Can you find a solution by only changing (moving) one stick?
From my point of view I think I systematically exhausted all possible configs and I couldn't find a solution, so I think there is no solution under condition of only moving one stick
I'm looking for a solution using "=" operator
As I said, I don't think it exists
So please prove I'm wrong
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$\begingroup$ If I understood you correctly, like write it in a way like 6+0<7, I'm afraid no also writing it using not equal operator is correct but I'm looking for solution with = operator $\endgroup$– jahlyCommented Jul 26, 2020 at 15:27
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$\begingroup$ You are probably wrong in believing we can prove you wrong. $\endgroup$– Florian FCommented Jul 26, 2020 at 20:14
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$\begingroup$ Do you have a translation of the text at the top of the picture? If so, can you tell us what it says please? If not, then how do you know what the conditions of the puzzle are? $\endgroup$– chasly - supports MonicaCommented Jul 27, 2020 at 10:16
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$\begingroup$ Translation: change just one stick to make equation correct, and yes I know as per just this statement any operator other than "=" is a ok but I added that extra condition is it possible using operator "=" $\endgroup$– jahlyCommented Jul 28, 2020 at 7:04
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3 Answers
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Hereis a possible solution to the problem:
answered Jul 26, 2020 at 18:27
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$\begingroup$ this doesn't solve the problem as it says MOVE a matchstick $\endgroup$ Commented Jul 26, 2020 at 20:38
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2$\begingroup$ @TruVortex_07 That's what my answer did. It moved one matchstick. $\endgroup$ Commented Jul 26, 2020 at 20:40
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$\begingroup$ your answer contains something that is not an actual number though $\endgroup$ Commented Jul 26, 2020 at 20:44
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2$\begingroup$ @TruVortex_07 They are still numbers. We just need to think outside of the box ;) $\endgroup$ Commented Jul 26, 2020 at 20:50
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This is from a programming perspective:
$6 + 0 \vDash 1$
Explanation:
$\vDash$ is entailment. $6 + 0 = 6$. It has a boolean value of True. Also, it entails ($\vDash$) $1$ which is also a boolean equivalent of True. So, in all the models where $6$ is true (obviously always), $1$ is true (again, obviously always). There is no model where either $6$ or $1$ or both are considered False. Both are always True. And the only number equating to false is $0$
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$\begingroup$ The OP said to use only the "=" operator. $\endgroup$ Commented Jul 26, 2020 at 21:07
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Here's a solution with a "little" cheating:
First:
Break the $7$ top match in two:
Then:
Place the two pieces like this:
And finally:
Look at your scheme upside down:
I know the first step is a bit far-fetched. I was just having a little fun :)
