Ball bearing gifts

Hey people can I make a small request? Don’t keep sending me the 12 balls problem to run. You know the one where one of them is either lighter or heavier and you have three weighings to figure it out

Hey people can I make a small request? Don’t keep sending me the 12 balls problem to run. You know the one where one of them is either lighter or heavier and you have three weighings to figure it out because not only is it the oldest one on the block which even newborns know about by now but it can also be Googled with one hand tied behind your front. But since you seem to be so much on the ball let me gift you with some more..


 You’re given two balls which are red in colour, another two ditto in green and finally ditto ditto yellow. Meaning six in all balls. However, for each colour one ball is heavy while the other’s light.

Conveniently though, all heavy balls weigh the same as do all the light ones. And even more conveniently you also have a beam balance at your unique disposal to do what you want with. So how many weighings would you require to identify the three heavy ones?

THROUGHPUT
(The smelly old problem was: “Find a famous Greek mythical act of heroism in the following nonsensical sentence: ‘Provide men the use so toleration finds residence free of misleading theories gift of decried species’.”)


After trying out various combinations like joining first letters, last letters, letters at different positions, etc. I indulged in various unequal cuts of the words from start as well as end. The sentence “PROMETHEUS STOLE FIRE FROM THE GODS” emerges after taking one or more starting letters from each word.

I don’t know if this is right. -- Saishankar Swaminathan, saishankar482@gmail.com
Taking at least one letter from the beginning of each of the sixteen words of the nonsensical sentence and grouping them together appropriately we get the famous Greek mythical act of heroism as “PROMETHEUS STOLE FIRE FROM THE GODS”. -- Narayana Murty Karri, _n_murty@yahoo.com
(The second one was: “?, LI, MA, NA, AM, NA, HG, YD, DE, IR, NI, NA, DA. And the hint was: Tianzhu honcho.” Incidentally only one person got this. Good for you NB!)


This is the answer of the Tianzhu honcho puzzle. The hint actually got me puzzled initially but everything fell into piece once it was cracked.

Tianzhu honcho literally translates to “India head” and the series is made up of the reversed last two letters of the surnames of the thirteen Indian Presidents in the reverse order: -- ie, Pranab MukherjEE, Pratibha PatIL, A P J Abdul KalAM and so on. -- Neethi Balagopal, neethibala@gmail.com


(The third puzzle was: “A cup of wine is suspended over another cup of equal capacity full of water. Through a very small hole in the bottom, wine drips into the water, and the mixture again drips out of a tiny hole in the lower cup at the same rate. When the wine cup is completely empty, what part of the content of the lower cup is water?”)


My thinking is that if the flow is sufficiently small (assumed constant and equal for cups),  the maximum percentage of water in the lower cup cannot exceed 1/e where ‘e’ is the base  of Napierian logarithm 2.718. Accordingly, the water content will be a maximum of 36.787%. If the cup capacities are say 100 cc with wine/mixture dripping out at a constant rate of say 1 cc/sec then final water content after 100 seconds would around 36.6%. -- Ajit Athle, ajitathle@gmail.com


Assuming that the hole in the cup is very small, so that the drops too are very small, the number of drops, n, will be very large. Thus, in the limiting case of n tending to infinity, (1 - 1/n)^n tends to 1/e, e being Euler’s number approximately equal to 2.71828. Consequently, we can say that the mixture in the lower cup, when all the wine in the upper cup has dripped into it, has 1/2.71828 parts, or rather, slightly more than one part in three, of water. -- Balagopalan Nair K, balagopalannair@gmail.com


The water will be approx 36.8% unless some miracle man is able to convert water into wine. -- Kishore Rao, kishoremrao@hotmail.com

BUT GOOGLE THIS NOW
1. Decapitate me and all becomes equal. Then truncate me and I become second. Cut me front and back and I become two less than I started.


2. A man with body surface area of 1 square metre, has to cover 100 metres in the rain on foot. There are 50 raindrops falling per cubic metre per second. Should he walk or run to avoid getting too wet?
Sharma is a scriptwriter and former editor of Science Today magazine.(mukul.mindsport@gmail.com)

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