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Can anyone solve this one?
1 + 11 + 111 = 1111
only move one matchstick on the left to make the equation true.
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1 + 11 + 11 ≠1111
igloo myrtilles fourmis
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I think, if I understand this correctly, that '1' here is not a digit, nor these actual numbers.. I think it's supposed to just represent a quantity of sticks. Using '|' instead of '1':
| + || + ||| = ||||
Let's move one of the sticks from the group of 3 (|||) and place it with the single stick (|) :
|| + || = ||||
Now we have four sticks on both sides, which is true. (2 + 2 = 4)
I don't see any other way this could make sense.
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If we move one of the sticks from the group of 3(III) and place it with the single stick (I):
II + II + II = IIII
we have 6 sticks on the left side , while we have 4 sticks on the other
So ..
polylog solution is wrong
Let me suggest this idea:
I + (II) + III = IIII
I + (+) + III = IIII
Converting the (II) into (+) , will give 1 + 3 = 4
Last edited by mahmoudaljamel (2006-10-09 21:00:21)
If you always do what you always did, you'll always get what you always got
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1 + 11 + 11 ≠1111
John E. Franklin's equation of matchsticks works, even when not referring to digits.
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How about 1 + 11 + 11 ≤ 1111?
Can a less-than sign have 2 horizontal lines at the bottom, rather than 2 angled lines and still be a proper symbol?
Cheating even more, you could say that 1 + 11 + 111 = 1N and define N as 123.
Why did the vector cross the road?
It wanted to be normal.
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Or, to save on sticks, X.
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That would involve moving more than one stick though. You could have H instead of N if you wanted, but I think that's the only variation.
I'm trying to find some solution that doesn't involve changing the = sign or using algebra but counting with a different base instead (probably 2). I haven't found one yet, but I think there's probably one there.
Why did the vector cross the road?
It wanted to be normal.
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If we move one of the sticks from the group of 3(III) and place it with the single stick (I):
II + II + II = IIII
we have 6 sticks on the left side , while we have 4 sticks on the other
So ..
polylog solution is wrong
Right, oops.
So it doesn't look like there is a way to do this by doing nothing but moving only 1 stick on only the left side, unless we are allowed to replace it with an operator or using it to cross out the equals sign or something.
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OMG - I solved this in like 30 minutes and I'm no rocket scientist... Ok, lets look at this using the KISS formula for those of you using trig to move one match stick to make the equation = 1111 correct. Really 2+2+2=1111, where do you come up with this stuff????
Let's look at this, 1+11+111 with 1's being matchsticks = 1111... I'm just going to cut to the chase here - move one matchstick from the "11" and put it on the middle matchstick of the "111" and then you'll end up with 1+1+1+1=1111 - I think the rocket scientist in mensal need to take the test again....Have a great day folks...hope you can sleep now...
Hi;
That is a one matchstick move but how does I + I + I + I = IIII make the equation correct?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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1 + 11 + 11 ≠1111
That's an inequation. That's not an equation.
mathsyperson,
yours is an inequality, so it's not allowed either.
If we move one of the sticks from the group of 3(III) and place it with the single stick (I):
II + II + II = IIII
we have 6 sticks on the left side , while we have 4 sticks on the other
So ..
polylog solution is wrongLet me suggest this idea:
I + (II) + III = IIII
I + (+) + III = IIIIConverting the (II) into (+) , will give 1 + 3 = 4
My calculator gives an error for
1 + + + 3, so I am voting against that.
bobbym,
that works in the Roman numeral system.
IIII also equals IV.
Last edited by reconsideryouranswer (2011-08-25 11:23:45)
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