27 replies to this topic

[DechWerks]

Rookie, I love your puzzles. But a note to help me read these better - when you say balance for this one (or similar ones in the future), can you state it as:
'You have one opportunity to use a balance that will return the accurate weight (in number of g).'
I was confused and thought you meant scale (as balance in English conveys a sense of being on a scale that is even...). This has confused me a minor bit in the past as when you state scale, I envision something returning numbers (rather than what you refer to as a balance)...

For amusement, I spent some time trying to figure out how you could put x bags (or coins) on one side vs another side to determine which single bag was the heavier in one go!

Thanks!

[rookie1ja]

Rookie, I love your puzzles. But a note to help me read these better - when you say balance for this one (or similar ones in the future), can you state it as:
'You have one opportunity to use a balance that will return the accurate weight (in number of g).'
I was confused and thought you meant scale (as balance in English conveys a sense of being on a scale that is even...). This has confused me a minor bit in the past as when you state scale, I envision something returning numbers (rather than what you refer to as a balance)...

For amusement, I spent some time trying to figure out how you could put x bags (or coins) on one side vs another side to determine which single bag was the heavier in one go!

Thanks!

for this particular puzzle, I used words "accurate weighing-machine" and thought that it would be clear that it returns numbers ... for some other puzzles I used words "pair of scales" where I meant two panes that can be brought to balance

I see your point ... I have replaced "balancing" with 'weighing"

[SmithTech]

I think your solution is overly complicated.
If you stick with Balancing, simply place one bag on the left side of the balance, then compare it to each of the remaining bags on the right side of the balance.
If the bag on the left is is heavier than any of the one compared to then it is fake and any that balanced evenly are also fake.
If the bag on the left is lighter than any of the ones compared to then is is real and any that balanced evenly are also real.
This solution answers both having only one fake bag as well as having multiple fake bags.

[SmithTech]

Also you ask "And what if you didn't know how many bags contain forgeries?
"
Your solution doesn't properly handle this as if you don't know how many bags there are, you can't possibly know which solution to use. And since you can only balance/weigh only once you can't try multiple solutions.

EDIT to my above post, change balancing to balancing each bag/coin only once.
Take 9 coins from the first bag, balance the first coin to one coin from the first of the remaining 9 bags.
balance the second coin to one coin from the second of the remaining 9 bags. etc..

This is the only solution that works.

[rookie1ja]

I think your solution is overly complicated.
If you stick with Balancing, simply place one bag on the left side of the balance, then compare it to each of the remaining bags on the right side of the balance.
If the bag on the left is is heavier than any of the one compared to then it is fake and any that balanced evenly are also fake.
If the bag on the left is lighter than any of the ones compared to then is is real and any that balanced evenly are also real.
This solution answers both having only one fake bag as well as having multiple fake bags.

the requested solution allows only one weighing (which is not the case of your procedure)

btw, you may use weighing machine (eg. a digital one with numbers) - so no pair of scales (that resolves your second query as well)

[SmithTech]

weighing only once could be taken to mean weighing each bag/coin only once, i did edit my post on that.

I must be missing something as I don't see how if you don't know how many bags are fake, the proposed solution works, can someone explain?

[rookie1ja]

as I already tried to explain ...

Spoiler for ...

Weighing I. - solution
If there is only 1 bag with forgeries, then take 1 coin from the first bag, 2 coins from the second bag ... ten coins from the tenth bag and weigh the picked coins. Find out how many grams does it weigh and compare it to the ideal state of having all original coins. The amount of grams (the difference) is the place of the bag with fake coins.
If there is more than 1 bag with forgeries, then there is lots of possible solution. I can offer you this one as an example: 1, 2, 4, 10, 20, 50, 100, 200, 500 and 1000.

try it ...
for 1 bag - imagine which bag is fake (eg. the third one) and you will see difference of 3 grams on the weighing machine (so the total weight won't be 1+2+3+4+5+6+7+8+9+10=55 ... it will be 1+2+3.3+4+5+6+7+8+9=55.3 ... the .3 indicates that the third bag contains forgeries)

for more than 1 bag - every combination of differences must be unique (so you can not achieve difference of X grams by more than 1 combination - eg. first and third bag and no other 2 or Y bags)

[Logic Master]

You are given one opportunity to use the balance. Your method requires 1 or more tries.

No, taking things off the scale doesn't require more tries, you have to remove the bags anyhow so why not use it to help with the answer

[sparrotic]

Weighing I. - Back to the Water and Weighing Puzzles
Imagine you have 10 bags full of coins, in each bag 1000 coins. In one bag, there are all coins forgeries. The original coin is 1 gram light, forgery is 1.1 gram. Balancing (Edit: Weighing) just once on an accurate weighing-machine, how can you identify the bag with forgeries? And what if you didn't know how many bags contain forgeries?

Spoiler for Solution

Weighing I. - solution
If there is only 1 bag with forgeries, then take 1 coin from the first bag, 2 coins from the second bag ... ten coins from the tenth bag and weigh the picked coins. Find out how many grams does it weigh and compare it to the ideal state of having all original coins. The amount of grams (the difference) is the place of the bag with fake coins.
If there is more than 1 bag with forgeries, then there is lots of possible solution. I can offer you this one as an example: 1, 2, 4, 10, 20, 50, 100, 200, 500 and 1000.

Wouldn't the bag of forgeries weigh 100 coins more than the bag originals (1000 x 0.1 = 100)

[rookie1ja]

Wouldn't the bag of forgeries weigh 100 coins more than the bag originals (1000 x 0.1 = 100)

yes it would ... but the question is how would you find that 1 bag if you could use an accurate weighing machine just once