Problem 281

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elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

Problem 281

Post by elendiastarman »

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... :oops:
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?
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jpatou
Posts: 9
Joined: Sat Mar 09, 2013 8:13 am

Re: Problem 281

Post by jpatou »

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
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mpiotte
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Posts: 1961
Joined: Tue May 08, 2012 5:40 pm
Location: Montréal, Canada

Re: Problem 281

Post by mpiotte »

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.
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jpatou
Posts: 9
Joined: Sat Mar 09, 2013 8:13 am

Re: Problem 281

Post by jpatou »

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 !
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jarlsberg
Posts: 2
Joined: Sun Jan 24, 2016 7:55 pm

Re: Problem 281

Post by jarlsberg »

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>
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RobertStanforth
Administrator
Posts: 2666
Joined: Mon Dec 30, 2013 11:25 pm

Re: Problem 281

Post by RobertStanforth »

Hi Jarlsberg,
I've edited your post to remove the values, and sent you a PM.
Robert
countyguy
Posts: 9
Joined: Mon Nov 30, 2020 6:23 pm

Re: Problem 281

Post by countyguy »

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?
DJohn
Posts: 90
Joined: Sat Oct 11, 2008 12:24 pm

Re: Problem 281

Post by DJohn »

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".
00gogo00
Posts: 8
Joined: Tue Dec 26, 2023 1:55 am

Re: Problem 281

Post by 00gogo00 »

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.
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byhill
Posts: 29
Joined: Mon Aug 01, 2022 3:13 am

Re: Problem 281

Post by byhill »

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?
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Incompleteusern
Posts: 9
Joined: Sun Aug 17, 2025 8:09 am

Re: Problem 281

Post by Incompleteusern »

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.
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