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# 4 unknown coins - another weighing puzzle

Started by k-man, Nov 27 2012 08:39 PM

14 replies to this topic

### #1

Posted 27 November 2012 - 08:39 PM

You have 4 coins that look identical, but you know nothing about their weights. They may all weigh the same or may have different weights or some of them may weigh the same and some be lighter or heavier - you get the idea. The difference (if any) is small relative to the weight of the coins

Goal: You need to organize the coins based on their relative weight, i.e. sort them by weight grouping identical coins together. All you have is a balance scale that tells you which side of the scale is heavier or if the sides are in balance.

Question 1: What's the minimum number of weighings required to achieve the goal?

Question 2: Provide a method that accomplishes the goal in the minimum number of weighings.

Have fun!!!

Goal: You need to organize the coins based on their relative weight, i.e. sort them by weight grouping identical coins together. All you have is a balance scale that tells you which side of the scale is heavier or if the sides are in balance.

Question 1: What's the minimum number of weighings required to achieve the goal?

Question 2: Provide a method that accomplishes the goal in the minimum number of weighings.

Have fun!!!

### #2

Posted 27 November 2012 - 09:22 PM

Just a question about the conditions. "The difference is small relative to the weight of the coins". I'm guessing that means something like "the difference between the lightest and the heaviest coin is less than the weight of the lightest coin."

Is that sort of what you mean?

(What an interesting weighing puzzle!)

Is that sort of what you mean?

(What an interesting weighing puzzle!)

### #3

Posted 27 November 2012 - 10:54 PM

Spoiler for If they are all different and no two weigh the same as the other two then...

### #4

Posted 27 November 2012 - 11:11 PM

Aye, Captain. You can interpret it that way. I put this condition there to exclude the possibility of one coin weighing as much as 2 others.

### #5

Posted 27 November 2012 - 11:19 PM

good point, Curr3nt!

Spoiler for Even with ties, that's enoughj

**Edited by CaptainEd, 27 November 2012 - 11:26 PM.**

### #6

Posted 27 November 2012 - 11:49 PM Best Answer

Spoiler for One less

### #7

Posted 28 November 2012 - 03:36 PM

Very good, Captain.

This is the same solution that I've found to be the best so far. Now, I've been trying to either find a way to do it in 4 weighings or prove that it cannot be done.

This is the same solution that I've found to be the best so far. Now, I've been trying to either find a way to do it in 4 weighings or prove that it cannot be done.

Spoiler for Thoughts...

### #8

Posted 28 November 2012 - 04:05 PM

K-man, you've clearly given it a lot of thought. I wonder if this number of weighings can resolve more than 4 coins.

### #9

Posted 28 November 2012 - 04:10 PM

Adding the fifth coin under the same conditions increases the number of possibilities dramatically, so 5 weighings will not cover 5 coins. I'll do some calculations and if I find it interesting enough, I'll post a new topic for 5 coins.

### #10

Posted 28 November 2012 - 04:19 PM

Good point. It would be nice to have a number of coins where there's a benefit to a multi-coin weighing. My solutions for single-coin weighings are 8 weighings for 5 coins, 11 weighings for 6.

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