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Math Help - find the least number n.

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    find the least number n.

    if 9^n -1 is divded by 2012, what is least value of n.
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    Re: find the least number n.

    Quote Originally Posted by saravananbs View Post
    if 9^n -1 is divded by 2012, what is least value of n.
    Did you mean that \displaystyle \begin{align*} 9^n - 1 \end{align*} is EVENLY divided by 2012?
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    Re: find the least number n.

    2012 divides 9^n -1 what is least value of n.
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    Re: find the least number n.

    Quote Originally Posted by saravananbs View Post
    2012 divides 9^n -1 what is least value of n.
    It doesn't for any \displaystyle \begin{align*} n \end{align*}.

    If \displaystyle \begin{align*} 2012 | \left(9^n - 1\right) \end{align*} then there is some value of \displaystyle \begin{align*} n \end{align*} such that \displaystyle \begin{align*} \frac{9^n - 1}{2012} = 1 \end{align*}

    \displaystyle \begin{align*} \frac{9^n - 1}{2012} &= 1 \\ 9^n - 1 &= 2012 \\ 9^n &= 2013 \end{align*}

    Clearly this means that 9 needs to be a factor of 2013. But for any number to be divisible by 9, its digits need to add to a multiple of 9.

    2 + 0 + 1 + 3 = 6, which is not a multiple of 9.

    So 2013 can not ever be a value for \displaystyle \begin{align*} 9^n \end{align*}, and therefore 2012 can not ever divide \displaystyle \begin{align*} 9^n - 1 \end{align*}.
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    Re: find the least number n.

    why can't
    9^n-1 / 2012 is equal to k (instead of 1)
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    Re: find the least number n.

    Quote Originally Posted by saravananbs View Post
    why can't
    Oh, you're right, you can let \displaystyle \begin{align*} \frac{9^n - 1}{2012} = k \end{align*}.

    I need to keep thinking...
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    Re: Find the least number n

    There may well be no solutions but this is what I've found so far.

    2012k=9^n-1. The RHS =(10-1)^n-1\equiv0\ \text{or}\ -2\mod{10} and so k must \equiv0,5\ \text{or}\ 4,9\mod{10}. But k must be even because 9^n-1=(8+1)^n-1 is divisible by 8 and 2012\equiv4\mod8. Hence k\equiv0\ \text{or}\ 4\mod{10}.

    Now 2012\equiv2+0+1+2=5\mod9, \therefore\ 2012k+1\equiv5k+1\equiv0\mod9 \implies k\equiv7\mod9. Combining this with k\equiv0\ \text{or}\ 4\mod{10} gives k \equiv 34\ \text{or}\ 70\mod{90}.

    So if there is a solution, we must have, at the very least, k \equiv 34\ \text{or}\ 70\mod{90}. In fact, it is clear that n\geqslant4 (as 9^n-1<2012 for n=1,2,3) so you should really start by looking at k \equiv 10960\ \text{or}\ 37204\mod{65610}.
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