I don't get where the + 99...9dk + part comes from especially.
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?
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discrete-mathematics
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099...9 = 100...0 - 1 – 2013-01-23
3 Answers
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$10^k=10000...0=9999999...9+1$
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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0But it doesn't show any 1. – 2012-12-17
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0Yes it does. $$d_k\cdot 1000 = d^k \cdot (1+999) = d_k + 999d_k$$ The 1 is wrapped into the $d_k$. – 2012-12-17
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0In my example, $d_k = 2$. – 2012-12-17
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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.