# A company has 10 machines that produce gold coins, One of the machines is producing coins a gram light.?

A company has 10 machines that produce gold coins, One of the machines is producing coins a gram light.

How do you tell which one is making the defective coins with only one weighing?.......

Relevance

Good job bugman. It's not really clear though.

So you take 1 coin from machine 1. 2 coins from machine 2. 3 coins from machine 3. etc....

You need to know how much everything would way if all the machines were working properly so you can find the flawed machine. We'll say that the gold coins when produced correctly weigh 2 grams. So, 10+9+8+7+6+5+4+3+2+1=55. 55x2grams= 105 grams. 105 grams is how much it should weigh.

Now, weigh all 55 coins. Let's say it measures 100 grams. That means 5 coins don't weigh enough because it is 5 grams shy of what 55 proper coins should weigh. Since you took the amount of coins that correspond with the machine number, you discover that machine 5 is defective.

104 grams = Defective Machine #1

103 grams = Defective Machine #2

102 grams = Defective Machine #3

101 grams = Defective Machine #4

etc....

• ?
Lv 4
4 years ago

Gold Coin Machine

Weigh one coin from the first machine, two from the second machine, three from the third machine and so on... (all at the same time)

The coins really should weigh 1+2+3+ ... +10 times whatever the weight per coin is.

But the weighing will be either 1 gram off, or 2 grams off, ...

This reveals the defective machine.

So if the weight of all the coins is 7 grams off, then it's machine #7.

• Anonymous
5 years ago

If there is only one machine producing with a different weight, the problem is easy: Put one coin from each machine on the scale. If the total weight is different from 100 gms, you know there is a machine not producing correctly. To find which machine, make a note of the difference. Let w be the total weight on the scale. Then d = w - 100 gms denotes the difference in weight of the coin with the different weight. Now empty the scale. Put 1 coin from machine 1 on the scale. Put 2 coins from machine 2 on the scale. ... Put 9 coins from machine 9 on the scale. Put 10 coins from machine 10 on the scale. Let u denote the total weight of the coins on the scale. There are10+9+...+2+1 = 55 coins on the scale. Let v = 55 * 10 gms = 550 gms. Then D = u - v denotes the difference in weight between 55 coins of 10 gms and he 55 coins on the scale. D/d = the number of the machine with the wrong weight. Note that d and D may be negative. If you suspect that there may be two or more machines with differing weights, you need more than one weighing. To see this, observe that if you have decided on a distribution of coins from each machine on the weight, you can always find a set of machines they'll give you a total weight of an average of 10gms per coin. In general, if you suspect that there are at most n machines with differing weights, you need n weighings to find them. I hope this answers your question.

• Anonymous

You sure it's not 2 weighings?

I can see this:

taking one coin aside, weighing 6 of the remaining 9 coins. If one side is lighter, take those 3 coins, leave one aside and weigh again. If even, the odd one out is light. If the first six are all equal, take the remaining 3 and weigh one each, same rule applies.

If each set of 3 is equal, it's the 1st coin you set aside.

Damn, go bugman, I was assuming a balance scale, not a gram scale, Sweet.

• 6 years ago

how one of the machines is producing one gram less?