Finding the nth Partial Sum for a Series Involving Floor and Mod Operators

[m84uily]

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!

 

[micromass]

Perhaps try a few examples of n to see what it actually is that you're doing??

 

[dodo]

I take that 'log' here means the base-10 logarithm. Is that so?

 

[m84uily]

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

 

[micromass]

What about a general number

$$n=a_1a_2...a_k$$

?? Can you find it for that??

 

[m84uily]

$$n=a_1a_2...a_k$$

I think it would be:

$$a_1a_2...a_k + a_1a_2...a_{k-1} + ... + a_1a_2 + a_1$$

 

[m84uily]

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.