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I'm going to setup a puzzle contest for my family and friends for Christmas. The puzzles are hard, but doable in 1-2 hours (along the lines of Sudoku, Crosswords and Logic Tables puzzles).

However, I have all of them ready, except for one. I would like to propose for one of them (randomly chosen) a near-impossible to solve puzzle. By near-impossible I mean:

  • It has a solution;
  • The solution is very hard to find, and impossible to solve in 1-2 hours for an intelligent person with access to the Internet and a computer;
  • The puzzle does not seem to be very hard (as I said before, something that appears to be solvable with some effort);
  • He/She will get the prize if stating that the puzzle is impossible to solve with the time and resources given to them.

Could you help with that? I will accept the most original puzzle satisfying these constraints in the next 24 hours. EDIT: Sorry guys, I got two answers that do not satisfy all constraints (both are far from near-impossible), so I'm going to leave the question open.

PS: Based on Ideas on puzzles for a scientific event, I came to the conclusion that this question is not off-topic (requesting puzzles).

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Chaotic
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2 Answers2

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Instead of taking "nearly impossible" puzzle, which can frustrate easily, I would suggest "open optimization problem". I.e. a task with many solutions, which can be ordered from worst to best. So everyone can enjoy it and find a solution, while it is still competitive.

For example:

Fastest way to collect an arbitrary army

Find a straight tunnel

Two spies throwing stones into a river

Knights and jokers

Maximize the number of paths

Or may be you can find something better with "optimization" tag.

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klm123
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1) What is the lowest number of cuts in a circle do you need to get 622 pieces? (They don't have to be all the same size and the cuts can be randomly placed) It didn't take 1-2 hours to solve but i had fun solving it :-)

2) Take the sum of 17 successive numbers. Like starting at 55 for example: $55+56+57+58...$ 17 times. Is this sum ALWAYS and I mean ALWAYS divisible by one of the 17 numbers? If or if not justify your answer.
This is also a very mathematical problem but this is everything what comes to my mind right now.

Hope you get the answers not too fast and enjoy my puzzles. Cheers :)

Arji
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    I don't think the second is near-impossible. If the middle of those 17 numbers is N then their average is also N, so the sum is 17N, so the sum is always a multiple of the middle number. Am I missing something? (The same thing works if you replace 17 with any other odd number.) – Gareth McCaughan Dec 24 '16 at 23:06
  • @GarethMcCaughan yes thats true! I guess that one was to easy – Arji Dec 24 '16 at 23:08
  • Is the answer to the first question 35 cuts? – wildBillMunson Dec 25 '16 at 03:37
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    Both are actually fairly easy, because the Lazy carterer's sequence will say that (n^2+n+2)/2 = total cuts, and you just have to solve an equation after that – bleh Dec 25 '16 at 21:37