= \;5k \;+\; 3\cdot2^n \;+\; 2^{n+1}


    \text{Factor: }\;N \;=\; 5k + (3 + 2)\!\cdot\!2^n \;=\;5k + 5\!\cdot\!2^n \;=\;\underbrace{5(k + 2^n)}_{\text{multiple of 5}}

    Therefore, N is divisible by 5.

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  5. #5
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    thank you
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