Weighing I.

[rookie1ja]

Weighing I. - Back to the Water and Weighing Puzzles

You have 10 bags with 1000 coins each. In one of the bags, all coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 gram.

If you have an accurate scale, which you can use only once, how can you identify the bag with the forgeries? And what if you didn't know how many bags contained counterfeit coins?

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.

Pls visit New Puzzles section to see always fresh brain teasers.

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.

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?

[Guest]

If we dont know how many bags has faulty coins, then min number of coins from bag i = 2 * number of coins in bag i-1.

Therefore, the sequence would be

1, 2, 4, 8, 16, 32, 64, 128, 258, 512.

Basically, represent the weight difference divided by 0.1g in binary numbers and positions of 1 will give you which bags have faulty coins.

[Guest]

Binary series is more precise.. which depicts.

2^0, 2^1, 2^2.... so on.

and whatever the extra weight it will be simple for the calculation..

100101010 format

[Guest]

This sequence won't work:

1, 2, 4, 8, 16, 32, 64, 128, 258, 512.

Here is why. If the 10th bag only is full of fakes you'll get an extra weigh amount of 51.2g.

And, if bags 2-9 are fake then you'll get (25.8+12.8+6.4+3.2+1.6+.8+.4+.2) 51.2g and then you won't be able to tell which situation you are in.

The total extra sum has to be distinct for any combo of fake bags.

[Guest]

There's a mistake on that sequence, so there's also a mistake on proving it wrong.

The sequence should be:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, and so on. This sequence should meet the mentioned conditions.

[Guest]

Place all bags on the scale, it should read 10,100g. Take off a bag until the 100 at the end goes away. That will be the bag with the forgeries. If you don't know how many bags contain the forgeries, then still place all bags on the scale and every time 1,100g is removed from the weight that bag will be full of forgeries.

[Guest]

Place all bags on the scale, it should read 10,100g. Take off a bag until the 100 at the end goes away. That will be the bag with the forgeries. If you don't know how many bags contain the forgeries, then still place all bags on the scale and every time 1,100g is removed from the weight that bag will be full of forgeries.

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

[Guest]

you know

all these weird riddles and stuff make my brain hurt like heck

can you make them a bit clearer and not so hard

[Guest]

It's a good one

[Guest]

You can always open the bag

[Guest]

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"

[Guest]

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.

[Guest]

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)

[Guest]

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 ...

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)

[Guest]

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

[Guest]

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?

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

[Guest]

The bags of forgeries would be 100 grams lighter than the bags of genuine coins. You could determine how many bags were forgeries by wieghing them.

harpsr

[Guest]

The solution is simple. Mark all the bags from 1 to 10. Take from the first bag only one coin, from the second take two and from the third take three and so on. At the end the result will be in the form sum of n natural numbers. Ideally the weight should be 55grams, but due to the faulty bag which ever it may be the extra portion will come into picture. For example if the first bag is the culprit then we get 55.1, if the second bag is culprit then we get 55.2 and so on.

[Guest]

You know, there is an even better solution that will make my wife happy without using a weighing machine.

1. Take all the bags of coins. Over the period of ten (or number of bags) days, give one bag to your wife and tell her to go mad, making sure that she spends every coin.

2. When the police call you to say that she has been arrested for using forged coins, you can simply take the number of days since you gave her the first bag, and you have your answer! (For more than one bag of fakes, repeat the same when she gets out of jail!)

SIMPLE!

[Guest]

The original solution is great, but heres a tip:

Don't get the coins muddled up.

You gotta do this:

(Bag number)

1 22 333 4444 55555 666666 7777777 88888888 999999999 10101010101010101010

(Scales)

If you get them mixed up then you won't know which bag they are from:

12567562498165846582364984651893658941651983456184365182469186549

Lol

[Guest]

If we dont know how many bags has faulty coins, then min number of coins from bag i = 2 * number of coins in bag i-1.

Therefore, the sequence would be

1, 2, 4, 8, 16, 32, 64, 128, 258, 512.

Basically, represent the weight difference divided by 0.1g in binary numbers and positions of 1 will give you which bags have faulty coins.

good answer

as the coins from each bag are powers of 2, the faulty bags would give extra weights which are powers of 2 which would set the

particular binary digit.