# Proof by Induction

1. Aug 23, 2009

### roam

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 $$n \in N$$ 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 $$k \in N$$ and suppose $$P(k)$$ is true, that is 81 | (10k+1-9k-10) is true. Then $$10^{k+1}-9k-10=81m$$ for some $$m \in Z$$. Then $$10^{k+1}=(81m+9k+10)$$.

So,

10k+2-10(k+1)-10=10 × 10k+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...

Last edited by a moderator: May 4, 2017
2. Aug 23, 2009

### rock.freak667

3. Aug 23, 2009

### roam

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

4. Aug 23, 2009

### snipez90

You have a few typos in your inductive step (should be 9(k+1), not 10(k+1)).

Note that
$$10^{(k+1)+1} - 9(k+1) - 10 = 10\cdot10^{k+1} - 9k - 9 - 10 = 10(10^{n+1} - 9n - 10) + 81n + 81$$

5. Aug 23, 2009

### roam

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

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

6. Aug 23, 2009

### mathie.girl

The above looks good, without some typos. Should be:
10k+2-9(k+1)-10=10 × 10k+1-9(k+1)-10
=10(81m+9k+10)-9k-9-10

Then keep going.

Last edited by a moderator: May 4, 2017
7. Aug 23, 2009

### snipez90

You always want to apply the inductive hypothesis as easily as possible. If the inductive step gives you
$$10^{(k+1)+1} - 9(k+1) - 10 = 10\cdot10^{(k+1)} - 9(k+1) - 10 ,$$
then the fact that you can pull out the very first 10 factor on the right hand side suggests multiplying the expression $$10^{(k+1)} - 9k - 10$$ in our inductive hypothesis by 10 and relating it to the inductive step somehow.

8. Aug 23, 2009

### rock.freak667

$$H_N: 10^{N+1}-9N-10=81m$$

x10

$$10^{N+2}-90N-100=810m$$

$$10^{N+2}-90N-90-10=810m$$

$$10^{N+2}-90(N+1)-10=810m \Rightarrow 10^{N+2}-9(N+1)-10=810m+81(N+1)$$

Left side happens to be HN+1 and the right side is divisible by 81 no matter what N is.....

so even if it is awkward, once you show it is divisible by 81, you have proven it.