# Tag Info

Accepted

### Why are all numbers from 1 to 2N covered by weights with powers of 3?

It helps to think about the scale not in terms of balancing two objects, but in terms of creating a weight difference between the two sides. (If you want to balance out an object, you simply put ...
• 148k

### Faulty Weight Scales

The solution is simple:
• 148k
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### 68 coins with 100 weighings

It seems to me that there's a simpler solution than the one accepted above. Step 1: Step 2: Step 3: The point here is that
• 120k
Accepted

Accepted

### Unknown weight of four identical objects

First weigh two of your objects against the other two. Whichever pair is heavier must contain the 11-oz object, since even $11+3>5+8$. Now you have two objects of which you know one weighs 11 oz. ...
• 117k
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• 16.7k
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• 23.8k

### 68 coins with 100 weighings

I see, it took me too long to fininsh my drawing, but let me present it as additional material to sousben's answer:
• 5,948
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### Weigh a scale with itself

You can: Note: Corrected after comment by Jaap Scherphuis.
• 2,590

### Twelve balls and a scale

Some of the existing answers to this ancient question are excellent, but there's one famous answer that I think deserves mention here. It comes from an article in Eureka, the annual magazine of the ...
• 120k
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• 344
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### Minimum number of turns

Oo, I've got this one; these come up a lot while organising board game tournaments. a) what is the minimum number of turns needed to determine the heaviest box? b) what is the min number of turns ...
• 78.1k
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• 33.7k
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### Unbalanced weight of boxes

I have found a set where the weight of the heaviest box is Here are the weights of the boxes General Strategy Minimality confirmed by Oray using computation.
• 138k
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### Is a fake coin lighter or heavier?

This is my first answer on the puzzling stack exchange and I'm really not used to explaining things like this so it's probably going to be rather convoluted. I'll probably come back and edit it later ...
• 341
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### 15 Balls Sorting

And now my computer generated and checked solution: Put the weights in a row and number the places from 1 to 15. Then do the following: I generated these command by an similar algorithm as Murch. ...
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### 30 fake coins out of 99 coins v2

You only need Assume the genuine coins weigh $x$ grammes. Credit to Hexomino and Jaap for the corrections!
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• 35.6k

### How can you get 13 pounds of coffee by using all three weights each trial?

Using all three weights in each trial is an interesting requirement, I haven't seen such a requirement before. It's not clear what exactly counts as a trial, in particular whether a trial may consist ...
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### A Dozen Golden Eggs

Firstly, use the balance scale once to compare the weights of If $A_1$ and $B_1$ both weigh the same, then both swapped eggs are in the same one of these two sets. In that case, use the balance scale ...
• 117k
Accepted

### Lots of Gold Golden Coins and a Scale

You are looking for 10 numbers such that no two subsets of the numbers sum to the same amount, such that the largest number is minimal. This is conjectured to correspond to sequence A005318 in OEIS, ...
• 33.7k

### 15 Distinct Weights' Sorting

I can do it in Step 1: use 5 weighings, each with 3 new balls, yielding 5 "stacks" that are sorted (imagine the balls stack like coins). Step 2: Make a state diagram covering all the cases where we ...
• 7,518
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### How to Balance?

Here is one way to do it
• 138k

### Lots of Gold Golden Coins and a Scale

edit: while the algorithm in this answer works, it is not optimal. See f"'s answer for one which provides a lower, and probably optimal, bound. I think you need coins per stack. Explanation: The ...
• 2,142

### 15 Distinct Weights' Sorting

Lemma: Fully ordering a pile of 5 stones takes 4 weighings (this is easy to prove but I can do so if necessary). Fully ordering a pile of 5 stones can be done as follows: a.) Assign each stone a ...

### 15 Balls Sorting

I can do better than Chelsea's and Penguino's 49 operations, as follows. (The following is pretty painful to read; sorry. For this reason it probably won't be much of a spoiler to anyone not going out ...
• 120k