Project Euler Forum

Problem 281 (View Problem)

Shouldn't f(2,1) = 2?

#1: Topping 1 on entire pizza.
#2: Topping 2 on entire pizza.



EDIT: OH it says slices, meaning 2 pieces. Never mind...

Hi,
I'm stuck with this problem. My results match the reults given but I can't validate. So I must have missed something. Can someone check these f(3,5)=50450 f(3,6)=953050 f(3,12)=94020327214187
Thanks

jpatou wrote:Hi,
I'm stuck with this problem. My results match the reults given but I can't validate. So I must have missed something. Can someone check these f(3,5)=... f(3,6)=... f(3,12)=...
Thanks

They are all incorrect.
Some of these numbers are small enough that I encourage you to write a small program to enumerate all possibilities, which will help you to find the solution.

Thanks very much. I was convinced that some of these were true (not all of these clearly). Indeed I'm quite sure of my results for n=2 (for instance f(8,2)=5108105520) and I don't see what is more difficult for n=3 or n=5. I'm going to look more deeply !

EDIT: I validated it ! Very nice problem !

Hi

I have a formula for f(m,n) which I have convinced myself is probably correct, but I seem to be getting the incorrect answer for the total sum. Is there anyone I can check with via PM? Alternatively, the values I get for the test cases requested by jpatou are:
f(8,2) =
f(3,5) =
f(3,6) =
f(3,12) = <answers removed by moderator>

Hi Jarlsberg,
I've edited your post to remove the values, and sent you a PM.
Robert

So, since f(2,2)=f(3,1)=2, as given in the problem, that means that we add 2 twice to the final sum, right?

countyguy wrote: Mon Nov 30, 2020 6:27 pm So, since f(2,2)=f(3,1)=2, as given in the problem, that means that we add 2 twice to the final sum, right?

Yes, at least twice (f might equal two for other values of m and n). I read it as "the sum of f(m,n), over all pairs of (m,n) that satisfy f(m,n) <= 10^15".

Would it be possible to replace the GIF in this problem with a chart of all 16 examples? It would be much more useful, and I believe there's precedent for similarly-sized example images elsewhere.

Not the dev team, but I'm able to download the gif and view each frame of the gif on my computer. The gif only has 16 frames so it is easy to browse. Does that help you?

It'd be lovely if the problem statement specified that the sum was with multiplicity, as I didn't know why I was messing up till I found this thread.