# 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 ...
• 143k

### Faulty Weight Scales

The solution is simple:
• 143k
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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
• 115k
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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. ...
• 115k
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### Are all balls the same weight?

• 23.6k
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In the worst-case scenario, it requires to locate the radioactive rods. Several answers already describe strategies for locating the radioactive rods. I will give another. Testing strategy: Start ...
• 13.9k

### 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:
• 5,676
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### Weigh a scale with itself

You can: Note: Corrected after comment by Jaap Scherphuis.
• 2,600
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### Thirty genuine and seventy fake coins

As many have already stated, the best you can do is Suppose the genuine coins have weight 0, and the other coins have weights of distinct powers of two. Then any try on the old balance will always ...
• 8,606
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Yes. Solution:
• 636

### 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 ...
• 115k
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### Brooklyn 99 riddle: Weighing Islanders

Divide them into 3 groups of 4 people. Put any two groups on each side of the see-saw. (First Use) Condition 1 If the see-saw balances, we are sure that the oddly wieghted one is in the other group ...
• 14.4k
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### Lots of Gold Stacks and a Balance Scale

The greatest X for which you can find the stack with the fake coins in 3 weighings is: Unfortunately, the strategy isn't as easy to describe as the one in my previous answer (you can read it in the ...
• 21.5k

### Spot the tumbler

Since the weighing scale can only be used once, I feel free to disassemble and modify it. Now, tie a ball from each tumbler to the 4 vertices of the weighing-scale, as shown in this picture: Now, ...
• 12.5k
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### Four coins (plus one) and a balance

How many weighings you need: Call the four coins A,B,C and D, and the true gold coin G. You start by weighing Furthermore, it can't be done in just one weighing. There are nine possible situations (...
• 31.7k
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• 33.4k
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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.
• 130k
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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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### 2016 coins and a balance

Fredo should weigh the single coin against nothing, then all 2016 coins together against nothing. If the coin is real, then 2016 times its weight will be within 99 grams of the total weight. Otherwise,...
• 33.4k
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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!
Accepted

• 35.4k

### 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 ...
• 4,854
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### Evaporating coins

You can guarantee finding and determining the weight of the fake coin for Any useful weighing will be of an even number of coins, with the same number on each side of the balance. If the result is ...
• 3,592
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### The Ebbozonian coin weighing puzzle

The answer is I ran a test with 6 coins, leaving 2 off each time. I got that to work. This made me think I could scale it up to 10 coins by doing 3 on each side. I tried that and got all the cases ...
• 7,320
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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. ...
• 989

• 141

### Trapped in my Cellar

An obvious way to select the weights is to have them be powers of 2, from 1 to $2^{99}$. Every positive integer has a unique binary representation, so no two different sets of weights can be have the ...
• 33.4k