# Help with nth partial sum?

1. Dec 22, 2011

### 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!

2. Dec 22, 2011

### micromass

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

3. Dec 22, 2011

### dodo

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

4. Dec 22, 2011

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

5. Dec 22, 2011

### micromass

Staff Emeritus

$$n=a_1a_2...a_k$$

?? Can you find it for that??

6. Dec 23, 2011

### m84uily

$$n=a_1a_2...a_k$$

I think it would be:

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

7. Dec 23, 2011

### 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.

Last edited: Dec 23, 2011