2

Here is a puzzle I made (I originally posted it here on the Puzzling Stack Exchange):

Suppose you constructed $m$ rows in the following way ($m,n$ are integers): $$\begin{align}&1,2,3,\ldots ,n \\ &n+1,\ldots , 2n \\ &2n+1,\ldots 3n\end{align}$$ $$\vdots$$ $$n(m-1)+1,\ldots, mn$$ In each row, you want to have an odd number of odd primes. If $m<13$, what prime number $n$ can you find that will form the highest possible prime value of $m$?

The answer is hidden in the puzzle...

I created such a puzzle because a user wanted to find puzzles that had the answer hidden in the puzzle itself. I will not tell you what the answer is; I only posted this as a question because I do have a question:

Question:

Suppose you did not know the answer, or at least did not know that the answer was hidden in the puzzle. Is there a way to find out the answer without brute-force/exhaustion? Otherwise the puzzle may seem a bit boring or tiring, which I don't want to make other users feel.

Also, there might be multiple values of $n$, so I fear this puzzle might be too broad, so I will just say now, the proposed solution is between $1$ and $20$ (inclusive, just to increase the potential possibilities at first glance!).

Apologies if this question is not appropriate for the site, as it is not a "typical" question like others. Also, I was not fully sure what tags were appropriate either, so sorry if they are not.

Thank you in advance.

Edit:

I now feel a little more relaxed about whether or not this is an appropriate question, after checking out this post.

Community
  • 1
Mr Pie
  • 9,459
  • 1
    While this is an appropriate question, you shouldn't use questions like "What are good math puzzles" as models for your own. They tend to be the exception, not the norm. – Rushabh Mehta Aug 24 '18 at 01:04
  • 1
    btw since you mention the intended range for $n$, I wonder if your intended solution is $n=13$? Note that $n=13$ only achieves $m=11$, while $n=4843$ achieves $m=12$. – stewbasic Aug 24 '18 at 01:08
  • @stewbasic yeah, the intended answer was $13$, but I made a mistake... I guess $4843$ is the real answer, then. – Mr Pie Aug 24 '18 at 01:14
  • @RushabhMehta okay, thank you for telling me. Should I discard the proof-verification tag, then? – Mr Pie Aug 24 '18 at 01:15
  • @user477343 Yes please! – Rushabh Mehta Aug 24 '18 at 01:17
  • 1
    @RushabhMehta Done! :) – Mr Pie Aug 24 '18 at 01:17
  • @stewbasic I edited the puzzle, now. $13$ is now a solution and $4843$ is not (since the latter is composite; $4843=29\times 167$) :) – Mr Pie Aug 24 '18 at 01:18
  • 2
    @user477343 I'm not convinced there's a better way than brute. Prime numbers are notoriously annoying – Rushabh Mehta Aug 24 '18 at 01:19
  • 1
    @user477343 This doesn't fix it; $n=165941$ is prime and achieves $m=12$. Naively I'd expect each integer to have a one in $2^{12}$ chance of achieving $m=12$. I suggest editing so that $n=13$ reaches the bound you've set (something like "$m+1<13$" but less contrived...) – stewbasic Aug 24 '18 at 03:17
  • @stewbasic Okay. Thanks for that. I will just include the two given solutions, and if somebody can find the solution $n=165941$ then they get a muffin :D – Mr Pie Aug 24 '18 at 09:12

0 Answers0