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
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$\begingroup$ Is the I made of 1 or 3 sticks? $\endgroup$– Omega KryptonDec 29, 2018 at 14:07
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$\begingroup$ It is only 1 stick $\endgroup$– MarcDec 29, 2018 at 14:09
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2$\begingroup$ Please add the source of the puzzle $\endgroup$– Dr XorileDec 29, 2018 at 14:32
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$\begingroup$ 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. $\endgroup$– MarcDec 29, 2018 at 14:47
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$\begingroup$ 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? $\endgroup$– eggyalDec 30, 2018 at 2:08
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7 Answers
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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: $VI=V+\sqrt{1}$
answered Dec 29, 2018 at 14:42
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Something like this:
VII = V | VII (because 7=5|7, using bitwise |)
or:
VI ≠ V + VI
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1$\begingroup$ Bitwise or is very clever. The ≠ trick doesn't work because you don't have a valid equation $\endgroup$ Dec 29, 2018 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
Ito the front of the secondVsuch that it forms an (italicised)Nand one gets:Which, in Arabic numerals, Bede (at very least) would have considered to be:VI = N + VI6 = 0 + 6
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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 $\endgroup$ Dec 29, 2018 at 14:44
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$\begingroup$ I also though about something like that. But then the 1 is not really a 1. $\endgroup$– MarcDec 29, 2018 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 initialVto create anX. You’d therefore have:XI = V + VIWhich, in Arabic numerals, is of course the valid equation:
11 = 5 + 6
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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)
answered Dec 29, 2018 at 14:53
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3$\begingroup$ Then you wouldn't have "a valid equation"! $\endgroup$ Dec 29, 2018 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 + VIIor11 = 5 + 7
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