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Posted on 3/6/14 at 4:53 pm to MIZ_COU
Got it.
1=111-11
Move the one stick from on top of the equal sign to the minus.
1=111-11
Move the one stick from on top of the equal sign to the minus.
Posted on 3/6/14 at 4:54 pm to CheeseburgerEddie
quote:
I also think my move one of the sticks to turn into does not equals should be a good solution.
i-ii=/=ii
Well, I like that one too, except that the majority won't recognize "=/=" as "not equal to".
Plus you broke the "don't cross the streams" rule that the OP neglected to mention.
Posted on 3/6/14 at 4:54 pm to Duke
No but good try
Posted on 3/6/14 at 4:55 pm to MIZ_COU
| -|| | = ||
very clever. I liked that one.
very clever. I liked that one.
This post was edited on 3/6/14 at 4:56 pm
Posted on 3/6/14 at 4:56 pm to finestfirst79
quote:
Well, I like that one too, except that the majority won't recognize "=/=" as "not equal to". Plus you broke the "don't cross the streams" rule that the OP neglected to mention.
Yeah actually you would have to move a number of sticks plus create a couple more to make this. Rememeber it's match sticks and you are then looking at the resulting image and reading it as an equation
Posted on 3/6/14 at 4:58 pm to CheeseburgerEddie
If that counts as zero moves, it's a horseshite question.
Spreading them out =/= move.
Spreading them out =/= move.
Posted on 3/6/14 at 4:58 pm to MIZ_COU
You only have to move 1 stick as "=/=" isn't the actual notation, it is just the only way to type it here.
The no crossing thing messes it up. But is my absolute value solution not correct?
The no crossing thing messes it up. But is my absolute value solution not correct?
Posted on 3/6/14 at 4:59 pm to Duke
no that's not the answer
plus i should have added no inequalities allowed but the actual answer is still valid
plus i should have added no inequalities allowed but the actual answer is still valid
This post was edited on 3/6/14 at 5:01 pm
Posted on 3/6/14 at 5:16 pm to MIZ_COU
Duke got it, unless there's another 1-move answer.
Posted on 3/6/14 at 5:18 pm to finestfirst79
oh, never mind. I see now.
This post was edited on 3/6/14 at 5:21 pm
Posted on 3/6/14 at 5:22 pm to CheeseburgerEddie
quote:
3)You have three opaque boxes. One box contains chocolate candies, another contains mint candies, and the last box contains a mixture of chocolate and mint. The boxes are labeled Chocolate, Mint and Mixed. None of the boxes are labeled correctly. You can take one candy out of each box (without looking directly into the box) and see what you get. What is the minimum number of boxes you have to open (and take one candy out) to assign correct labels to all boxes?
If the mint has a strong smell, I think you can solve this without actually having to remove any candies at all. Two of the boxes smell of mint. The other is the chocolate box. It gets the Chocolate label. Whichever box had the Chocolate label gets whatever label was on the other mint-smelling box. The remaining box gets the label that was on the one now labeled Chocolate.
If smell is not an option, take a candy from the box labeled Mixed. Since it can't be mixed, whatever you draw out is the label it should receive (either Mint or Chocolate). Then whichever box had the label that is now on the box originally marked Mixed gets the label from the 3rd box, which then receives the Mixed label.
Posted on 3/6/14 at 5:23 pm to InThroughTheOutDore
quote:
If smell is not an option, take a candy from the box labeled Mixed. Since it can't be mixed, whatever you draw out is the label it should receive (either Mint or Chocolate). Then whichever box had the label that is now on the box originally marked Mixed gets the label from the 3rd box, which then receives the Mixed label.
thats the one.
Now throw one back at us.
This post was edited on 3/6/14 at 5:27 pm
Posted on 3/6/14 at 5:51 pm to finestfirst79
No Duke did not get it no one has come close
Posted on 3/6/14 at 5:52 pm to MIZ_COU
Why is my absolute value response not valid?
ETA: I'll think on this one some more at home. See if I can find an acceptable answer for you.
ETA: I'll think on this one some more at home. See if I can find an acceptable answer for you.
This post was edited on 3/6/14 at 5:56 pm
Posted on 3/6/14 at 5:55 pm to CheeseburgerEddie
It's just not correct, and now we are giving out too many hints.
Also I do now remember the original puzzle said no inequalities when I read it, but no matter. No valid argument can be made that the inequality answer is as good of an answer as the actual answer
Also I do now remember the original puzzle said no inequalities when I read it, but no matter. No valid argument can be made that the inequality answer is as good of an answer as the actual answer
This post was edited on 3/6/14 at 5:57 pm
Posted on 3/6/14 at 5:56 pm to MIZ_COU
quote:
I - III = II
III = III
Eta nvm
This post was edited on 3/6/14 at 5:57 pm
Posted on 3/6/14 at 5:58 pm to UMRealist
You got rid of a matchstick. Which at the least would have to count as a move. So no
Duke's answer, while good, can also not be argued to be as good or better than the correct answer
and all of these things are hints
Duke's answer, while good, can also not be argued to be as good or better than the correct answer
and all of these things are hints
This post was edited on 3/6/14 at 6:08 pm
Posted on 3/6/14 at 6:14 pm to CheeseburgerEddie
Answering the OP before I read the thread to see if I got em right:
1)Start the 4 and 7 min hourglass at the same time. Once the 4 ends, flip it. Once the 7 ends, flip it. When the 4 ends again, at the 8 minute mark, the 7 will have elapsed 1 minute. Flip the 7 back over to countdown the minute and end at 9 minutes.
2)1. Flip two on, wait a while, then flip one of the two back off. Enter the room. If the light is on, the remaining on switch controls the light. If the light is off, feel it. If it is warm to the touch, the switch you flipped twice controls it. Otherwise, the third switch controls it.
3)1. The chocolate box contains either mint or mixed. If you take one out and it is mint, then you know that the mixed box cannot contain mint or mixed, and must contain chocolate, and then the chocolate box must contain mixed. Same logic applies for every other combination of boxes.
4)9,2,2. For the sum to not be enough information, their must be another combination of ages that adds up to the same number. For 3 numbers that multiply to 36, the only combinations that add to the same number are 9,2,2 and 6,6,1. Since only the 9,2,2 combination has a single oldest, it must be that one.
1)Start the 4 and 7 min hourglass at the same time. Once the 4 ends, flip it. Once the 7 ends, flip it. When the 4 ends again, at the 8 minute mark, the 7 will have elapsed 1 minute. Flip the 7 back over to countdown the minute and end at 9 minutes.
2)1. Flip two on, wait a while, then flip one of the two back off. Enter the room. If the light is on, the remaining on switch controls the light. If the light is off, feel it. If it is warm to the touch, the switch you flipped twice controls it. Otherwise, the third switch controls it.
3)1. The chocolate box contains either mint or mixed. If you take one out and it is mint, then you know that the mixed box cannot contain mint or mixed, and must contain chocolate, and then the chocolate box must contain mixed. Same logic applies for every other combination of boxes.
4)9,2,2. For the sum to not be enough information, their must be another combination of ages that adds up to the same number. For 3 numbers that multiply to 36, the only combinations that add to the same number are 9,2,2 and 6,6,1. Since only the 9,2,2 combination has a single oldest, it must be that one.
This post was edited on 3/6/14 at 6:17 pm
Posted on 3/6/14 at 6:15 pm to MIZ_COU
I don't see how it can be any lower than 1 matchstick moved
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