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
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 $11/1=11^1$
Using
Roman numerals,
Move the last two matchsticks on the right, changing them to V.
IIII = IV
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.
$ VII = 111 $
Roman $7$ = Binary $7$
2. and another one...
$$ | -11 | = 11 $$ $$ abs(-11) = 11 $$
3. ...and another one...
$ II = 10 $
Roman $2$ = Binary $2$
4. ... and another "new" concept...
$$-iiii = -1$$ with the imaginary unit $i^2 = -1$ $$-i*i*i*i = -1$$
5. ...although the "game is over"...
...a new combination from known principles...
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...)
$ XI = +11 $
Roman $11$ = $+11$
7. ... yet another unconventional "rot90" version...
Roman $2$ = 90° rotated $2$
8. ... another one, maybe a bit too nerdy or ... ?
$ 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...
$ 11 = \omega $
Binary $3$ = 90° clockwise rotated 3
10. ... sorry, again... a simple one which is not yet listed in any of the answers...
$ II - 1 = +1 $
Roman $2$ - $1$ = $+1$
11. ... and a last(?) one...
$ II + 1 = 11$
Roman $2$ + $1$ = Binary $3$
12. ... some other bases for new options...
(base 4) $11 = 5 $ (rotated by 90°)
$11_4 = 5 $
13. ... and combined with earlier ones...
$ VIII = 11 $ (base 7)
Roman $8 = 11_7 $
hmmm, I couldn't resist...
14. ...for your info(rmatics)...
$ 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...
$ 71 - 1 = 11 $
$ 71_{10} - 1 = 11_{69} $ (base 69, ok very special)
16. ...seven & eleven fo(u)r your convencience ...
$ 11 - 7 = 11 $
$ 11_{10} - 7_{10} = 11_3 $
$ 4 = 4_{10} $
17. ... and finally(?) a nicer one...
(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.
31. - 33.
... 111-Hattrick... some nice $111$-solutions should be mentioned:
$$111_3 = 1D_{16} \rightarrow 13_{10} = 13_{10} $$ $$111_5 = 1F_{16} \rightarrow 31_{10} = 31_{10}$$ $$111_{15} = F1_{16} \rightarrow 241_{10} = 241_{10}$$
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.
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°.
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
|_| |\| = |
A simple answer to this could be:
| I | = | i |
_____ | | | | | | | | | --- | | | | | | --- | | | | __|__ | | | |
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.