# The Best Math Or Logic Riddle Puzzle Ever

## Best logic or math riddle or puzzle

Here is a logic puzzle that this author would bet you won't solve quickly. It is not for the faint of heart or those that give up quickly. While the answer is given below, I ask that you try, and try hard, to solve the riddle before examining the answer. While the riddle is based on both math and logic, it is primarily a logic riddle and should be treated as such.

### The logic riddle

You begin with 12 coins and the knowledge that one of these coins is counterfeit. The coins look identical and the only difference between them all is that the counterfeit coin is of a different weight. You don't know whether it is lighter or heavier than the rest of the coins, just that it will weigh differently.

Your only tool is a set of old fashioned scales; the type shown below below with two pans hanging from a balancing arm. When more weight is placed on one pan than the other that pan will sink while the lighter one will rise.

You have just 3 weighings to find the counterfeit coin. As you discover which one is counterfeit you should also be able to tell if it is heavier than the rest or lighter. Good luck!

** STOP HERE! **Below is a picture of the scales to be used, followed by the solution. The objective, of course, is to find the solution before reading how to solve the riddle. When you give up a week from now, come back and scroll slowly down to the next section; the first few sentences will give a valuable hint. Of course, if you figure it out come back and let us all know in the poll at the bottom that it happened, or leave your online name in the comment section.

## Solution to the Logic Riddle

Let's start by numbering the coins from 1 to 12, just so that they can be identified in this text. The actual solution doesn't require this, but it certainly makes it easier to describe the actions to be taken in solving the puzzle.

Being by weighing coins 1,2,3 and 4 against 5,6,7 and 8.

**STOP HERE** if you are looking for a hint. That's it - weigh four coins against four coins. If you have given up and are looking for the answer, continue reading.

There are three possibilities, call them case 1, case 2 and case 3.

Case 1 is if both sides weight the same; all the coins from 1 to 8 have the same weight and none are counterfeit. Scroll down for the second weighing in case 1.

Case 2 is if the left side, with coins 1,2,3 and 4 are heavier and that side sinks. This means either that one of these coins is heavy (and counterfeit) **or** that one of the coins 5,6,7 or 8 is lighter and thus counterfeit. Scroll down to see the second weighing for case 2.

Case 3 is the mirror image of case 2; the right side sinks. The solution for case 3 is analogous to case 2 and is left to the reader. The steps will be the same, you must simply visualize a different weight.

## Case 1 - Both sides on weighing #1 were the same

The obvious conclusion is that one of coins 9,10,11 or 12 is counterfeit. For your second weighing, put coins 9, 10, and 11 on the left side and 1,2 and 3 on the right side. There are once more three possibilities; case 4 where both sides are equal, case 5 where the left side goes down because it is heavier and case 6 where the left side goes up because it is light. Bear in mind here that the right side contains *only* legitimate coins; if the left side goes down it isn't because a light coin is on the right side, it is because a heavy coin is on the left.

In case 4, the solution is obvious; the bad coin is #12; the only one not proven to be of equal weight with all the others. For a third weighing that coin can be measured against any other coin to determine if it is light or heavy.

For case 5, scroll down to find the third weighing if you haven't figured it out already.

Case 6 is again analogous to case 5 and is left to the reader to find.

## Case 2 - the left side of weighing #1 was heavy

This is the trickiest of all the possibilities. The author assumes you haven't figured it out - if you had you wouldn't be here! Lets look at it.

For the second weighing, put coins 1,2,3 and 5 on the left pan and 9, 10, 11 and 4 on the right side. Remember that we already know that 9, 10, and 11 are all good coins.

There are (as always) three possibilities. Case 7 is that both sides weigh the same, Case 8 is that the left side goes down. Case 9 is the right side goes down.

Case 7 tells us that the bad coin is number 6, 7, or 8 **and** that it is lighter than all the rest (remember weighing #1 where the side with those coins went up). All other coins have been proven to be of equal weight. For the third weighing, weight 6 against 7; the one that goes up is the bad coin while if they are equal the final answer is coin 8.

Case 8 tells us that the bad coin is #1, 2 or 3 **and** that it is heavy. Were the problem in either coin 4 or 5 the scales would go the other way. For the third weighing test coin 1 against coin 2. If one of them is heavy it is bad while if they are equal in weight the bad coin is number 3.

Case 9 is that the right side of weighing 2 goes down. This can only happen if coin 4 is heavy or coin 5 is light. For the third weighing test coin 4 against coin 1 (a known good coin). If unequal coin 4 is bad and heavy, if equal coin 5 is bad and light.

## Case 5 - the left side of weighing #2 is heavy

The conclusion to date is that the bad coin is number 9, 10 or 11 **and** that it is heavy. That side of the scales went down; it must be heavy.

As has already been seen, the solution to this scenario is to weigh #9 against #10; if one is heaver that is the bad coin, while if they are equal the bad coin is #11.

==============================================================

This finishes the solution to one of the very best math and logic puzzles or riddles. For those readers interested in mathematical problems, you might find Zeno's Paradox of interest. This was found in about 400 BC and has baffled thinkers ever since.

## Winning poll - no fibbing now!

## Did you find the answer?

**© 2011 Dan Harmon**

## Comments

I actually figured it out in a couple of minutes, but that is only because I have been doing puzzles like this for many years. Since it was a good hub with a good problem, I voted it up. Thanks, I like stuff like this.

You should include a flowchart at the bottom. It would make the explanation easier to follow. Great puzzle.

Ummm...huh?? I have no left brain. lol. Great riddle, though. Voted up!

Your poll doesn't allow the option, "I had to read all of the solution several times before it hit me!" :) Although maths and physics were always my two strongest subjects and should give me a good grounding/chance in problems of this type, they never have - I'm hopeless with these conundrums, however many I try to solve. I don't honestly believe I would have figured this out in ten years of constant pondering. Good question!

I don't know why, but I came up with the answer right away. It just seemed logical. This is a great hub. I'm sending it to my husband for him to look at.

Very clever puzzle - I got as far as finding the bad coin (#12 in your solution) if both sides weighed the same both times. I needed some help to sort it out if one side was light or heavy on the second weighing, but didn't need the first hint. Thanks for the work-out! ...and Zeno's Paradox isn't really a paradox, is it? I will go read your hub and find out. :D

I have seen this puzzle before and I admit that I already knew the answer. However, when I first found the answer, it was ALSO from peeking. These types of questions are often used in interviews for those who apply at places like Google. I am complete rubbish at figuring these out. :P I think I've only figured out one of these types of puzzles in my life.

Great hub, by the way! I love to attempt to solve these kinds of things (even though I eventually give up and peek. :P)

Cool riddle! I've got another answer:

1. Split the 12 into two groups of 6 and weigh these two groups. Now you know which group of 6 the counterfeit is in.

2. Split the group of 6 that contains the counterfeit into two groups of 3. Now weigh these two groups of 3.

3. Now you've narrowed the counterfeit down to one of three coins. Choose any two of these coins to weigh.

If one is lighter, it's the counterfeit.

If they're the same, the one not weighed is counterfeit.

Voila! I think :)

Ah, I thought the counterfeit was lighter! Damn. thanks :)

Great Hub, and share worthy. Would be an awesome exercise in a classroom setting, especially if you could actually supply the coins and scales. Voted up and shared.

Aaargghhh, I hate you!! Only joking. I find these type of puzzles frustrating but can't help but sit and figure it out. I managed after reading the hint, and got it very quickly after that. What a great hub, and very well explained. Voted up and funny...there was no 'frustrating' option ;-)

Hi Wilderness ~ This is a super puzzle. I took out my coins and got some inkling how to get started. Your explanation really sorted it out well. First thing I thought of when chose this hub to read was 'Car Talk' which I also love. Blessings, Debby

I love logic puzzles, but I usually don't actually take the time to figure them out. For this one, I immediately knew the general procedure to follow, but I didn't have the patience to work it all out. But it was a fun hub!

I feel like a genius! :) This was really neat and I'm about to share it.

Sending this to two members of my huge family who adore puzzles. I'm sure Im dyslexic because i can NOT do anything so clever - though this hub, I do admire!

I'm useless at this kind of thing! Great fun though - will share it for sure - could be a good end of term challenge for the kids too.

I've seen this before, but I didn't remember it until I read your first instructions. Fun stuff!

Dan, this was a good hub for math puzzles. It does boggle your mind. Voted up for interesting!

29