Question

Prove that 81|(10^(n+1) -9n -10) for every non negative integer n value.

Answer 1

So I'm guessing this is a proof by induction problem. I showed the base case of zero, but for (n+1) we have

81| 100(10^n) -9n -19

And I'm not sure how to set it up so that it's clear 81 divides that.

Answer 2

ok you want to use induction, thats probably easiest

Answer 3

If there's some other way I'm open to it, but the way the problem was written it seems like that's what'd be easiest.

Answer 4

Oh no, the original case was

81|(10^(k+1) -9k-10)

Answer 5

ok i see

Answer 6

The goal is going to be

81| (10^(k+2) -9(k+1)_10)

Answer 7

correction*

assume that : 81 | 10^(k+1) -9k -19

<many steps later>


it is true that: 81 | 10^(k+2) -9(k+1) -19

Answer 8

-10 but yeah haha

Answer 9

So any ideas how to manipulate it so it's clear 81 divides it?

Answer 10

Haha it's fine. I've done the same thing plenty of times.

Answer 11

:)

Answer 12

well now that you understand the problem, you are 3 quarters there

Answer 13

@perl it should be 10^(k+2) - 9(k+1) - 10.

Answer 14

\[10^{n+1} -9n-10 = (1+9)^{n+1}-9n-10 = 1+9(n+1)+ 9^2M - 9n-10 = 9^2M\]

Answer 15

OK, I found another way to prove this...

Answer 16

$$ 10^{n+1} -9n-10 = (1+9)^{n+1}-9n-10 \\= 1+9(n+1)+ 9^2M - 9n-10 = 9^2M$$

Answer 17

yes :)

Answer 18

Assume that \(10^{k+1} - 9k - 10\) is divisible by 81.\[n = k+1:\]\[10\cdot 10^{k+1} - 9(k + 1) - 10\]\[= 10\cdot 10^{k+1} - 9k - 19\]\[= 10\cdot 10^{k+1} - 90k - 100 + 81 k - 81 \]\[= 10(10^{k+1} - 9 k - 10) + 81 (k - 1)\]

Answer 19

Okay. I think I'm following what you did Ganeshie. I get what to do now, thank you.

Answer 20

These come in handy when you seek an alternative for induction proofs like these :
```
(1+x)^n = 1+xn + x^2(stuff)
(1+x)^n = 1+x(stuff)
```

Answer 21

if you use ganeshie's argument in your inductive step, but then you won't need to cite the inductive hypothesis

Answer 22

Alright then. I never would've seen something like that on my own, thanks a ton.

Answer 23

ganeshie's argument is a constructive approach, you are actually showing how to produce a factor of 81 for any nonnegative integer of the form 10^(n+1) -9n -10

Answer 24

you can solve for k in terms of n .
10^(n+1) -9n -10 = 18k

Answer 25

so it stands as a proof by itself

Answer 26

Right. It's pretty creative. I'm not sure how I would think of something like that on an exam

Answer 27

But it was super helpful here, thanks a ton.

Answer 28

I don't think it is easy to think of it on one's own.
One needs others to show that it can be done like that... Once you see it can be done like that, it becomes easy to apply it in exam

Answer 29

Induction proofs are awesome because you always know what do do : where to start and how to proceed etc..

Answer 30

Induction works almost every time.
Just try to express the \(n+1\)-case in terms of the \(n\)-case, and you'll see how things begin to converge.

Answer 31

Right. A lot of the other problems have been pretty straight forward. This was the first one that required creativity like that. Again, thanks for pointing out the method.