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Can You Solve the Counterfeit Coin Puzzle?

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Mathematicians have long plagued humankind with a style of puzzle in which you must weigh a series of items on a balance scale to find one oddball item that weighs more or less than the others. They're known collectively as balance puzzles, and they can be maddening...until someone comes along and trots out the answer.

Within the world of balance puzzles, the 12-coin problem is well-known (there's also a nine-coin variant, and a horrendous 39-coin variant). There is in fact a generalized solution for such puzzles [PDF], though it involves serious math knowledge.

In the video below, we are presented with a version of the 12-coin problem in which we must determine a single counterfeit coin in a dozen candidates. The problem is, we're only allowed the use of a marker (to make notes on the coins) and three uses of a balance scale. Here are the detailed conditions:

1) All 12 coins look identical.

2) Eleven of the coins weigh exactly the same. The twelfth is very slightly heavier or lighter.

3) The only available weighing method is the balance scale. It can only tell you if both sides are equal, or if one side is heavier than the other.

4) You may use the scale no more than three times.

5) You may write things on the coins with your marker, and this will not change their weight.

6) There's no bribing the guards or any other trick.

So how do we solve this specific case? Watch the video to find out.

For a bit more on this puzzle, check out this TED-Ed page.

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See If You Can Solve This Tricky Coin-Flipping Riddle
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Make sure your head is in working order before trying to solve this riddle from TED-Ed, because it's a stumper.

Here's the scenario: You're an explorer who's just stumbled upon a trove of valuable coins in a remote dungeon. Each coin has a gold side and a silver side, each with an identical scorpion seal. The wizard who guards the coins agrees to let you have them, but he won't let you leave the room unless you separate the hoard into two piles with an equal number of coins with the silver side facing up in each. You've just counted the total number of silver-side-up coins—20—when the lights go out. In the dark, you have no way of knowing which half of a coin is silver and which half is gold. How do you divide the pile without looking at it?

As TED-Ed explains, the task is fairly easy to complete, no psychic powers required. All you need to do is remove any 20 coins from the pile at random and flip them over. No matter what combination of coins you choose, you will suddenly have a number of silver-side-up coins that's equal to whatever is left in the pile. If every coin you pulled was originally gold-side-up, flipping them would give you 20 more silver-side-up coins. If you chose 13 gold-side-up coins and seven of the silver-side coins, you'd be left with 13 silver coins in the first pile and 13 silver ones in your new stack after flipping it over.

The solution is simple, but the algebra behind it may take a little more effort to comprehend. For the full explanation and a bonus riddle, check out the video from TED-Ed below.

[h/t TED-Ed]

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No One Can Figure Out This Second Grade Math Problem
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Angie Werner got a lot more than she bargained for on January 24, when she sat down to help her 8-year-old daughter, Ayla, with her math homework. As Pop Sugar reports, the confusion began when they got to the following word problem:

“There are 49 dogs signed up to compete in the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?”

Many people misread the problem and thought it was a trick question: if there are 36 more small dogs and the question is how many small dogs are competing, then maybe the answer is 36?

Wrong!

Frustrated by the confusing problem, Angie took to a private Facebook group to ask fellow moms to weigh in on the question, which led to even more confusion, including whether medium-sized dogs should somehow be accounted for. (No, they shouldn’t.) Another mom chimed in with an answer that she thought settled the debate:

"Y'all. A mom above figured it out. We were all wrong. If there is a total of 49 dogs and 36 of them are small dogs then there are 13 large dogs. That means 36 small dogs subtracted by 13 large dogs then there are 23 more small dogs than large dogs. 36-13=23. BOOM!!! WOW! Anyone saying there's half and medium dogs tho just no!"

It was a nice try, but incorrect. A few others came up with 42.5 dogs as the answer, with one woman explaining her method as follows: "49-36=13. 13/2=6.5. 36+6.5=42.5. That's how I did it in my head. Is that the right way to do it? Lol I haven't done math like this since I was in school!"

Though commenters understandably took issue with the .5 part of the answer—an 8-year-old is expected to calculate for a half-dog? What kind of dog show is this?—when Ayla’s teacher heard about the growing debate, she chimed in to confirm that 42.5 is indeed the answer, but that the blame in the confusion rested with the school. "The district worded it wrong,” said Angie. “The answer would be 42.5, though, if done at an age appropriate grade."

Want to try another internet-baffling riddle?


Here's the answer.

[h/t: Pop Sugar]

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