# You have 30 gold coins but one is counterfeit and lighter than a real coin. The only equipment you have is…?

Use lateral thinking.

You have 30 gold coins but one is counterfeit and lighter than a real coin. The only equipment you have is a simple pair of balance scales. What is the least number of weighings necessary to discover the counterfeit coin?

PLEASE!!!

Explain the problem.

Explain the method you used to reach your answer and show working out.

State the answer.

HEHE TRY THIS PEOPLE. =D

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Since you know that one coin is lighter, that gives you 30 possible outcomes: coin 1 is light, coin 2 is light, etc.

Each time you weigh, you have 3 possible outcomes. Either left side is lighter (in which case the light coin is there), the right side is lighter, or they are equal (in which case the light coin is not being weighed). Since you have 3 possible outcomes, In one weighing you could tell between 3 coins. If you did two weighings, you could tell between 3*3 or 9 coins. Three weighings would give you between 27, and four would give you between 81. Since 27 is not enough, you need FOUR weighings.

Having said this, these would be the optimal weighings:

1. Measure 10 on each side. Keep the 10 that contain the light coin.

2. Measure 3 on each side. Keep the 3 or four that contain the light coin.

3. Measure 1 of the remaining coins of each side. If you had 3 coins going into this step, you will know which is light after this weighing. If you went in with 4, and they are still balanced:

4. Measure the remaining two coins against each other to tell which is the light one.

Minimum is 3. But it can also be 4. Read below to understand.

Here is how.

You devied the pile into two piles of 15 each. weigh them…. chose the lighter one. Devied it into three piles of 5. let they be a,b,c.

Weigh a and b.

if a<b then a contains the fake. similarly if b<a.

But if b=a then c contains the lightest.

Let the five remaining coins be 12345.

Weigh 1 and 2.

Case 1:

1<2 or 1>2

1 or 2 is the fake one.

Case 2:

1=2.

In this case wigh 3&4.

if 3=4 then 5 is the fake one. Otherwise either 3 or 4 is the fake one.

Thus depending on weather its case one or case 2 we get 3 or 4 tries.

Don’t know if this would be quickest but I’d split them 15:15 and weigh, discarding the heavier set as all genuine. I’d then weight 7:7 with one left out. If they are the same then the remaining coin is counterfeit, if not I’d take the discard the heavy 7, and add the lighter 7 to the remaining 1. That 8 i’d spit 4:4, and again retain the lighter 4. Split those 2:2, keeping the ligher pair and then weigh the final coins to find out which is the counterfeit.

So you’ll either get lucky and have the answer after 2 weigh ins or you’ll have to weigh them 5 times.

You could just put the coins on the scales two at a time (one in each pan) until it no longer balances, and then count that as one continuous weighing, but these problems usually count adding or removing something from the scales as another weighing. So assuming we can’t do that…

Divide the coins into three piles of ten. Weigh two of the piles on the balance - if one of them is lighter than the other, the counterfeit is in that pile; if they both weigh the same, the counterfeit is in the pile you didn’t weigh. Whatever the result, you’re down to 10 coins. Divide that pile into 3, 3 and 4. Put the two groups of three coins on the balance, and again whichever is the lightest has the counterfeit (or if they balance, it’s among the four). If we have three coins left, only one more weighing is needed - one in each pan, and the third not weighed. If we have four coins, two more weighings may be needed - two in each pan to eliminate a further two genuine coins, then the two remainder can be compared. By this method, only four weighings are needed at the most.