find the least number n.

**saravananbs** · May 13th 2012, 05:34 AM

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

**Prove It** · May 13th 2012, 06:04 AM

Originally Posted by saravananbs

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?

**saravananbs** · May 13th 2012, 06:09 AM

2012 divides $9^n -1$ what is least value of n.

**Prove It** · May 13th 2012, 06:23 AM

Originally Posted by saravananbs

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 $\displaystyle \begin{align*} 9^n - 1 \end{align*}$ .

**saravananbs** · May 13th 2012, 06:30 AM

why can't

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

**Prove It** · May 13th 2012, 06:36 AM

Originally Posted by saravananbs

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...

**Sylvia104** · May 14th 2012, 02:28 AM

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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