False coin

1st Problem

We have 3 coins, and a twin-pan balance without weights. With the balance we can determine whether the weight of the coins put into the two pans are equal or not. One of the coins is false, the height is different from the weight of the genuine coin. The weight of the genuine coins is equal. Which one the false coin and lighter or heavier than the genuine one?

Solution:

Mark the coins as 1, 2, and 3. Put the coin 1 to the left side pan and the coin 2 to the right side pan.

If the balance sheet shows that the weights are equal, then the coin 3 is false. But we do not know yet that it is lighter or heavier than the genuine coin. We need one more measuring. Now we know that the coin 1 is genuine, we can compare it with coin 3. The second measuring will tell us that the false is heavier or lighter than genuine coin.

If the balance sheet shows that the weights are equal, then the coin 3 is genuine. Another measuring is needed to determine the coin 1 or the coin 2 is the false coins.
Assume now that the coin 1 is lighter than the coin 2 and replace the coin 2 with the genuine coin 3. If the weight is equal, then the coin 2 false, and it is heavier than the genuine coin. If the coin 1 lighter than the coin 3, then the coin 1 false, and lighter than the genuine coin. The coin 1 can not be heavier than the coin 3, because it would be the 3 coins of different weights.

Note: In all cases, two measurings are needed to determine which is the false coin, and to find lighter or heavier than the genuine.

2nd Problem

There are 12 coins and the same balance as it is in the first Problem. Weight of genuine coins is the same, but there is one of 12 coins which is false, with a different weight. We don’t know if it is heavier or lighter than the genuine coin. Determine with maximum 3 measurings which is the false coin, and that it is lighter or heavier as genuine coin.

Solution:

Mark the coins as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

The first measuring compares coins 1, 2, 3, 4 with coins 5, 6, 7, 8.

- If they are of equal weight, they are 8 coins for which we know is genuine.

From the remaining four coins select 3 and compare them with 3 genuine coins:
coins 9, 10, 11 with coins 1, 2, 3.

If they are equal, then the false is the coin 12. The third measuring: compare coin 12 with a genuine coin, and we get to know it lighter or heavier than genuine.
If, however, they are not equal, we can immediately know that the false coin is lighter or heavier one. Select two of the three coins, compare them and we will find the false.

- If the result of the first measuring is: not equal, we know that the coins 9, 10, 11, 12 are genuine. 2 measurings remained to find the false coin from the remaining 8 coins.

Suppose that the coins 1, 2, 3, 4 are lighter than the coins 5, 6, 7, 8 at the first measuring.

Compare now the coins 1, 5, 9, 11 with coins 2, 3, 6, 7 with the balance!

Three cases must be look over.

a) If 1, 5, 9, 11 and 2, 3, 6, 7 is equal, they are all genuine coins and the coin 4, or the coin 8 can be the false coin. We, however, know that coin 4 is lighter than coin 8, so it is enough to compare to a genuine one coin of them and we know that which is false.

b) If 1, 5, 9, 11 heavier than 2, 3, 6, 7, this case the coin 6, coin 7 and coin 1 can not be false, because by the first measuring it would be heavier and by the second measuring it would be lighter than the genuine coin. Therefore, the false is the coin 5, and it is heavier than genuine, or the false is the coin 2 and lighter than the genuine coin, or the false is the coin 3 and lighter than the genuine coin. It is enough to compare coin 2 and coin 3. If one of them is lighter than it is the false if they are equal then the coin 5 is the false.

c) Finally, when 1, 5, 9, 11 lighter than 2, 3, 6, 7, then coins 2, 3 and 5 are genuine. The coin 1 can be false, and lighter than genuine, or the coin 6 can be false and heavier than genuine, or the coin 7 can be false and heavier than genuine. We have to compare coin 6 and coin 7 to find the false one.

So, 3 measuring were to be sufficient to find the false coin and tell that its weight is lighter or heavier than the genuine.

Note: The next Problem could be very interesting after a detailed examination of previous two Problems

3rd Problem

Given 10 boxes, each with 10-10 coins and a single-pan balance. We can measure the weight of coins with the balance. We know that genuine coins are in nine boxes, each genuine coins weight is 10 pounds. However there is one box with false coins, each weight is 9 pounds. What is the minimum number of measuring, which allows you to be able to choose the box containing the false coins?

Solution:

Number the boxes in mind: 1, 2, 3, … 10. Take 1 coin from the box 1, 2 coins from the box 2, 3 coins from the box 3, … and finally 10 coins from the box 10. Place the coins into the pan and check the weight. If in each box would be only genuine coin, the weight would be: 10 +20 +…+ 100 = 550 (pounds).
If the false coins are in the box 1 the weight sum is lesser with 1 pound as we placed 1 false coin to the pan.
If the false coins are in the box 2 the weight sum is lesser with 2 pounds as we placed 2 false coins to the pan.
…
If the false coins are in the box 10 the weight sum is lesser with 10 pounds as we placed 10 false coins to the pan.
Thus, a single measuring has been enough to find the box containing the false coins.