47 votes
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 ...
Deusovi's user avatar
  • 145k
43 votes

Faulty Weight Scales

The solution is simple:
Deusovi's user avatar
  • 145k
42 votes
Accepted

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
Gareth McCaughan's user avatar
38 votes
Accepted

Faulty Weight Scales

Just browsing's user avatar
35 votes
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. ...
Rand al'Thor's user avatar
29 votes
Accepted

Are all balls the same weight?

Building on Lawrence’s answer:
Hugh Meyers's user avatar
  • 23.7k
25 votes

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:
A. P.'s user avatar
  • 5,686
24 votes
Accepted

Weigh a scale with itself

You can: Note: Corrected after comment by Jaap Scherphuis.
CG.'s user avatar
  • 2,590
21 votes

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 ...
Gareth McCaughan's user avatar
17 votes
Accepted

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 ...
ffao's user avatar
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16 votes
Accepted

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 ...
Bass's user avatar
  • 76k
15 votes
Accepted

Determine your weight from 2 different answers from 2 weighting scales

f'''s user avatar
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15 votes
Accepted

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.
hexomino's user avatar
  • 133k
15 votes
Accepted

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 ...
AngelPray's user avatar
  • 341
14 votes
Accepted

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,...
f'''s user avatar
  • 33.6k
14 votes
Accepted

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. ...
Etoplay's user avatar
  • 997
14 votes
Accepted

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!
Alexander Geldhof's user avatar
14 votes
Accepted

Sorting marbles based on weightings

JMP's user avatar
  • 35.6k
14 votes

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 ...
Magma's user avatar
  • 5,024
13 votes
Accepted

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 ...
Joel Rondeau's user avatar
  • 7,540
13 votes
Accepted

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 ...
Zandar's user avatar
  • 3,592
13 votes

Weigh a scale with itself

hpp3's user avatar
  • 141
12 votes
Accepted

212 weights of 1 gram

Let's try looking at it from the biggest weights first. Let $w$ be the biggest weight of one such system, and let $n$ be the number of $w$-weights in the system. Let the remainder $r = 212 - n \...
The Dark Truth's user avatar
12 votes
Accepted

Rank the Fencers

There are $5!=120$ possible orderings of the fencers, so we need $\log_2(120)\approx 6.9069$ bits of information. Each duel provides at most $1$ bit of information, so at least $7$ duels will be ...
Julian Rosen's user avatar
  • 14.2k
12 votes
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, ...
f'''s user avatar
  • 33.6k
12 votes
Accepted

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 ...
Rand al'Thor's user avatar
11 votes

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 ...
user3294068's user avatar
  • 7,438
11 votes
Accepted

Find a stone which is not the lightest one

Hermes's user avatar
  • 695
11 votes
Accepted

How to Balance?

Here is one way to do it
hexomino's user avatar
  • 133k
10 votes

2016 coins and a balance

The answer is: Determined as follows: Next:
Hugh Meyers's user avatar
  • 23.7k

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