Guess the faulty machine

[ashitpro]

There are 7 machines, each of them produces the coin of weight 2 grams.
Someday somehow one machine gets faulty and started producing coin of weight 1 gram. All coins looks similar in dimensions.
You can take any number of coins from any machine.
Condition is that, whatever coins taken from 7 machines would be weighted only once.
Can somebody find the faulty machine.

[Nepomuk]

There are 7 machines, each of them produces the coin of weight 2 grams.
Someday somehow one machine gets faulty and started producing coin of weight 1 gram. All coins looks similar in dimensions.
You can take any number of coins from any machine.
Condition is that, whatever coins taken from 7 machines would be weighted only once.
Can somebody find the faulty machine.

Let's say, the machines are called m0 - m6. Take 2^0 = 1 coin from m0, 2^1 = 2 coins from m1 and so on. So if all machines work correctly, you should have 1*2 + 2*2 + 4*2 + 8*2 + 16*2 + 32*2 + 64*2 = 127*2 = 254 grams.

Say, m3 is faulty. Then you'll have 1*2 + 2*2 + 4*2 + 8*1 + 16*2 + 32*2 + 64*2 = 127*2 - 8 = 246 grams. So basically, you take those coins, weigh them and the difference between the weight and 254 grams gives you the faulty machine (2^X difference -> mX is faulty).

Where's the next question? ^^

Greetings,
Nepomuk

[sitko]

Wouldn't it make more sense, to make each subgrouping get bigger by a factor of 10, instead of a factor of 2?

Then you'd get a lot more coins that you could spend after the problem is solved.

:)

[Banfa]

Wouldn't it make more sense to use an arithmetic progression rather than a geometric one.

That is rather than 2^0, 2^1, 2^2, 2^2 ... take 1, 2, 3, 4, 5, 6, 7, then the number of grams you are short is the machine number + 1.

[sitko]

Wouldn't it make more sense to use an arithmetic progression rather than a geometric one.

That is rather than 2^0, 2^1, 2^2, 2^2 ... take 1, 2, 3, 4, 5, 6, 7, then the number of grams you are short is the machine number + 1.

No, I know that wouldn't work. It would be more confusing to figure out specifically which subgroup of coins was the culprit.

as if 6 was the bad one, it would weigh the same as a good 3, ...maybe it would work...I'll let smarter people chime in...

:)

[jkmyoung]

Banfa's correct. Using powers of 2 just complicates things.

I think the assumption missed is that all the coins you get from the 7 machines are weighed altogether.