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 ...
- 143k
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
- 115k
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. ...
- 115k
29
votes
Accepted
28
votes
Accepted
Radioactive Rods
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
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:
- 5,676
25
votes
Accepted
23
votes
Accepted
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
22
votes
Accepted
19
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 ...
- 115k
18
votes
Accepted
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
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 ...
- 21.5k
15
votes
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
15
votes
Accepted
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
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.
- 130k
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 ...
- 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,...
- 33.4k
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 ...
- 4,854
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 ...
- 3,592
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 ...
- 7,320
13
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.
...
- 989
13
votes
12
votes
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
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 \...
- 5,856
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 ...
- 13.9k
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