Summation of Number of Digits

Question

I have seen this problem a while ago and I wasn't able to find the answer.

$$ \begin{align*} P_n & = nd(7^n) \\ S & = \sum_{n=1}^{3981} P_n \end{align*} $$

Where $nd(n)$ is the number of digits of $n$, such that:
$$ \begin{align*} nd&(5) & = 1 \\ nd&(10) & = 2 \\ nd&(5123) & = 4 \end{align*} $$

How to find the value of $S$ without use of a computer/calculator?

I could expand $P_n$ as $$\lfloor n\log_{10}7\rfloor + 1$$ but then I have no idea what to do.

Have you looked for a pattern in the number of digits in a number of the form $7^n$? — recursive recursion, Mar 02 '14 at 21:17
the answer I got was $6697021$, but that was with a computer... — recursive recursion, Mar 02 '14 at 21:22
@recursive: "Have you looked for a pattern...": Have you looked for a pattern? — TonyK, Mar 02 '14 at 21:24
The real exact answer seems to be $6,700,385$. Gaffney's estimate (with the $3981/2$ correction term) approximates this as $6,700,382$. Not a bad approximation, I'd say. — TMM, Mar 02 '14 at 21:46

score 2 · Answer 1 · answered Mar 02 '14 at 21:31

2

Here's my solution. I don't know what the application is, so it may not be ideal. If you take the greatest integer signs off of your expression for $P_n$, then sum, you will get:

$$ 3981+\log_{10} 7\cdot\sum_{n=1}^{3981}n=3981+\left(\log_{10} 7\right)(3981)(3982)/2 $$

Since we're not taking greatest integers, this will over-estimate the real answer, by pretty close to $3981/2$. If you subtract that off, you should be within a few numbers of the answer.

answered Mar 02 '14 at 21:31

T.J. Gaffney

1,568

Similarly, you could count the number of "heptal digits" in each number (which is just $n + 1$), take the sum, and scale by $\log_{10} 7$ to get an estimate. – TMM Mar 02 '14 at 21:39

Summation of Number of Digits

1 Answers1