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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?

Arturo Magidin
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Ronnie
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    For the record, this was cross-posted on SO: http://stackoverflow.com/questions/4481022/sum-of-999999-30-9s – moinudin Dec 19 '10 at 00:55
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    possible duplicate of N= 9 + 99 + 999 + ... 999...99 – Aryabhata Dec 19 '10 at 01:31
  • and how would i explain this in words? – Ronnie Dec 20 '10 at 01:27
  • @Ronnie: Could you be more specific about what you are having trouble expressing in words? – Jonas Meyer Dec 20 '10 at 01:43
  • as in how would i tell a fellow student or a teacher that the answer is 28 1's 080 – Ronnie Dec 20 '10 at 02:32
  • By explaining to them one of the proofs you've been given? If you understand the mathematics, you should be able to use words to explain the steps. If you don't, will you please ask more focused questions about what you don't understand? – Jonas Meyer Dec 20 '10 at 02:35
  • No i understand the trend and will be able to give an answer if you told me whats 9+99+...to whatever digit. But in words, how would i explain it. Would i be like adding 9+99+999+... is equal to 10 to whatever power -1.... – Ronnie Dec 20 '10 at 02:39
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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 02:42

2 Answers2

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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.

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

davidlowryduda
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Bill Dubuque
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