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lol. Slight of hand mathematics bothers me. Maybe because I have a math degree, I don't know.
The equation I gave you is faulty. You owe your parents $49 each which adds up to $98. All in all you now owe your parents $98. The pair of jeans is $97 plus the dollar you kept is $98. $98 plus the $2 you gave your parents is equal to $100.

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The equation I gave you is faulty. You owe your parents $49 each which adds up to $98. All in all you now owe your parents $98. The pair of jeans is $97 plus the dollar you kept is $98. $98 plus the $2 you gave your parents is equal to $100.

Oh, I know how it works. I'm a mathematician. I just don't like sleight of hand mathematics because there are so many cool actual math puzzles.
 
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Like what?

You have twelve coins that are identical, except one of the weighs slightly more or slightly less than the rest--you don't know which. With no more than three separate weighings on a balance scale, determine which coin is different and whether it weighs more or less. If this is not possible, explain why.
 
You have twelve coins that are identical, except one of the weighs slightly more or slightly less than the rest--you don't know which. With no more than three separate weighings on a balance scale, determine which coin is different and whether it weighs more or less. If this is not possible, explain why.
I could figure out whether it weighs more or less but I wouldn't be able to tell you which one based on the amount of weighings I get. Not that great at math.
 
Okay, I finally got on my computer. Here's the answer:

It helps if you label the coins, but you don't have to. Take 8 of the 12 coins and put 4 on each side of the scale. Either one side will be heavier, or they will balance even. If one side is heavier, take 3 coins from the heavier side and remove them. Put 3 coins from the lighter side on the heavier side and replace them with 3 of the unweighed coins.

If the heavier side is still heavier, then either the original coin on the heavier side is heavier or the original coin on the lighter side is lighter. Balance them against each other to find out.

If the side that was originally heavier is now lighter, that means that one of the 3 coins that was moved from the lighter side was the light coin. Weigh two of them against each other. Whichever is lighter is the coin; if they balance, the odd coin out is the light one.

Finally, if the sides balance, then you know that one of the 3 coins from the originally heavy side is the heavy coin. Do the same as above.

Now, if you do the original weighing and they balance, then you can take all 8 of them off the balance. Take 3 of the unweighed coins and put them on one side, then take 3 of the coins you know are identical and put them on the other side. If it balances, then you know the remaining coin is light or heavy. Weigh it against one of the identical coins to find out which.

If it doesn't balance, then you know that one of those 3 coins is either heavy or light. Weigh two against each other. If they balance, then the third is either heavy or light (obviously if the 3 coins were heavy, then it's heavy; vice versa for light). If they don't balance, then you know the one that goes the same way as the 3 coins is the odd one out and is either heavy or light (if the 3 coins were heavier, then it is heavy; if the 3 coins were lighter, then it's light).

The formula for the maximum number of coins you can do this for using n weighings when you only know that one is different is (3^n - 1)/2. For 3 weighings, that gives us 13 coins, so 12 should work. Though it's not guaranteed.

Alternatively, the formula for how many weighings it will take for n number of coins is log_3(2*n + 1), which gives us 2.9 ~= 3.
 
Okay, I finally got on my computer. Here's the answer:

It helps if you label the coins, but you don't have to. Take 8 of the 12 coins and put 4 on each side of the scale. Either one side will be heavier, or they will balance even. If one side is heavier, take 3 coins from the heavier side and remove them. Put 3 coins from the lighter side on the heavier side and replace them with 3 of the unweighed coins.

If the heavier side is still heavier, then either the original coin on the heavier side is heavier or the original coin on the lighter side is lighter. Balance them against each other to find out.

If the side that was originally heavier is now lighter, that means that one of the 3 coins that was moved from the lighter side was the light coin. Weigh two of them against each other. Whichever is lighter is the coin; if they balance, the odd coin out is the light one.

Finally, if the sides balance, then you know that one of the 3 coins from the originally heavy side is the heavy coin. Do the same as above.

Now, if you do the original weighing and they balance, then you can take all 8 of them off the balance. Take 3 of the unweighed coins and put them on one side, then take 3 of the coins you know are identical and put them on the other side. If it balances, then you know the remaining coin is light or heavy. Weigh it against one of the identical coins to find out which.

If it doesn't balance, then you know that one of those 3 coins is either heavy or light. Weigh two against each other. If they balance, then the third is either heavy or light (obviously if the 3 coins were heavy, then it's heavy; vice versa for light). If they don't balance, then you know the one that goes the same way as the 3 coins is the odd one out and is either heavy or light (if the 3 coins were heavier, then it is heavy; if the 3 coins were lighter, then it's light).

The formula for the maximum number of coins you can do this for using n weighings when you only know that one is different is (3^n - 1)/2. For 3 weighings, that gives us 13 coins, so 12 should work. Though it's not guaranteed.

Alternatively, the formula for how many weighings it will take for n number of coins is log_3(2*n + 1), which gives us 2.9 ~= 3.
For some reason I wasn't thinking of a two sided scale. I was thinking of one single scale.
 
For some reason I wasn't thinking of a two sided scale. I was thinking of one single scale.

Oh, haha. Yeah you could still do it. Not sure if you could do it in 3 though. Be curious to see what the formula would be for a single scale.
 
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You start weighing 4 against 4, say ABCD-EFGH and IJKL are to the side. If they weigh the same, you weigh ABC-IJK and then you'll know whether the off-coin is IJK or L. If it's IJK you do IAD-JBC and you'll know which it is and heavier or lighter. This last step is the 3-coin solve

If the 4-4 is not equal you then do BCE-DFI. If they're equal you do the 3-coin solve with AGH. If the balance changes direction, it's E or D and you use your last weigh to figure out which. If it's still unbalanced in the same way, you do the 3-coin solve with BCF.

Didn't cheat, scout's honor

Edit: forgot 2 words
 
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You start weighing 4 against 4, say ABCD-EFGH and IJKL are to the side. If they weigh the same, you weigh ABC-IJK and then you'll know whether the off-coin is IJK or L. If it's IJK you do IAD-JBC and you'll know which it is and heavier or lighter. This last step is the 3-coin solve

If the 4-4 is not equal you then do BCE-DFI. If they're equal you do the 3-coin solve with AGH. If the balance changes direction, it's E or D and you use your last weigh to figure out which. If it's still unbalanced in the same way, you do the 3-coin solve with BCF.

Didn't cheat, scout's honor

Edit: forgot 2 words

Good job!
 
@JoaoMoutinho @libertyyne @TheBiologist @Gilakend @freak7 @Matthew9Thirtyfive bump!

6f576af010c40135a228005056a9545d
 
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No, that was a question lol. Your answer looks as though you don't need to be switching things across the scales like I did.
Oh, I see what you mean. I thought you were asking if you should post a simpler solution haha. Nah, my solution and your solution are very similar. There's still switching going on. Not really any other way to do it (read: I mean, there are multiple solutions, but all of them involve switching things around).
 
Oh, I know how it works. I'm a mathematician. I just don't like sleight of hand mathematics because there are so many cool actual math puzzles.
65-43=21. You can only move two numbers to make it right.
P.S. How do you say parabola.
 
65-43=21. You can only move two numbers to make it right.
P.S. How do you say parabola.

Technically there is no way to rearrange the numbers to make that correct; however, if you don't consider adding operations to be a violation of the rules, then you can move the 3 to the exponent of 4 and the 2 to the exponent of 1:

65 - 4^3 = 1^2
65 - 64 = 1.

It's puh-RAH-boe-luh
 
Technically there is no way to rearrange the numbers to make that correct; however, if you don't consider adding operations to be a violation of the rules, then you can move the 3 to the exponent of 4 and the 2 to the exponent of 1:

65 - 4^3 = 1^2
65 - 64 = 1.

It's puh-RAH-boe-luh
yep. Heard someone say it differently and wondered if that was an American way of pronouncing it or something.
 
Where do they put the emphasis? If you say parabola fast, the “bo” part can sound like “buh.”
It's a US/UK pronunciation difference. I can't post the link, but the Cambridge dictionary provides audio files of the pronunciation both ways.
 
Good. It's been a stressful year, but I can finally relax. How you been?
I bet with all those classes you have to make good grades on to stay in.

almost done with my neurosurg residency... haha nah. I have been great, brotha. Took me a while to figure out things, but I'm dead set on getting on with a fire department. Maybe PA scchool in the future. I know a captain who just retired and went off to that, but I'm sure I will be happy with just being a firefighter.
 
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I bet with all those classes you have to make good grades on to stay in.

almost done with my neurosurg residency... haha nah. I have been great, brotha. Took me a while to figure out things, but I'm dead set on getting on with a fire department. Maybe PA scchool in the future. I know a captain who just retired and went off to that, but I'm sure I will be happy with just being a firefighter.

My bro in law is a firefighter/paramedic and loves it.
 
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My bro in law is a firefighter/paramedic and loves it.
Yeah, it's a great job. I love the brotherhood, station life, the diversity of things, events to attend (charities), fighting fire, and of course badge bunnies. Jk on the last part haha.
 
Yeah, it's a great job. I love the brotherhood, station life, the diversity of things, events to attend (charities), fighting fire, and of course badge bunnies. Jk on the last part haha.

Lol yeah. Be careful with that last part. They had a female firefighter who slept with every dude in the house, watched porn with some of them, etc., then she turned around and sued the department for sexual harassment.
 
Hi everyone
You every have one of those moments when you're talking to someone- and you blurt out something that comes into your head and end up interrupting them? Bc I just had that while talking to a prof and I can't stop thinking about it.
She knows me well and she was talking to my friend ( who was with me) and I zoned out and blurted out my question and she just laughed an answered it but aaahhhhhhhhhhh
I mean I'm normally really polite and 1 moment of rudeness ( that's more awkward than straight up rude) won't be held against me forever but aaaAAAHHHHHHHH
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