# find the least number n.

• May 13th 2012, 05:34 AM
saravananbs
find the least number n.
if $9^n -1$ is divded by 2012, what is least value of n.
• May 13th 2012, 06:04 AM
Prove It
Re: find the least number n.
Quote:

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?
• May 13th 2012, 06:09 AM
saravananbs
Re: find the least number n.
2012 divides $9^n -1$ what is least value of n.
• May 13th 2012, 06:23 AM
Prove It
Re: find the least number n.
Quote:

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*}.
• May 13th 2012, 06:30 AM
saravananbs
Re: 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 AM
Prove It
Re: find the least number n.
Quote:

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...
• May 14th 2012, 02:28 AM
Sylvia104
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 $k \equiv 10960\ \text{or}\ 37204\mod{65610}.$ (Thinking)