The fake gold coin problem

Logic Level 1

Let's say you have 10 bags with 10 gold coins each except for in one bag there are all 10 fake coins. What is the minimum number of weighs you need to find which bag is fake.

Note:
1. You can take out coins from the bag.
2. You are given the weight of both the real coin and fake coin.
3. You are given a digital balance.

1 6 2 7 5 4 3

5 solutions

Richard Grayson
Jun 30, 2015

Let us name the bags a,b,c,d and so on.

Take out 1 coin from bag A, 2 coins from bag B, 3 coins from bag C and so on.

Let us assume that a real gold coin weighs 10 g and a fake gold coin weighs 9 g.
Weigh all the coins at once.

Now comes the interesting part,
after you take out all the coins, you would have 55 of them.

If there were no fake coins, the whole set of coins would weigh 550 g (which is obviously not the case here).

Now, if the first bag (A) contained the fake gold coins, the whole set of coins would weigh 549 g (as 54 coins weigh 10 g and one coin weighs 9g)

Going by the same logic,
If the second bag (B) contained the fake gold coins, the whole set would weight 548g. And so on...

Good answer, but the problem should specify that the weight of a real coin and the weight of a fake coin are both known. Otherwise you would more weighs.

Andrés Castillo - 5 years, 11 months ago

Yes, Information required if it is a digital or traditional balance used.
you can't just imagine that you are using digital balance. and even though how do you know about each coin weights 10 gms. With only info given, it is intrinsic that traditional balance and it will take 3 times to find out.

Ànoop Kumar - 5 years, 11 months ago

Thank you sir!

Richard Grayson - 5 years, 11 months ago

Good Naman that was the ideal solution

Bhavik Bansal - 5 years, 11 months ago

It's awfully written. The problem says that the weights are given.
So it should ask what's the minimum number of weighs you would need. In this case, the answer is 1, cuz you might get lucky on the first try.

So none of that is needed !!!!!

Luis Rivera - 4 years, 11 months ago

Exactly right! They ask for the MINIMUM number which is obviously ONE.

Jo Ro - 4 years, 11 months ago

Ahmed Obaiedallah
Jul 8, 2015

Well it's safe to say that this problem needs to be modified as it misses a valuable information which is the weight of both the gold coin and the fake coin .. other than that, that one weighing won't provide you with enough info to determine which bag is the fake one

Imagine that the weight difference between the coins is X

Now think of 2 cases

$X_1=2 \space grams$

$X_2=10 \space grams$

And consider after you took

1,2,3,4,5,6,7,8,9,10 coins of each bag respectfully

You found out that the weight is down by 10 grams

For $X_1$ it will be the 5th bag

For $X_2$ it will be the 1st bag

Ahmed good job I agree with you. The relationship of good to bad coins is not for certain thus assuming that it is 9 g to 10 g is not allowed. assume its not 9g -10g but rather 8g to 10 g your assumption of 9-10 says with a weight of 548 g says the answer is bag B when in reality a difference of 2g says it is bag A. You can still do it in 2 just knowing the weight of good to bad coins are not equal. 5 bags to 5 bags. since it says the min number of weighs you can assume you grab the correct 5 bags weather the bad coins weigh more or less than the good coins. take and weigh 2 bags to 2 bags they are equal and again it says minimum thus the bad bag is the one that did not get weighed

Eugene Childress - 4 years, 1 month ago

Anurag Malakar
Jul 2, 2015

There are 10 bags having 10 coins each. Let's name the bags B1, B2, ..., B10

Take out 1 coin from B1

Take out 2 coins from B2

Take out 3 coins from B3

(And so on and so forth)

Finally,

Take out 10 coins from B10

Now, let the weight of a real gold coin be X. And let the weight of a fake gold coin be X1.

If there were no fake coins in any of the bags, the total weight would be: (1+2+...+10)*X

But we know that one of the bags have fake coins

So the weight would turn out to be: (1+2+...+10) X - K X + K*X1

Where, 1<= K <=10

So, weighing the selected coins once, we will find out the value of K

And hence, The Kth bag contains all the fake coins, as all the coins taken out of that bag weighed X1 each.

I'm struggling to understand why you're taking 1 from b1, 2 from b2. Could you help me understand?

Zubayr Bhyat - 4 years, 10 months ago

Tootie Frootie
Jul 5, 2015

easy logic !! if you take a bag ...you will get the actual weight ... condition 1 = if others are heavier so cheers you've got your bag .. condition two = if it's heavier ...like all others ....you've got one actual weight ...and can estimate others ...(as the material of fake coins is not specified)

Mohamed Yossry
Jul 2, 2015

if you take one of each bag you will sense one is different as all the fake ones are in one bag, then you do 1 weight just to check

Only if world was so easy.

Really its hard to sense some grams difference

Bhavik Bansal - 5 years, 11 months ago

with multiples it gets easier

Mohamed Yossry - 5 years, 11 months ago

The fake gold coin problem

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