Coins Coins and 12 Coins?
Suppose you have 12 coins, one of which is counterfiet ( that is, it is either heavier or lighter). Using a balance, determine in three weighings, which coin is the counterfeit and if it is heavier or lighter
Suppose you have 12 coins, one of which is counterfiet ( that is, it is either heavier or lighter). Using a balance, determine in three weighings, which coin is the counterfeit and if it is heavier or lighter
Mena M
Favorite Answer
1- put 3 coins on each side of the balance ... leave the other 6 away
if u got 1 heavier ... then the coin is one of the 6 on the balance (and u know which 3 are heavier),
else it's the other 6.
now u got 6 coins with the wanted coin between them (the worst case is that it's from the other 6 that u didn't balanced) .
2- take 2 of the coins and leave them away ... put the other four, 2 in each side
if u got 1 heavier ... then the coin is one of the 4 on the balance (and u know which 2 are heavier),
else it's the other 2.
a - if it's in the other 4 coins ... take "one of the light side with 2 of the heavy side" versus 3 of the other coins ...
if they're balanced, then the forth one is the lighter one.
if the 3 other coins were heavier, then the one taken of the light side is the lighter one
else ... u got 2 coins , one of them is heavier ... u got a 50% chance to know the coin
b- if it's in the other 2 coins ... put one of them versus one of the known coins.
if it's not balanced ... u got the coin and u will know if it's heavier or lighter.
if balanced, then u got the coin and u got a 50% chance to know if heavier or balanced
?
You can't do it in three unless you know whether it is lighter or heavier. Imagine there are two coins on the scale and they are unbalanced. You have no idea which one is bad. If you have already determined the rest are good, you can replace one of the coins on the scale with a good one. Either the pair will balance, in which case the one in the hand is bad and you will know if it is lighter or heavier because you took it off the high side or low side. If they don't balance, you know the one you didn't remove is bad and whether it is lighter or heavier because it is high or low.
But that is two weighings. To get from 12 to 3 requires at least two weighings, making 4 total. For example, dividing the 12 into groups of 4 and balancing two groups tell whether to the two on the scale are balanced or not, but not whether one is heavier or lighter. Another weighing can determine than but still leaves 4 to judge from which requires 2 more weighings.
Bomba
It takes four weighings for 12 objects (coins)
1. weigh 3 & 3 on with 6 off.
2a. If balanced replace the balls on with the six off for 3 & 3 on
you will have a heavy side and a light side and six good (G) off
3a. take off a heavy (H)and a light(L) and swap two on the scale.
If balanced the H,L are the ones off scale.
If it reverses, the H,L are the two you swapped on scale
If no change the H,L are the ones not moved on scale
4a. weigh the H against a G. If balance, it is the L. If not, the H
Now go back to the first weighing . If the initial 3 & 3 do not balance, then the six G are off and you already have a heavy and light side as before at the weighing 2. Just continue as before with two more weighings totalling 3.
But remember that it took 4 weighings to cover the worst case which was the first one.
If someone can really do this in three, I would like to see it. So far I have not seen it.
Anonymous
This is a classic problem, and has been around for years,. I do not remember all the details, but you start by weighing two groups of 4 coins against each other. If they balance, all eight are good and one of the remaining four is bad. Weigh three of the possible bad coins against three good ones. If these balance, the remaining coin is bad; if they don't, you have learned whether the bad coin is too heavy or too light. Weigh two of the suspect coins against each other; if they balance, the third suspect is bad; if they don't, you can tell which of the two groups is bad, and one more weighing will tell you which of the group. This is not complete, but you get the idea. The first response is faulty in that it assumes that the bad coin is heavier than the others.
Sparky
too complicated to explain because of the variables, but do this:
start with 4 on each side. If balanced, you know the 1 of the 4 left is counterfeit. If not, take 2 from one side and move them to the other and reweigh. add 2 known good the replace the ones you took. If weight stays the same, it is in the 2 you didn't move, if it changes, it is 1 of the 2 you moved.
Now weigh one of the last 2 with a good one. If equal, the one you didn't weigh last is the counterfeit, if unequal, the direction of the scale that the bad one made it will tell you if it is heavier or lighter.