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Something rather funny happened at our Puzzlers' meet the other day.

"Oh hey buddy, can you hand me exactly 18 matchsticks? I wanna show you this new matchstick puzzle I came up with." Matt the Matchstick Puzzler asked.

"Sure, here you go" I handed him the matchsticks, curious.

Matt laid them down on the table in front of us, forming a equation. "You have to move the least number of matches to make this into a correct equation."

"This looks interesting!" I remarked. "I like how blatantly wrong this equation is; I mean, of course one prime number times another bigger prime number cannot be a perfect cube!"

"Pfft, too easy", snarked Morgan the Modulo Man, standing next to us both. "The answer is zero matches; the equation is already true modulo 5."

Matt was annoyed. "Shut up Morgan, the whole 'look-modulo-this' or 'look-in-that-base' hack stopped being funny a long time ago."

Trying to change the topic, I asked, "On question though, if you were going for seven-segment display style, why are all the ones made of one matchstick each?"

"Probably so that we could do this", Evan the Engineer swooped in, and moved one match from the right side of equation to the left.

"Come on!" Matt replied, "I mean, this is a number, but really? Plus this isn't even true!"

"Eh, close enough for me." Evan shrugged.

"Ooh, maybe this works?" I picked up the match Evan had moved, and put it somewhere else.

"How exactly? One side is bigger than the other by more than three hundred! In fact, the difference is a perfect square!" Matt was confused.

I explained, "Yes, unless you read the second number as Roman numerals!"

"You can't go reading random numbers in Roman when everything else in the equation is decimal!" Matt cried.

"Yeah, that's just stupid" Morgan commented. "However, you can do this!" He now moved another match from the right side to the left.

"How on earth is that supposed to help?" Matt was visibly irritated at this point.

"Because it's now true modulo 83, see?" Morgan said.

At this point Phil the Physicist came from the other side of the table, looked at the matches for a bit, and moved two matches. "There you go, a correct equation! And a famous one to boot!"

"Ahh, that's clever, why didn't I think of that?" Evan was impressed.

"No, no, no, you can't just find weird lateral-thinking solutions to these and feel clever, that's cheating!" I had never seen Matt so furious. "Here, let me show the actual answer." He moved four matches to get to the initial state, and went, "See, you just take these two matches making the multiplication sign, and put them here, and here."

"Ah, I see!" I exclaimed. "We were so caught up with changing the numbers, none of us thought to change the operation!"

As interesting as the exchange was, I can't remember the actual puzzle for the life of me!

Can you figure out what the initial puzzle was and what solutions everyone came up with from the above conversation?

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  • $\begingroup$ Gb or pyrne, vs gur grezf ner n k o = p, jura Rina fnvq gung gur rdhngvba vf nyernql gehr zbqhyb 5, qvq ur zrna guvf: n, o, p ner nyernql va zbqhyb 5 naq urapr, guvf znxrf vg gehr BE guvf: vs lbh pbaireg n, o naq p gb zbqhyb 5 gur rdhngvba jvyy or gehr? $\endgroup$ – John Brookfields Jun 23 '20 at 6:18
  • $\begingroup$ @John the second one. For example, it could have been 6x7=2. $\endgroup$ – Ankoganit Jun 23 '20 at 6:21
  • $\begingroup$ Oh, I thought modulo in the sense base-5. Now, I get it. $\endgroup$ – John Brookfields Jun 23 '20 at 6:32
  • $\begingroup$ Cyrnfr purpx gur sbyybjvat yvax: (onfr64) nUE0pUZ6Yl9jLKA0MJWcov5wo20irRbmpJWUIIZ= $\endgroup$ – John Brookfields Jun 23 '20 at 7:21
  • $\begingroup$ @John Looks good so far! $\endgroup$ – Ankoganit Jun 23 '20 at 7:24
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The answer is:

$3\times 11 = 8$

Reason:

The total number of matches used for numbers alone is 14 ($=$ and $\times$ consume 4 matches in total. So $18-4 = 14$). Also, $3\times 11 = 33 \equiv 3 (\text{mod }5)$ and $ 8 \equiv 3(\text{mod }5)$. Also, we can remove one match from $8$ to make it $9$ and put it in $11$ to make it $111 = 3$ (Roman) and hence $3\times 111 (=3) = 9$. Also, $3\times 111 = 333$ and it is greater the right hand side by $324 = 18^2$ Taking one match from $9$ to make it $3$ and putting it in LHS $3$ to make it $9$ we get $9\times 111 (\equiv 28 \text{ (mod }83)) = 9\times 28 \equiv 3 \text{ (mod }83)$. Now the equation looks like this: $$9\times 111 = 3$$. When the physicist comes from the other side of the table he must read $E = 111 \times 6$. All I knew is that it must be made to $E=mc^2$. Or we can make it as: enter image description here i.e. $E = \text{Nu} \times h$. But I don't know if it could be done. Or it could be $E = w\times h$ where $w = mg$ is the weight of the particle (object) and $h$ is the height at which the particle or object is located. Thus it is the potential energy of the object. Finally from the initial state, we get, $3 - 11 = -8$ (Taking the two matches from $\times$ and placing one in the same place as $\times$ and another in front of $8$

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  • $\begingroup$ On the right track, but some more thinking is needed on the last equation! $\endgroup$ – Ankoganit Jun 23 '20 at 8:32
  • $\begingroup$ Yeah @Ankoganit. I will do it. Only the last two remaining. $\endgroup$ – John Brookfields Jun 23 '20 at 8:37
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    $\begingroup$ Maybe for Evan's equation you have to rot13(chg gur fvatyr zngpufgvpx ba gbc bs gur gjb 'barf' gb znxr cv) which is pretty close and related to a famous joke. $\endgroup$ – hexomino Jun 23 '20 at 16:28
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    $\begingroup$ ^ I agree with @hexomino - the step you're missing is this. $\endgroup$ – Stiv Jun 23 '20 at 22:25
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    $\begingroup$ It's also possible that the Physics equation is this, i.e. rot13(S = zN (Arjgba'f Frpbaq Ynj) jurer S naq N ner haqreyvarq (va erq sbe rzcunfvf) orpnhfr gurl ner irpgbef.) - Wiki page for reference $\endgroup$ – Stiv Jun 23 '20 at 22:37

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