Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
For the record, this was cross-posted on SO: – marcog Dec 19 '10 at 0:55
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
@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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.