Just move two matchsticks to find the equality in the equation below:
Note: There are two reasonable answers.
$\begingroup$
$\endgroup$
9
-
$\begingroup$ Can I add a match, and only move one? $\endgroup$– MithicalJul 14, 2016 at 9:48
-
$\begingroup$ @Mithrandir just by moving. $\endgroup$– OrayJul 14, 2016 at 9:48
-
$\begingroup$ I can do it by moving one... $\endgroup$– WillJul 14, 2016 at 9:55
-
3$\begingroup$ Is making a "non-equal" reasonable? :) $\endgroup$– YousendJul 14, 2016 at 12:33
-
27$\begingroup$ Move $0$ matches: read it in base $45$. $\endgroup$– Jonathan AllanJul 14, 2016 at 13:44
|
Show 4 more comments
11 Answers
$\begingroup$
$\endgroup$
10
Moving two:
Rotate table 180°Walk around the table and:
(matches moved: horizontal bar in the 4, horizontal bar in left 7)
-
$\begingroup$ yes this is what I was looking for and a proper answer :) $\endgroup$– OrayJul 14, 2016 at 10:36
-
-
4$\begingroup$ Can you explain what you see on this picture. Because I see
111 = c * 1. $\endgroup$– talexJul 14, 2016 at 12:35 -
3
-
14$\begingroup$ Isn't rotating the table moving all the matches? $\endgroup$ Jul 14, 2016 at 13:18
$\begingroup$
$\endgroup$
1
Another way moving $2$:
i.e. $\frac77=1^4$
Or similarly:
i.e $1\times1=1^4$
answered Jul 14, 2016 at 14:01
-
5$\begingroup$ Or alternatively, you could use the two sticks to extend the "1" downward, on the RHS, making the 4 look like a proper superscript. $\endgroup$ Jul 14, 2016 at 16:24
$\begingroup$
$\endgroup$
3
Yet another way moving $2$:
i.e $7\times7=IL^I$ using Roman numerals on the right hands side: $IL^I=49^1$
answered Jul 14, 2016 at 14:18
-
1
-
3$\begingroup$ The charts I've seen suggest that subtraction by prefix is only applicable for producing 4, 9, 40, 90, 400, 900, or similar larger values using over-bar notation. The value 49 is properly written as 40+9, (XLIX), not 50-1 (IL). Similarly, 1999 is 1000+900+90+9, (MCMXCIX), not 2000-1 (MIM). $\endgroup$– supercatJul 14, 2016 at 18:18
-
1$\begingroup$ @supercat yeah you're correct, it's not in the recognised "standard" form of XLIX; but it wasn't standardised until long, long after the Roman empire. IL is nothing other than 49 (in fact there is also evidence of double subtractive notation being used too, such as XIIX for 18, which is even more confusing). $\endgroup$ Jul 15, 2016 at 5:00
$\begingroup$
$\endgroup$
0
Well you can do that by
How
The green line is the moved matchsticks. My drawing is not good, though.
$\begingroup$
$\endgroup$
Although, I personally feel that Will seems to have the best solution. We may even do this by removing 2 matches.
$\begingroup$
$\endgroup$
5
Moving either the leftmost or the rightmost matches making up the X to between the two digits 1 and 4 gives 2 different equations using boolean algebra
7>7 = 1 > 4
7<7 = 1 > 4
in both cases
the boolean value of both LHS and RHS simplify to false resulting in the equations simplifying to false = false
i.e. with both sides being equal.
-
1$\begingroup$ Welcome to puzzling.SE. Although your answer might seem ok as
lateral-thinking, but you need to "find the equality" :) $\endgroup$ Jul 15, 2016 at 14:57 -
1$\begingroup$ I think you need to reread my answer - it is expressed as a logic equation i.e. both the LHS and the RHS are equal To clarify taking the first example LHS expression is 7>7 - this simplifies to FALSE RHS expression 1>4 - this also simplifies to FALSE This means the whole equation simplifies to FALSE = FALSE - and that is the equality $\endgroup$ Jul 15, 2016 at 16:51
-
$\begingroup$ I read you answer thrice before editing. There is no clear evidence that you used any function to calculate the Boolean value of the expressions of both LHS and RHS, which would thus make them equivalent, which is not obvious in your case. And as the saying goes, "The absence of evidence isn't the evidence of absence itself." $\endgroup$ Jul 15, 2016 at 17:02
-
$\begingroup$ Also, you need to re-read your answer(under "in both cases") and look for the explanation I put in so, False=False is True :) $\endgroup$ Jul 15, 2016 at 17:05
-
$\begingroup$ I agree with this answer, $False=False$ and $x>y$ is a boolean expression $\endgroup$ Jul 16, 2016 at 23:46
$\begingroup$
$\endgroup$
Moving only ONE match:
And rotating 180 degrees:
We get 61 = LXI (61 in roman)
$\begingroup$
$\endgroup$
1
I feel like the simplest answer is just to
move the two matches in the equals sign to make it a greater than sign : 7x7 > 14
-
9
$\begingroup$
$\endgroup$
3
So how about just:
Make use of the fact that they are matches!
Take the top match of the first digit, making the 7 into a 1.
Add the match you took to somewhere in the last digit.
Then take top of the second digit, making also that 7 into a 1.
But before you also add that match to somwhere in the last digit,
LIGHT THE MATCH, and when the fire is out, you will have:
.
. _ _
. | \/ | _ | |_| INTO | \/ | _ |
. | /\ | _ | | | /\ | _ |
-
-
$\begingroup$ @wizzwizz4 what do you think I missed from the tour? $\endgroup$– nl-xJul 18, 2016 at 7:28
-
$\begingroup$ Nothing in particular, just that you hadn't read the tour. I have even less reputation than you here, so you must be doing something better than me! :-) $\endgroup$ Jul 18, 2016 at 17:14
$\begingroup$
$\endgroup$
1
Without attempting to draw a picture ...
If I take 2 matches from the X and move them to the =, I get 7/7 ≠ 14
-
5$\begingroup$ Yes, but the question is "to find the equality". Yours is an inequality. $\endgroup$– M OehmJul 14, 2016 at 13:52
$\begingroup$
$\endgroup$
7*7 = 49, we can count in ${\mathbb Z}_{10}$ ( or in normal language "only save last digit" ), and then only move the 2 sticks in the "1" in 14 to make a 9 out of the 4.
