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 ...
43
votes
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
38
votes
Accepted
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. ...
29
votes
Accepted
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:
24
votes
Accepted
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 ...
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 ...
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 ...
15
votes
Accepted
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.
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 ...
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,...
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.
...
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!
14
votes
Accepted
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 ...
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 ...
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 ...
13
votes
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 \...
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 ...
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, ...
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 ...
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 ...
11
votes
Accepted
11
votes
Accepted
10
votes
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