Recreational Math Puzzle
Straits Times has posted Cheryl’s Birthday Problem – Part Two. However, it is actually identical to an old math problem that has existed for decades.
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Straits Times has posted Cheryl’s Birthday Problem – Part Two. However, it is actually identical to an old math problem that has existed for decades.
You are given 12 balls which are identical in appearance except that one of them is of different weight from the others. You are not told whether the odd ball is heavier or lighter.
Using at most 3 weighings on a scale balance (similar to the one depicted below), determine which is the odd ball and also determine whether it is heavier or lighter than the other balls.
5 pirates of different ages have a treasure of 100 gold coins.
On their ship, they decide to split the coins using this scheme:
The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it.
If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.
As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.
Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?
9724-0163 (Mr Ng)
Ph.D. in Mathematics;
B.Sc. First Class Honours in Mathematics