The fake coin is indistinguishable from the rest except that it is either heavier or lighter, but you don't know which.
Can you determine which is the fake coin and whether it is lighter or heavier using a balance scale and only 3 weighings?
Direct Link: www.brainbashers.com?ZLKY
Hint: Finding the correct weighings requires some very careful thinking.
One solution is to label the coins with the letters FAKE MIND CLOT and weigh the coins in the following three combinations:
MA DO -- LIKE
ME TO -- FIND
FAKE -- COIN
Logic will now allow you to find the fake coin based on the three results. Bearing in mind we don't know whether the fake coin is lighter or heavier.
For instance, if the results were left down, balanced, left down, we could work out which coin is fake in the following way:
From the middle weighing we know that the coins METOFIND are all normal. So one of the coins ACKL is fake. Therefore looking at these coins one at a time in the other two weighings, we can see that:
A - appears on the left twice and could be fake.
C - appears only once, therefore can't be fake (otherwise the first weighing would be balanced).
K - appears on opposite sides, so it can't make the left side go down both times.
L - appears only once, therefore can't be fake (otherwise the third weighing would be balanced).
Therefore the only possibility is A, which must be heavier. Any other combination of ups and downs will allow you to use the same logic to find the fake coin.
Answer: 6423: 109191 / 17. The numbers 1234567890 have been replaced by 7182435069 respectively.
Below is a very special grid, around each shaded number are 8 white squares. However, each white square should have a number from 1 to 7. Once filled in, these 8 numbers will sum to the shaded number. In addition, once completed correctly, no row nor column will contain a duplicate number within a white square. For example, the top row may be 5 6 4 2 3 1 7, etc. This is a very difficult puzzle, and many people resort to using a computer to help them, which is a challenge in itself.
Hint: This puzzle almost certainly requires trial-and-error to solve.
There are a number of correct solutions, this is just one of them.