Project Euler Forum

Is the answer to S(4, 10^2) really 229?
I believe M4(10^2) = 228, and M3(10^2) can't be 1, doesn't it? Isn't it 173?
Doesn't the sum equals 401?

DeatH_StaR wrote: Sat May 20, 2023 5:30 pm Is the answer to S(4, 10^2) really 229?
I believe M4(10^2) = 228, and M3(10^2) can't be 1, doesn't it? Isn't it 173?
Doesn't the sum equals 401?

Thanks for your question; the problem development team are looking into this.
In the meantime, do you perhaps mean $M_4(10^2) = 238$ rather than 228?

Upon further investigation, I can confirm that the answer to $S(4, 100)$ really is 229. Feel free to PM me if you believe you have more 3- and 4-Markov numbers than this sum affords.

RobertStanforth wrote: Sat May 20, 2023 9:14 pm Upon further investigation, I can confirm that the answer to $S(4, 100)$ really is 229. Feel free to PM me if you believe you have more 3- and 4-Markov numbers than this sum affords.

The key part of the question statement is "any part of a solution", so that, for example, if $(a,b,c)$ is a solution with $a,b\leq N < c$ then $a$ and $b$ count towards $M_3(N)$, but $c$ doesn't. Is that right?

yourmaths wrote: Sun May 21, 2023 12:49 am The key part of the question statement is "any part of a solution", so that, for example, if $(a,b,c)$ is a solution with $a,b\leq N < c$ then $a$ and $b$ count towards $M_3(N)$, but $c$ doesn't. Is that right?

Yes, the other numbers are allowed to exceed the bound (and while they are themselves $k$-Markov numbers, they are not included in $M_k$ if over the bound).

And of course, make sure you are only counting each $k$-Markov number once (for each $M_k$), even if it is present in multiple solutions, or present multiple times within the same solution.