## Help with nth partial sum?

Given some natural number n find the nth partial sum for:

$$\displaystyle\sum_{k=0}^{\lfloor log(n) \rfloor} \lfloor \frac{n}{10^k} \rfloor$$

I find this question really difficult! If anyone could help, it would be greatly appreciated. Thanks in advance!

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Perhaps try a few examples of n to see what it actually is that you're doing??
 I take that 'log' here means the base-10 logarithm. Is that so?

## Help with nth partial sum?

It is indeed the base 10 logarithm.

For n = 1: $$\displaystyle\sum_{k=0}^{0} \lfloor \frac{1}{10^k} \rfloor = \lfloor 1/1 \rfloor = 1$$

For n = 21: $$\displaystyle\sum_{k=0}^{1} \lfloor \frac{21}{10^k} \rfloor = \lfloor 21/1 \rfloor + \lfloor 21/10 \rfloor= 21 + 2 = 23$$

 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus What about a general number $$n=a_1a_2...a_k$$ ?? Can you find it for that??
 $$n=a_1a_2...a_k$$ I think it would be: $$a_1a_2...a_k + a_1a_2...a_{k-1} + ... + a_1a_2 + a_1$$
 Something else that could possibly be used is: $$\displaystyle\sum_{k=0}^{\lfloor log(n) \rfloor} \lfloor \frac{n}{10^k} \rfloor = \displaystyle\sum_{k=0}^{\lfloor log(n) \rfloor} \frac{n}{10^k} - \displaystyle\sum_{k=0}^{\lfloor log(n) \rfloor} \frac{n}{10^k} mod 1$$ Where mod represents the remainder operator. Here the first sum is quite easy to figure out, however the summation involving mod is equally as difficult as the original, to me.