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Use some creative thinking to make the equation below correct. You must move 2 matches. You cannot alter/modify the = sign. You cannot remove any matches or put one over another match (to make 2 matches look like 1).

Text version

1 1 1 1 = 1 1 1

enter image description here

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11 Answers 11

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One easy way:

$1\times 11 = 11$

    o       o    o            o    o
    | o   o |    |            |    |
    |  \ /  |    |   ------   |    |
    |   \   |    |            |    |
    |  / \  |    |   ------   |    |
    | /   \ |    |            |    |
    |       |    |            |    |
 

Other possibilities (sorry, no ascii art):

$11/11 = 1^1$ and $11/1=11^1$

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    $\begingroup$ reminds me of candybox $\endgroup$
    – justhalf
    Oct 30, 2023 at 13:09
  • $\begingroup$ @justhalf Sorry, I don't follow. Is that a reference to something? $\endgroup$
    – loopy walt
    Oct 30, 2023 at 13:11
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    $\begingroup$ Well, your ASCII-art matches look like the lollipops here. (Request features until the map is unlocked. Then visit the shop. Don't plan on doing anything useful today ...) $\endgroup$
    – M Oehm
    Oct 30, 2023 at 13:32
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Using

Roman numerals,

Move the last two matchsticks on the right, changing them to V.
IIII = IV

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    $\begingroup$ I would argue this doesn't follow the spirit of roman numerals, but nearly every roman numeral clock would disagree, so +1 $\endgroup$ Oct 31, 2023 at 17:18
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    $\begingroup$ @JoelRondeau the clock on Elizabeth Tower, Westminster (aka Big Ben) is a notable exception. $\endgroup$ Oct 31, 2023 at 17:40
  • $\begingroup$ Checked our local one (Allen-Bradley Clock Tower), but unfortunately, it has no numbers. $\endgroup$ Oct 31, 2023 at 17:55
  • $\begingroup$ @JoelRondeau sundials too, but there could be many of those that use IV. There are several theories about why IIII came to be used but now it's tradition. $\endgroup$ Oct 31, 2023 at 19:20
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Slide two matches down to get:

1^111=1^11

Left side is $1^{111}=1$
Right side is $1^{11}=1$

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    $\begingroup$ Nice. Variant $111^1=111$ moving one from the right side to the left as exponent $1$ and replacing it with the one from left side. $\endgroup$
    – z100
    Oct 31, 2023 at 17:47
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+150
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Yet another a bit "special" solution, but I would say it is not against the (current) rules, it's just a matter of interpretation.

1.

enter image description here

$ VII = 111 $
Roman $7$ = Binary $7$

2. and another one...

enter image description here

$$ | -11 | = 11 $$ $$ abs(-11) = 11 $$

3. ...and another one...

enter image description here

$ II = 10 $
Roman $2$ = Binary $2$

4. ... and another "new" concept...

enter image description here

$$-iiii = -1$$ with the imaginary unit $i^2 = -1$ $$-i*i*i*i = -1$$

5. ...although the "game is over"...

...a new combination from known principles...

enter image description here

Can be interpreted as $|1| = |-1|$ or $|i| = |-i|$ or $|1| = |-i|$ or $|i| = |-1|$

6. ... and a slightly "odd" one ... (maybe I should stop now...)

enter image description here

$ XI = +11 $
Roman $11$ = $+11$

7. ... yet another unconventional "rot90" version...

enter image description here

Roman $2$ = 90° rotated $2$

8. ... another one, maybe a bit too nerdy or ... ?

enter image description here

$ 11 = 11 | 11 $
in many programming languages $|$ is a symbol for bitwise inclusive OR
so, this can be interpreted as
binary $3$ = binary $3$ OR binary $3$
or
decimal $11$ = decimal $11$ OR decimal $11$

9. ... another combination of principles from above...

enter image description here

$ 11 = \omega $
Binary $3$ = 90° clockwise rotated 3

10. ... sorry, again... a simple one which is not yet listed in any of the answers...

enter image description here

$ II - 1 = +1 $
Roman $2$ - $1$ = $+1$

11. ... and a last(?) one...

enter image description here

$ II + 1 = 11$
Roman $2$ + $1$ = Binary $3$

12. ... some other bases for new options...

enter image description here

(base 4) $11 = 5 $ (rotated by 90°)
$11_4 = 5 $

13. ... and combined with earlier ones...

enter image description here

$ VIII = 11 $ (base 7)
Roman $8 = 11_7 $

hmmm, I couldn't resist...

14. ...for your info(rmatics)...

enter image description here

$ 7 \wedge 11 = 1 $
$\wedge$ is used as binary exclusive OR (XOR)
$7_{10} \wedge 11_5 = 1$
$ 7_{10} \wedge 6_{10} = 1 $
$ 111_2 \wedge 110_2 = 1 $

15. ... a special move...

enter image description here

$ 71 - 1 = 11 $
$ 71_{10} - 1 = 11_{69} $ (base 69, ok very special)

16. ...seven & eleven fo(u)r your convencience ...

enter image description here

$ 11 - 7 = 11 $
$ 11_{10} - 7_{10} = 11_3 $
$ 4 = 4_{10} $

17. ... and finally(?) a nicer one...

enter image description here

(binary) $ 1111 = F$ (hexadecimal)
$1111_2 = F_{16}$
$15 = 15$

18.- 30. ...a Roman's ocean of elevens...

Actually, $11_n = (n+1)_{10}$. Hence $11$ can be interpreted as any decimal number $>=3$ depending on the base $n$. So, this boils down to which numbers can be created on the other side having 5 matches with exactly 2 moves.
You can do this for example with the following Roman numbers:
$$ VIII, XIII, XIV, XVI, XIX, XXI, LIV, LVI, LIX, LXI, CII, DII, M $$ Actually, $VIII$ has been used already above, but you could also use some less common writings of roman numbers, e.g. $IIC$.
For illustration, just the last 3 versions.

enter image description here

31. - 33.
... 111-Hattrick... some nice $111$-solutions should be mentioned:

enter image description here

$$111_3 = 1D_{16} \rightarrow 13_{10} = 13_{10} $$ $$111_5 = 1F_{16} \rightarrow 31_{10} = 31_{10}$$ $$111_{15} = F1_{16} \rightarrow 241_{10} = 241_{10}$$

34. - $\infty$

For the arrangement $11111_m = 11_n$ you can basically find an infinite number of suitable bases for $m$ and $n$.

Now, I guess it is enough... unless somebody wants more.

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  • $\begingroup$ So many interesting ideas! $\endgroup$ Oct 31, 2023 at 19:54
  • $\begingroup$ Disappointed you didn't use chinese: 一 one · 二 two · 八. eight and 十 ten. $\endgroup$
    – Florian F
    Nov 2, 2023 at 12:52
  • $\begingroup$ @FlorianF haha, good point. I did not check other numerals. Maybe you want to post an answer combining these for getting equality? $\endgroup$
    – theozh
    Nov 2, 2023 at 13:47
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Perhaps

|||-|=||

You take the

right-most match (from the right side of the equal sign) and place it before the left-most match (of the left side of the equal sign). The third match on the left of the equal sign you just rotate it 90°.

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  • $\begingroup$ You should state that you are interpreting these as Roman numerals. $\endgroup$ Oct 31, 2023 at 20:34
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1N = N
N is a variable
1 * N = N

another answer

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Going French

UN = 1
Two matches from the right hand side moved to the left hand side to create U out of the first 11 and N out of next 11
|_| |\| = |

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A simple answer to this could be:

| I | = | i | _____ | | | | | | | | | --- | | | | | | --- | | | | __|__ | | | |

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    $\begingroup$ Where did the eighth match come from? $\endgroup$ Nov 1, 2023 at 13:50
  • $\begingroup$ @WeatherVane Strictly speaking, the problem only forbade removing matches, not adding them. $\endgroup$ Nov 2, 2023 at 11:45
  • $\begingroup$ @eyeballfrog the extra match must have been removed from somewhere else :) And as for the possibility of snapping a match into two, the rule forbids making 2 matches look like 1, so by inference that would include making 1 match look like 2. $\endgroup$ Nov 2, 2023 at 11:47
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A method I'm surprised hasn't been posted yet:

1 + 1 - 1 = 1

One match has been moved over the second from the left to make a + while the other makes a -

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    $\begingroup$ well, how many matches do you need to move in order to get this layout from the original arrangement (including original distances between the matches)? $\endgroup$
    – theozh
    Nov 2, 2023 at 8:46
  • $\begingroup$ True—I assumed the author's exact spacing was not intentional, and that the general layout was the main puzzle. I think it's fair to judge this as not conforming the original rules $\endgroup$ Nov 2, 2023 at 16:24
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| | - | = | / |

2 - 1 = 1/1 (to fill up 30 characters!)

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1 x 11, move one match from the right and place it diagonally. Then, place a match from the left on top of the one you just moved to form an "x" shape between the first match and the last two matches. If done correctly, it should look like 1 x 11.

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    $\begingroup$ This solution was posted yesterday $\endgroup$ Oct 31, 2023 at 17:43
  • $\begingroup$ Sorry. I didn't look at the answers that were previously posted and I just solved it myself. If you want me to delete my response I will. $\endgroup$ Nov 2, 2023 at 17:28

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