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Math Help - 6^n + 4 is divisible by 10

  1. #1
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    6^n + 4 is divisible by 10

    How can I prove that 6^n + 4 is divisible by 10? I want to write a direct proof.
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    How can I prove that 6^n + 4 is divisible by 10? I want to write a direct proof.
    Do it by mathematical induction .
    What is the base case?
    Suppose you know that $6^K+4$ is divisible by ten.

    Then $6^{K+1}+4=6^{K+1}+4\cdot 6-4\cdot 6+4=6(6^k+4)-4(5)$.
    Is this divisible by ten?
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    Re: 6^n + 4 is divisible by 10

    I should not use mathematical induction because the question is in an exercise on direct proof and mathematical induction is introduced later.
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    Re: 6^n + 4 is divisible by 10

    For a direct proof:

    Assume $n$ is a positive integer (as it is false otherwise).

    $\displaystyle \begin{align*}6^n + 4 & = 5\left(1+\sum_{k=0}^{n-1} 6^k\right) \\ & = 10\left(1 + \sum_{k=1}^{n-1} 3^k2^{k-1} \right)\end{align*}$
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  5. #5
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    Re: 6^n + 4 is divisible by 10

    What kind of expansion is done for 6^n? I don't think I've seen it before.
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    What kind of expansion is done for 6^n? I don't think I've seen it before.
    Use geometric sums:

    $\displaystyle \dfrac{a^n-1}{a-1} = \sum_{k=0}^{n-1}a^k \Longrightarrow a^n = 1 + (a-1)\sum_{k=0}^{n-1}a^k$

    Plugging in for $a=6$, we have: $\displaystyle 6^n = 1+5\sum_{k=0}^{n-1}6^k$
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    Re: 6^n + 4 is divisible by 10

    Thanks.
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    Re: 6^n + 4 is divisible by 10

    The question said to consider the units digit. There is a hint: what does 6x6 have as a units digit?
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    The question said to consider the units digit. There is a hint: what does 6x6 have as a units digit?
    If $6^n$ has a unit digit of $6$ what is the unit digit of $6^n+4~?$

    Look at this table.
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    Re: 6^n + 4 is divisible by 10

    It would be necessary to prove that 6^n has a unit digit of 6.
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    It would be necessary to prove that 6^n has a unit digit of 6.
    I am pretty sure that would require induction, so I agree, that is not a terribly useful tip.
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    It would be necessary to prove that 6^n has a unit digit of 6.
    @Stuck Man, I hope that you will tell us all what your instructor expected of you in an answer. Please.

    Here are some comments I would ask you to consider.
    You posted this question in a number theory forum. It is unthinkable that someone doing number theory would not have induction as a tool. As posed then this is basic algebra.

    If I am correct about that, then I suspect the hint you were given, $6\cdot 6=~?$, is right on. At a very basic level, any integer ending in a six when multiplied by six results in an integer also ending in six. Now I admit that without induction there is no rigorous proof for that. But without induction available a rigorous proof is not called for.
    Last edited by Plato; March 20th 2014 at 03:38 PM.
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    Re: 6^n + 4 is divisible by 10

    The question is "By considering the units digit prove that 6^n + 4 is always divisible by 10 where n is a member of N." The topic of the exercise is direct proof. Proof by induction is later in the chapter.
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    Re: 6^n + 4 is divisible by 10

    Quote Originally Posted by Stuck Man View Post
    The question is "By considering the units digit prove that 6^n + 4 is always divisible by 10 where n is a member of N." The topic of the exercise is direct proof. Proof by induction is later in the chapter.
    There is a direct but very long proof.

    $Step\ 1:\ Prove\ that\ 5m\ ends\ in\ 5\ if\ m\ is\ odd.$

    $Step\ 2:\ Prove\ that\ 4^k\ is\ even.$

    $Step\ 3:\ Prove\ that\ 5^p = (4 + 1)^p\ is\ odd.$

    $Step\ 4:\ Prove\ that\ 5^q = 5 * 5^{(q - 1)}\ ends\ in\ 5.$

    $Step\ 5:\ Prove\ that\ 6^r\ is\ even.$

    $Step\ 6:\ Prove\ that\ 6^s = (5 + 1)^s\ ends\ in\ 1\ or\ 6.$

    $Step\ 7:\ Prove\ that\ 6^n\ does\ not\ end\ in\ 1.$ This step is the only one that is even a bit tricky.

    $Step\ 8:\ Prove\ that\ 6^n + 4\ ends\ in\ 0\ and\ so\ is\ divisible\ by\ 10.$ Trivial

    I must admit that I have not worked out each step in detail, but I believe that you never need to use induction or rely on any theorem that itself is proved by induction. Step 6 does rely on contradiction. Can you use that?

    AFTERTHOUGHT: I am not sure, but I think I see a way to avoid step 7. I'll think about it more as I walk the dog.
    Last edited by JeffM; March 21st 2014 at 11:43 AM. Reason: Afterthought
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    Re: 6^n + 4 is divisible by 10

    OK. I am back from letting the dog do canine things and think there is a way to prove this exercise without utlilizing reductio ad absurdum. I am not 100% sure. I was a history major, and my proofs are undoubtedly not up to modern standards of rigor.

    Step 5 $\implies last\ digit\ of\ 6^n \in\ Set_A = \{0,\ 2,\ 4,\ 6,\ 8\}.$

    Step 6 $\implies last\ digit\ of\ 6^n \in\ Set_B = \{1,\ 6\}.$

    So Step 7 reduces to $last\ digit\ of\ 6^n \in Set_A \bigcap Set_B = \{6\}.$

    Because I have never understood those mathematicians who deny the law of the excluded middle and consequently reject proofs by reductio, I don't know whether this is considered a direct argument or not, but it is not, at least not superficially, a proof by contradiction.

    Hope this all helps.
    Last edited by JeffM; March 21st 2014 at 02:12 PM.
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