Proof by Induction

1. The problem statement, all variables and given/known data

http://img200.imageshack.us/img200/7097/99175506.gif [Broken]


2. Relevant equations


3. The attempt at a solution

For [tex]n \in N[/tex] let P(n) be the statement: "81 | (10ⁿ⁺¹-9n-10)"

Base Case: when n=1: 10ⁿ⁺¹-9n-10 = 81 = 81 × 1

So P(1) is true

Inductive Step: let [tex]k \in N[/tex] and suppose [tex]P(k)[/tex] is true, that is 81 | (10^k+1-9k-10) is true. Then [tex]10^{k+1}-9k-10=81m[/tex] for some [tex]m \in Z[/tex]. Then [tex]10^{k+1}=(81m+9k+10)[/tex].

So,

10^k+2-10(k+1)-10=10 × 10^k+1-10(k+1)-10

=10 × (81m+9k+10)-10k+10-10

= 10 × (81m+9k+10-k)

I'm stuck here and I don't know how to factor things out and bring the 81 forth to get the expression in the form: "81(something here)" to show that it's divisible by 81 for all n. Any help is appreciated.

P.S.
I might be able to do it this way:

10 × (81m+9k+10-k) = 810m+90k+100-10k
=> 81(10m+(90/81)k+(100/81)-(10/81)k)

But I don't feel that this is the right way...

 

It's a bit awkward... are you sure that there is not a better way?

 

How did you get from [tex]10\cdot10^{k+1} - 9k - 9 - 10[/tex] to [tex]10(10^{n+1} - 9n - 10) + 81n + 81[/tex]? Could you please explain :)

Yes that was a typo but I thought it would be a good way to replace 10^k+1 by 81m+9k-10, since [tex]10^{k+1}-9k-10=81m \Rightarrow 10^{k+1} = 81m+9k+10[/tex]