# Could anyone explain my prof's proof here for the induction of every natural number is congruent modulo 9 to the sum of its digits?

I don't get where the + 99...9dk + part comes from especially.

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99...9 = 100...0 - 1 –  vonbrand Jan 23 '13 at 2:55

$10^k=10000...0=9999999...9+1$

thus

$$10^kd_k=(9999999...9+1)d_k=999...9d_k+d_k \,.$$

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Look at an example and maybe it becomes clear:

$$2000 = 2\cdot 1000 = 2\cdot(1+999) = 2+2\cdot 999$$

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But it doesn't show any 1. –  Doug Smith Dec 17 '12 at 18:20
Yes it does. $$d_k\cdot 1000 = d^k \cdot (1+999) = d_k + 999d_k$$ The 1 is wrapped into the $d_k$. –  Arkamis Dec 17 '12 at 18:21
In my example, $d_k = 2$. –  Arkamis Dec 17 '12 at 18:22

He is just saying that $$10^k=(10^k-1)+1=(999\ldots 9)+1$$ For example, $10^3=999+1$. He is also factoring out $d_k$ in this step.

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