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I was given the question:what is 9+99+999+9999+...+999..99(30 digits) After noticing a trend, I came with the conclusion that the answer would be 28 1's 080. Can anyone confirm my answer and give a reason as to why?

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For the record, this was cross-posted on SO: stackoverflow.com/questions/4481022/sum-of-999999-30-9s – marcog Dec 19 '10 at 0:55
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possible duplicate of N= 9 + 99 + 999 + ... 999...99 – Aryabhata Dec 19 '10 at 1:31
    
and how would i explain this in words? – Ronnie Dec 20 '10 at 1:27
    
@Ronnie: Could you be more specific about what you are having trouble expressing in words? – Jonas Meyer Dec 20 '10 at 1:43
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@Ronnie: It's not; each summand is equal to a power of $10$ minus 1, but the entire sum is not a power of 10 minus 1. In words, you would say what the argument is: Notice that $10^k$ is a $1$ followed by $k$ zeros, so $10^k - 1$ is $k$ 9's. So you can replace each summand with a power of 10 minus 1; then you can reorder the sum so that you add all powers of 10 first, and subtract all the 1s later; then you can figure out what the sum of the powers of 10 is; etc. – Arturo Magidin Dec 20 '10 at 2:42
up vote 16 down vote accepted

Note that $$\underbrace{99\cdots 9}_{k\text{ digits}} = 10^k - 1.$$ So your sum is the same as $$(10-1) + (10^2-1) + (10^3-1) + \cdots + (10^{30}-1),$$ which is equal to $$(10 + 10^2 + 10^3 + \cdots + 10^{30}) - 30.$$ The first sum is easy to do, the difference is easy to do, and it gives your answer.

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I particularly like this approach since I think it could be use for general problems with repeating digits. – Quixotic Dec 19 '10 at 10:31
    
Thank you very much – Ronnie Dec 20 '10 at 2:50

HINT $\ $ Exploit the linearity of $\Sigma\:$: $\rm\ \Sigma\ (f(k)+c)\ = \ \Sigma\ f(k) + \Sigma\ c\ $ to reduce to a geometric sum.

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