Counterfeit Coin

One of twelve coins is counterfeit: it is either heavier or lighter than the rest. Is it possible to identify the counterfeit coin in three weighings on a beam balance?

Solution:

This solution is first presented by "BRIAN D. BUNDY":

We may label the coins 1, 2, ..., 12 so that we can distinguish between and identify them using these labels.

1.            1. Weigh 1, 2, 3, 4 against 5, 6, 7, 8.

2. They balance, so 9, 10, 11, 12 contain the odd coin. Weigh 6, 7, 8 against 9, 10, 11.
1. They balance, therefore 12 is the odd coin and so weigh 12 against any other to discover whether it is heavy or light.
2. 9, 10, 11 are heavy and so they contain an odd heavy coin. Weigh 9 against 10. If they balance, 11 is the odd heavy coin, otherwise the heavier of 9 and 10 is the odd coin.
3. If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light coin.
3. 5, 6, 7, 8 are heavy and so either they contain an odd heavy coin or 1, 2, 3, 4 contain an odd light coin. Weigh 1, 2, 5 against 3, 6, 10.
1. They balance, so the odd coin is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy coin.
2. 3, 6, 10 are heavy, so the odd coin can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light coin.
3. 3, 6, 10 are light, so the odd coin is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.
4. If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

Four Words

Four words related to basic gardening have been written in a row with neither spaces nor vowels. What are these four words?

PRNNGSLWDRSCP

Pruning
Soil
Weeder
Scoop

Shape and Size

With 10 sticks divide this figure into four parts that are all equal in shape and size!

1/3 Probability by One Coin

At a restaurant, how can John choose one out of three desserts with equal probability with the help of one coin?

He tosses the coin twice; where he supposes HT, TT and TT correspond to his three choices and if HH he repeats.

Which is the odd one out?

Which is the odd one out?

C.

C has three white and one black on the left and three black and one white on the right. The others are the opposite way round.