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Stuck toward the end of the proof:

Prove: That $5\cdot10^n + 10^{n-1} + 3$ is divisible by $9$:

If $n=1$ then $5\cdot10^1 + 10^{1-1} + 3= 5\cdot10+10^0+3=54 $

$9$ surely divides $54$.

Assume, If $k$ is a natural number such that $9/5\cdot10^k + 10^{k-1} + 3$

then show that $$9/5\cdot10^{k+1}+5\cdot10^k + 3$$ $$9/10\cdot(5\cdot 10^{k+1}+5\cdot10^k + 3)$$ $$9/5\cdot10^{k+2}+5\cdot10^{k+1} + 10\cdot3$$

Stuck here I need to get each term to be divisible by $9$. I am stuck trying to get this.

Any help would be appreciated.

OLE
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1 Answers1

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The difference \begin{align*} 5\cdot 10^{k+1} +10^k + 3 - (5\cdot 10^{k} +10^{k-1} + 3) &= 5\cdot 10^k(10-1) + 10^{k-1}(10-1) \\ &= 9(5\cdot 10^k + 10^{k-1}) \end{align*} is a multiple of 9. Since by inductive assumption, $5\cdot 10^{k} +10^{k-1} + 3$ is a multiple of 9, it follows that $5\cdot 10^{k+1} +10^k + 3 $ is also a multiple of 9.

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    Aside from subtraction, is there anyway I can manipulate what I wrote out to get something similar to your final statement. – OLE Oct 27 '16 at 04:46
  • You can expand $10^{k+1}$ and $10^{k+2}$ using binomial theorem expressing them as $(9+1)^{k+1}$ and $(9+1)^{k+2}$. This will give the result directly (without induction). –  Oct 27 '16 at 04:53
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    how did you get $$51*10^{n-1}$$ – OLE Oct 27 '16 at 05:00
  • $5 \cdot 10^{n} + 10^{n-1} = 5 \cdot 10 \cdot 10^{n-1} + 10^{n-1} = 51 \cdot 10^{n-1}$ –  Oct 27 '16 at 05:03
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    So in my original proof when I multiplied by 10 I did not have to multiply the other side by 10? I found a similar proof that does so and it has thrown me off – OLE Oct 31 '16 at 03:40