Hi, maybe some of you guys know Project Euler already, if not, please follow the link and read a bit about it. It is a collection of interesting mathematical problems. This is my take on the first problem:

Find the sum of all the multiples of 3 or 5 below 1000

While this is a one line problem in python:

solving this analytically is challenging.

This is how far I got:

\begin{align} W &- \textrm{maximal value, here } 1000 \\ m_j &= \lfloor \log_j W \rfloor \\ i, j &- \textrm{exponents, here }3, 5 \\ k_i, k_j &\in N; \textrm{exponents for }i, j; k_x <= m_x \end{align}

wherefore $i^{k_i} \cdot j^{k_j} < W$ is always valid.

As $\log$ is a strict monotonic function, we can change this to

\begin{align} i^{k_i} \cdot j^{k_j} &< W \\ k_i \cdot \log_i \left(i\right) + k_j \cdot \log_i \left(j\right) &< log_i \left(W\right)\\ k_i &< log_i \left(W\right) - k_j \cdot \log_i \left(j\right) \\ m_{i} \left( j \right) &= \lfloor log_i \left(W\right) - k_j \cdot \log_i \left(j\right) \rfloor \end{align}

So the needed value can be defined as:

\begin{align} L &= \sum_{a=0}^{m_j} \sum_{b=0}^{m_i\left(a\right)} j^a \cdot i^b \\ &= \sum_{a=0}^{m_j} j^a \cdot \sum_{b=0}^{m_i\left(a\right)} i^b \end{align}

Now, apply this summation formula:

$$\sum_{i=0}^n k^i = \frac{k^{n+1}-1}{k-1}$$

which results in

\begin{align} L &= \sum_{a=0}^{m_j} j^a \cdot \frac{i^{m_i \left(a\right)+1}-1}{i-1} \\ &= \frac{-1}{i-1} \cdot \sum_{a=0}^{m_j} j^a + \frac{i}{i-1} \cdot \sum_{a=0}^{m_j} j^a \cdot i^{m_i \left(a\right)} \\ &= \frac{-1}{i-1} \cdot \left(\frac{j^{m_j + 1}-1}{j-1} \right) + \frac{i}{i-1} \cdot \sum_{a=0}^{m_j} i^{a \cdot \log_i \left( j \right) + m_i \left(a\right)} \ \end{align}

well... and this is how far I got. I could reformulate it to:

$$L = \frac{-1}{i-1} \cdot \left(\frac{j^{m_j + 1}-1}{j-1} \right) + \frac{i \cdot W}{i-1} \cdot \sum_{a=0}^{m_j} i^{- {log_i \left(W\right) - a \cdot \log_i \left(j\right)}}$$

with $\{ \}$ as the frac function. I have no idea how to proceed from here. :(

So if you see an error or have an idea, please write me!

I also found a different solution here: Muvik.de but I really like to know how to solve my problem!