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The riddle is as follows:

The figure from below represents a set of roman numerals wrote in such a way that the sum is not correct.

The numbers are arranged as:

50+12=51

Sketch of the problem

How many matches minimum should me moved to make operation correct?

The alternatives given in my workbook are as follows:

  1. 2
  2. 4
  3. 1
  4. 3

I found this puzzle in my riddles book Logical fun and challenges from 2000's. It seems to be an adaptation from an old IQ test, can't say which exactly.

This problem has an official solution from my workbook, and it is presented in the figure from below.

Sketch of the solution

Therefore it indicates that the answer is 2, which to me, doesn't make any sense.

The reason for that is I can't recall any weird roman numeral with the shape indicated in the official answer. I'm referring to the letter T. Could it be that I have the wrong interpretation?

Is there a way to get this right?

Can someone help me here? Perhaps with a guideline for which matchstick someone should look for initially, or which part of the operation should be changed?

For example, I noticed that a sum usually (not always) should be changed for another operation. Sometimes swapping sum by a subtraction is right because the number on the left side would require less movements. I think it also helps to count the number of matchsticks, so that you can get a hint on what is the greatest number which can be formed by using these given the operations you can make. I mean multiplication. I also noticed that, there isn't a way to make a division with the matchsticks because there isn't a set of dots which you could use to make that operation. Although you could make the division L shape which you put below the divisor or dividend in the long hand division algorithm, but this of course requires moving additional matchsticks, and there is finally multiplication which sometimes could help.

So overall, does a rule of thumb for for this exist? And more importantly, is the solution method accurate?

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  • $\begingroup$ Of course, you can just rot13(zbir bar zngpu naq ghea gur "cyhf" vagb n "zvahf" naq gur "rdhnyf" vagb n "abg rdhnyf"), which is probably less wild than the intended answer anyway. $\endgroup$ – wimi Feb 19 at 10:41
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To me it looks like

A dynamic equation where $L$ and $L1$ represent lengths/positions, $V$ represents velocity and $T$ represents time. For a body travelling at constant velocity $V$, its position, $L1$, at time $T$ would be given by in terms of its initial position, $L$, as $$L1 = L + VT$$ Although this specific way of writing this equation is a little unfamiliar, that may be due to the age of the source.

If it puts your mind at ease, I've found another way to do it moving two matches

enter image description here

As samm82 points out in the comments

If you don't like the idea of removing the matches, you can put each match beside an $LI$ to keep the equation correct.

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  • $\begingroup$ Your alternative answer technically removes two matches instead of moving them, but if that's an issue, you can just rot13(chg gur rkgen zngpurf orfvqr gur svsgl-barf gb ghea gurz gb svsgl-gjbf) $\endgroup$ – samm82 Feb 18 at 22:23
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    $\begingroup$ @samm82 yes, true. I tend to be very liberal with the idea of moving. I've moved the matches into my pocket for later use. $\endgroup$ – hexomino Feb 18 at 22:24
  • $\begingroup$ @hexomino Thanks for that. I'm already accepting your answer. I must confess that I would never imagined that the equation was a dynamical equation I agree with you that it seems a very unfamiliar way to put it, specially the subindex one next to L. Your second solution seems to be more in tone with the way these answers are presented. But there is still pending the second part of my question, is there any guideline for solving these problems, are my observations correct?. $\endgroup$ – Chris Steinbeck Bell Feb 18 at 22:24
  • $\begingroup$ @ChrisSteinbeckBell Ah, yes, sorry I forgot to address that part of your question. From my experience, there is no guideline. These matchstick type problems often encourage you to think creatively - there is usually no one "right" answer. Whenever this kind of question comes up on PSE there are usually seven or eight very clever and different ways of solving the problem (I'm not sure if I've even seen the type you've presented before). $\endgroup$ – hexomino Feb 18 at 22:29
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I'm pretty sure hexomino's ingenious answer is not what the book intends, and that the mysterious thing is actually a square root sign on a Roman "I". (Which is pretty stupid, because of course no one ever used square root signs with Roman numerals.)

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  • $\begingroup$ Ooooh, I think you might be right, actually, would never have seen that as a single entity. $\endgroup$ – hexomino Feb 18 at 22:32
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    $\begingroup$ If this is the intended rationale, it feels pretty misleading to have the bar touch the 1 and not the "v". Either way, the answer from the book is absolutely wild when @hexomino's alternative is much more intuitive $\endgroup$ – samm82 Feb 18 at 22:37

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