Project Euler Forum

I think, I don't get the point. Can't we define for F(6) the set S = {1, 4, 5, 6} and hence the union of S, 2S and 3S becomes {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18}, which is intersected with {1, ..., 6} again {1, 2, 3, 4, 5, 6}. Overall we then would have F(6) = 6 and more generally F(N) = N. Am I missing something???

The set $S=\{1,4,5,6\}$ is NOT 123-Separable.
$2S = \{2,8,10,12\}$ and $3S = \{3,12,15,18\}$ are not disjoint as $12\in 2S$ and $12\in 3S$.

Yes, thanks for the clarification! In my mindset I forgot to intersect 2S and 3S...

I'm currently getting the wrong answer for F(10^16) and trying to track down where things are coming apart in my algorithm. Unfortunately, I am getting the right answer for F(6) and F(20). Would it be OK to post my results for, and ask for confirmation of, some moderately high F(n)'s?

Unfortunately, this isn't the right place for asking that. These threads are meant to clarify possible confusion and inaccuracies in the problem statement. Often, the examples given in these problem are carefully crafted to verify basic understanding without giving too much away.

Have you tried writing a simpler (but slower) solution to verify your optimized one?

No worries -- I can pursue that approach.