(1) I own no birds less than eight feet high; There are no birds in this aviary that belong to anyone but me; No ostrich lives on coconuts; No birds, except ostriches, are eight feet high.
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(2) All the human race, except my footmen, have a certain amount of common sense; No one, who lives on barley sugar, can be anything but a mere baby; None but a hopscotch player knows what real happiness is; No mere baby has a grain of common sense; No engine-driver ever plays hopscotch; No footmen of mine is ignorant of what true happiness is.
(3) The only animals in this house are cats; Every animal is suitable for a pet, that loves to gaze at the moon; When I detest an animal, I avoid it; No animals are carnivorous, unless they prowl at night; No cat fails to kill mice; No animals ever take to me, except what are in the house; Kangaroos are not suitable for pets; None but carnivores kill mice; I detest animals that do not take to me; Animals that prowl at night always love to gaze at the moon.
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Dear
MS, |
There are N persons in the queue having either 50 or 100 bucks. The total number of events will be N + 1 (considering all having 100 bucks to all having 50 bucks). The cashier will not face a problem of change if at least half the persons have 50 bucks. If N is odd then the probability will be (N + 1)/2(N + 1) = 0.5. If N is even than the probability will be (N + 2)/2(N + 1). The value will depend on the number of persons but always a value more than 0.5. So, the minimum probability will be 0.5. Sanjeev Kumar, sanjeevnpc@rediffmail.com
Lucky-That-Time-Dept:
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Golden-Sections-Dept: Regarding the logic behind the series 1, 2, 3, 5, 10, 25, after trying several half-baked possibilities, I have stumbled onto this: First, take the Fibonacci sequence f(i) = 0, 1, 1, 2, 3, 5, 8, 13, 21, where i = 0 to infinity. Then the numbers in the series (let's call it a(n) where n = 1 to infinity) in question can be generated as follows: a(1) = 1 = 0^1 + 1^0 = 0 + 1 a(2) = 2 = 0^2 + 1^1 + 1^0 = 0 + 1 + 1 a(3) = 3 = 0^3 + 1^2 + 1^1 + 2^0 = 0 + 1 + 1 + 1 a(4) = 5 = 0^4 + 1^3 + 1^2 + 2^1 + 3^0 = 0 + 1 + 1 + 2 + 1 a(5) = 10 = 0^5 + 1^4 + 1^3 + 2^2 + 3^1 + 5^0 = 0 + 1 + 1 + 4 + 3 + 1 a(6) = 25 = 0^6 + 1^5 + 1^4 + 2^3 + 3^2 + 5^1 + 8^0 = 0 + 1 + 1 + 8 + 9 + 5 +1. Or, writing as a closed-form formula a(n) = Summation(f(i)^(n-i)). By this logic, some of the next numbers in the series are: 79, 318, 1637, 10753, etc. Kumar Vijay Mishra, Colorado State University, USA.
The sequence 1, 2, 3, 5, 10, 25 is part of A(x) = Sum_{n>=0} x^n/(1 - Fibonacci(n)*x). I don't understand this one bit but if you don't believe me then look up A135961 at On-Line Encyclopaedia of Integer sequences. The next few terms are: 79, 318, 1637, 10753, 89872. Write any sequence and you'll almost always find some logic to arrive at the next term -- much like reading and understanding quatrains written by Nostradamus where one can read any meaning into any of them.
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Change-Of-Heart-Dept:
In order to avoid the change problem, for every man having Rs 100 there should be one with Rs 50. Therefore there won't be any problem if out of those N people number of people carrying 100 note is from 0-N/2. Hence I guess in half of the cases there won't be any problem. So probability is 0.5
Tarun, nsit.tarun@gmail.com
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Endgames
When the following problem was first run there was no Internet (how did people live then?) and yet some people still managed to cheat the answer out. Amazingly, even with the Net around today, it's still only just as difficult as it was then. Here it is, all the way from 1986: If the end of the world should come on the first day of a new century, what are the chances that it will happen on a Sunday?
(Originally submitted by Mitali Chatterjee, Calcutta, India)
