# Thread: Finding whether a coin is fake

1. There are 101 coins out of which 50 are fake and are 1 gram lighter than the original ones.One coin is chosen randomly.How can you find whether it is fake or original by a machine that shows the difference in weights of what is placed on the two pans?

2. Request for clarification:
- How many weightings are we allowed to perform?
- When you say the machine shows the difference in weights, does it also give which one is lighter, or just the absolute difference in weight?

Oh,sorry...We are allowed only one weighing and the machine does not show which is heavier or lighter,it just gives the difference....

Looks like you have something in your mind...

5. Could you post the answer now Mathmaniac? It's been over a week and I don't think anyone is going to be answering, and I frankly have no idea but I am interested in the solution.

Let x be the weight of a lighter coin, then the total weight of all coins is 50x + 51(x + 1) = 101x + 51

Suppose you pick a light coin, place it on 1 side , place all remaining coins on the other side so now on one side you have 49x + 51(x + 1) and on the other side you have x. The difference is

49x + 51(x + 1) - x = 48x + 51(x + 1) = 99x + 51

Suppose you pick a heavier coin, place it on 1 side, place all remaining coins on the other side so now on one side you have 50x + 50(x + 1) and on the other side you have (x + 1). The difference is

50x + 50(x + 1) - (x + 1) = 50x + 49(x + 1) = 99x + 49

So... When the machine gives you the difference. subtract 49 then divide by 99. If you get a whole number then you picked the heavier coin, otherwise you picked the lighter coin.

Note*This method works pretty well only if the coins have whole number weights. Further consideration of what may happen if the coins don't have whole number weights is giving me a headache so i stop here.

Another very useful hint:

Parity seems to be very simple but it is very useful in many problems like this.

I wonder why the MHB fails to give the answer.

8. Originally Posted by mathmaniac
Another very useful hint:

Parity seems to be very simple but it is very useful in many problems like this.

I wonder why the MHB fails to give the answer.
I think i got it using the hint. The weight of the coin is inconsequential, what matters is the +1 difference in weight. The scale will always read integer so if you read even integer you have heavy coin set aside, if you read odd integer you have light coin set aside.

I wouldn't be surprised if someone gets an easier way.