3 hours ago
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Earlier today I asked you these three puzzles. Here they are again with solutions.
1. Battleships
You are an admiral in the Navy, in charge of an important mission. You have two choices.
a) To send a single ship whose chance of success is P per cent.
b) To send two ships, each of whose chance of success is P/2 per cent. At least one ship needs to be successful for the mission to be a success.
Which is the better option?
Solution. a) Send a single ship.
Intuition might tell you that sending the two ships is better, since two chances are better than one. However, it is easy to see that this is not the case if P = 100 . In this case, a) guarantees success, but b) only gives 75 per cent of success. (Since the chance of failure of both ships is 50 per cent x 50 per cent = 25 per cent.)
It turns out that for all values of P, sending one ship is better. Let p be the probability of success. (i.e P/100). If two ships each have probability p/2, the chance that both fail is (1–p/2)2, so the chance at least one succeeds is 1 – (1–p/2)2 = p – (p2)/4, which is always less than p, the chance of the single boat succeeding.
2. The two oracles
Before you are two oracles, Randie and Rando, who will answer yes or no to any question you ask.
Randie answers yes or no at random to all questions.
Rando decides at random whether to tell the truth or to lie for each question, and then answers the question accordingly.
Is there a way to tell them apart? If so, what is it?
Solution. Yes, you can do it!
The trick is to realise that it is possible to ask Rando questions that you know it will always answer “YES”.
Such as: “Are you answering this question truthfully?” Both a liar and a truthteller give the same answer.
So ask this question until you get a No. Once you do, you know it’s Randie. If you don’t, you can be pretty sure it is Rando.
3. Bad maths
Johnny’s homework is to calculate 5548-5489. The answer is 59. He figured that the 548 cancelled out, leaving 59.
He tested the technique again. He typed in a subtraction of the form XXYZ – XYZW where X, Y, Z and W were distinct digits, and found out it was indeed XW!
How many of the digits in the new calculation are the same as the digits in the old one? (i.e does X = 5, Y = 4, Z = 8 or W = 9)
Solution. Z and W are 8 and 9
The caclulation can be broken down as follows:
1100X +10Y + Z – 1000X – 100Y – 10Z – W = 10X + W
which reduces to
90X – 90Y = 9Z + 2W
We deduce that W must be divisible by 9, so it is either 0 or 9. Also, 9Z + 2W must be divisible by 10.
If W = 0, then Z = 0, which is a contradiction since they question states they are distinct.
So W = 9. Then 9Z + 18 is divisible by 10, which means Z = 8. We are left with
90X – 90 Y = 90. Thus
X = Y + 1, which has several solutions. But W = 9, and Z = 8.
Today’s puzzles are taken from an excellent new compendium of brainteasers, Mathematical Puzzles and Curiosities, by Ivo David, Tanya Khovanova and Yogev Shpilman. I have edited the wording of the puzzles and the solutions to make them fit the column.
I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
