if $\displaystyle 9^n -1 $ is divded by 2012, what is least value of n.

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- May 13th 2012, 05:34 AMsaravananbsfind the least number n.
if $\displaystyle 9^n -1 $ is divded by 2012, what is least value of n.

- May 13th 2012, 06:04 AMProve ItRe: find the least number n.
- May 13th 2012, 06:09 AMsaravananbsRe: find the least number n.
2012 divides $\displaystyle 9^n -1$ what is least value of n.

- May 13th 2012, 06:23 AMProve ItRe: find the least number n.
It doesn't for any $\displaystyle \displaystyle \begin{align*} n \end{align*}$.

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

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

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 $\displaystyle \displaystyle \begin{align*} 9^n \end{align*}$, and therefore 2012 can not ever divide $\displaystyle \displaystyle \begin{align*} 9^n - 1 \end{align*}$. - May 13th 2012, 06:30 AMsaravananbsRe: find the least number n.
why can't

Quote:

9^n-1 / 2012 is equal to k (instead of 1)

- May 13th 2012, 06:36 AMProve ItRe: find the least number n.
- May 14th 2012, 02:28 AMSylvia104Re: Find the least number n
There may well be no solutions but this is what I've found so far.

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

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

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