Second whirled war.... and a dash of spooky fun!

Looks like I’ve been setting a few difficult ones of late considering some problems had to be held over for more than three issues and even then hints had to be given.

Looks like I’ve been setting a few difficult ones of late considering some problems had to be held over for more than three issues and even then hints had to be given. So in all fairness here’s something so ridiculously simple garbed in something so ridiculously difficult that it’s fun.
It concerns the year 1944 and some leaders involved in the Second World War. Take Churchill for instance. He was born in 1874, his age in 1944 was 70, he took office in 1940 and was there for four years. If you add up the numbers they come to 3888. Now take Hitler. Born: 1889; age in ’44: 55; took office 1933, years there: 11. Total 3888. You can do the same with Roosevelt, Stalin and Mussolini too. Spooky or what?

THROUGHPUT
(The Ramsay Locks problem was admittedly a tough one because of its probabilistic nature. However one person did get it somewhat correct, though Google might have something to do with it since no one ever bothers to rewrite. They just cut and paste!)
The question does not take into account the actual maximum number of attempts. We only consider the most probable maximum number of attempts but then what prevents the computer from getting sick or tired. For the purpose of the question, we undertake, maximum time undertaken to open one door. The maximum number of attempts that can be made are 100*99*98*97 = 94109400, assuming that there is a wrong attempt before and after the door is opened. Thus for 16 doors (64/4 = 16), it will take 1505750400 attempts. Thus, a maximum of 25095840 minutes, 418264 hours, 17428 days or 48 years. -- Shashi Shekher Thakur, shashishekher@yahoo.com
(The second problem was: “The two halves of a chapatti are always of unequal thickness; one thin like paper, the other thicker. It’s the same thing with deep-fried poories. Ever wondered how to make them with halves of equal thickness?”)

This can be done if the chapattis are kneaded well, water is fed into dough very slowly and the dough turns soft and smooth. Then it can be rolled out very thin. While puffing, the water in between the layers doesn’t climb up through the dough to reach the final thin layer -- the skin which has formed due to earlier heated side. But we can’t ever get 100% equal layer. -- Narayanan P S, narayananpsn@gmail.com
(The last one was about calculating the number of animals from five given clues.)
Assuming there is one freak beast in each category (and not all are freaks), there are a total of 15 mammals, 25 birds, 19 snakes, 21 spiders, and 17 insects. -- Ravi Nidugondi,
ravi.nidugondi@gmail.com
 Since there are three extra heads (2 of a snake and 1 of a bird), it emerges that there are only 97 animals. After working on the various combinations, accounting for the freak animals and satisfying all the given conditions, the final grouping  of 97 animals is as follows: There are 15 mammals (15 heads and 59 legs); 25 birds (26 heads and 50 legs); 19 snakes (21 heads and zero legs); 21 spiders (21 heads and 167 legs) and 17 insects (17 heads and 100 legs). Only this combination meets all the requirements. -- Narayana Murty Karri, k_n_murty@yahoo.com (Yes, J Vaseekhar Manuel, orcontactme@gmail.com; and Balagopalan Nair K, balagopalannair@gmail.com, you also got it correct.)
(And finally someone got the held over chess problem with the modified rules where White and Black both get two first moves each to begin with.)
Let White move one knight out and back in again. The board returns to the starting position but White has become, in effect, Black now since it’s Black’s turn to go first! If Black repeats the same move to get back to being Black and they do it three times then the original White can assure a draw! -- Altaf Ahmed, ctrlaltaf@yahoo.in

BUT GOOGLE THIS NOW
1. If pi is transcendental then any given digit should occur in it. So too any given pair of digits. And generalising, any given sequence of digits. Meaning any given message can be found in pi. Pi contains all possible information in the universe. Where’s the flaw here?
 2. In the beaker, oil is floating in water. But for all floating bodies, a portion of the body will be below the water surface (depending on its density). So why does the whole oil float? – (Submitted by Prof S Manikutty, manikuti@iima.ac.in)

Sharma is a scriptwriter and former editor of Science Today magazine.(mukul.mindsport@gmail.com)

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