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I got a puzzle from a book that asks to move one matchstick to get a valid equation. My "solutions" feels like cheating... Does anyone have a good idea? The original equation, layed out with matches, is

VI = V + VII

the original picture. Sorry for the low quality

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  • $\begingroup$ Is the I made of 1 or 3 sticks? $\endgroup$ – Omega Krypton Dec 29 '18 at 14:07
  • $\begingroup$ It is only 1 stick $\endgroup$ – Marc Dec 29 '18 at 14:09
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    $\begingroup$ Please add the source of the puzzle $\endgroup$ – Dr Xorile Dec 29 '18 at 14:32
  • $\begingroup$ I don't have a source. It's a picture a friend showed me. And she made it a couple of days agoe in a mountain hut. $\endgroup$ – Marc Dec 29 '18 at 14:47
  • $\begingroup$ There appear to be further matches above/below the equation shown, but which don’t appear to relate to other equations... could they be relevant? $\endgroup$ – eggyal Dec 30 '18 at 2:08
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Move the right most I so that it forms a square root sign with the right most V. So it looks something like this: $VI=V+\sqrt{1}$

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  • $\begingroup$ That is a nifty idea. $\endgroup$ – Marc Dec 29 '18 at 14:53
  • $\begingroup$ Really nice one :) $\endgroup$ – ABcDexter Jan 6 at 18:31
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Something like this:

VII = V | VII (because 7=5|7, using bitwise |)

or:

VI ≠ V + VI

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    $\begingroup$ Bitwise or is very clever. The ≠ trick doesn't work because you don't have a valid equation $\endgroup$ – Dr Xorile Dec 29 '18 at 14:50
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According to Wikipedia (emphasis added):

Zero

The number zero does not have its own Roman numeral, but the word nulla (the Latin word meaning "none") was used by medieval scholars in lieu of 0. Dionysius Exiguus was known to use nulla alongside Roman numerals in 525.[32][33] About 725, Bede or one of his colleagues used the letter N, the initial of nulla or of nihil (the Latin word for "nothing"), in a table of epacts, all written in Roman numerals.[34]

Therefore:

Move the final I to the front of the second V such that it forms an (italicised) N and one gets:

VI = N + VI

Which, in Arabic numerals, Bede (at very least) would have considered to be:

6 = 0 + 6

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  • $\begingroup$ Awesome first answer. $\endgroup$ – WAF Dec 30 '18 at 9:20
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How about this?

From: the right stroke of the right most V
To: intersect with the I on the right.
Outcome: VI=V+IxI (take x as multiplication sign)

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  • $\begingroup$ These puzzles allow for some lateral thinking, but usually a slanted line cannot be a 1 $\endgroup$ – Dr Xorile Dec 29 '18 at 14:44
  • $\begingroup$ I also though about something like that. But then the 1 is not really a 1. $\endgroup$ – Marc Dec 29 '18 at 14:49
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Take the matchstick that makes the final I, snap it in two, and place each half beneath the initial V to create an X. You’d therefore have:

XI = V + VI

Which, in Arabic numerals, is of course the valid equation:

11 = 5 + 6

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0
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How about:

moving the second I in VII so it turns the equal sign equal to a less than or equal sign (VI is less than or equal to V + VI)

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    $\begingroup$ Then you wouldn't have "a valid equation"! $\endgroup$ – Dr Xorile Dec 29 '18 at 15:07
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Partial answer-ish

Looking at the first V, if you take the right match and cross it over the first match, you get XI = V + VII or 11 = 5 + 7

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    $\begingroup$ That equation isn't correct, though. $\endgroup$ – Emmabee Dec 29 '18 at 17:11
  • $\begingroup$ Yeah, still working on it... $\endgroup$ – Chris Happy Dec 29 '18 at 17:12

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