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Math Help - Prove that 9 is a factor of...

  1. #1
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    Question Prove that 9 is a factor of...

    Prove that 9 is a factor of:
    P(n): 10^(^n^+^1^) + 3 * 10^n + 5 for all positive integer n

    Using the induction method,
    I have proven P(1) to be true

    P(k): 10^k (10+3) + 5 = 9a
    P(k+1): 10^(^k^+^1^) (10+3) + 5 = ?

    How do I continue? I am new at doing proofs
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  2. #2
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    Do you have to use induction? Sometimes that's tedious for problems like this. I would try to rewrite 10 as 9+1 and then use the binomial theorem.
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  3. #3
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    Yes, unfortunately it is required to use the induction method.
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  4. #4
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    Quote Originally Posted by mushroom View Post
    Prove that 9 is a factor of:
    P(n): 10^(^n^+^1^) + 3 * 10^n + 5 for all positive integer n

    Using the induction method,
    I have proven P(1) to be true

    P(k): 10^k (10+3) + 5 = 9a
    P(k+1): 10^(^k^+^1^) (10+3) + 5 = ?

    How do I continue? I am new at doing proofs
    10^{k + 1} (10 + 3) + 5

    = 10 \cdot 10^k (10 + 3) + 5

    Now use the P(k) step in the form of 10^k(10 + 3) = 9a - 5
    = 10 (9a - 5) + 5

    And I'm sure you can take it from here.

    -Dan
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  5. #5
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    Hello, mushroom!

    \text{Prove that 9 is a factor of: }\:10^{n+1} + 3\!\cdot\!10^n + 5\,\text{ for all positive integers }n

    \text{Verify }S(1)\!:\; 10^2 + 3\!\cdot10 + 5 \:=\:135,\,\text{ a multiple of 9 . . . True!}


    \text{Assume }S(k)\!:\;10^{k+1} + 3\!\cdot\!10^k + 5 \:=\:9a\,\text{ for some integer }a


    \text{Add }(9\!\cdot\!10^{k+1} + 27\!\cdot\!10^k)\,\text{ to both sides:}

    . . 10^{k+1} + {\bf9\!\cdot\!10^{k+1}} + 3\!\cdot\!10^k + {\bf27\!\cdot\!10^k} + 5 \;=\;9a + {\bf9\!\cdot\!10^{k+1}} + {\bf27\!\cdot\!10^k}

    . . . . . . . . 10^{k+1}(1 + 9) + 3\!\cdot\!10^k(1+9) + 5 \;=\;9a + 9\!\cdot\!10^{k+1} + 27\!\cdot\!10^k

    . . . . . . . . . . . . 10^{k+1}\cdot 10 + 3\!\cdot\!10^k\!\cdot\!10 + 5 \;=\;9\left(a + 10^{k+1} + 3\!\cdot\!10^k\right)

    . . . . . . . . . . . . . . . . 10^{k+2} + 3\!\cdot\!10^{k+1} + 5 \;=\;\underbrace{9\left(a + 10^{k+1} + 3\!\cdot\!10^k\right)}_{\text{a multiple of 9}}

    We have proved statement S(k+1).

    The inductive proof is complete.

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