Kids tough ....and silence reigns!

Three chess players agree to play a series of games for a prize which would go to the first player to win two consecutive games.

Published: 02nd December 2018 05:00 AM  |   Last Updated: 01st December 2018 12:16 PM   |  A+A-

Warning: this is only for children below the age of 12. Check out the following list of words: AT, HAT, MAT, MAD, SCRAP, PET, BIT, RID, STRIP, SLIM, PIN, QUIT, SIT, ON, HOP, SLOP, COP, TAP, CAP, BREATH, CUB, TUB, US, HUG, RAG, Adding just one letter to each could make the word longer, different in sound, in pronunciation, meaning (or both) even though the letter itself is not pronounced ever, in any one of them.

Throughput

(The problem was: “You’re camping at night. You have a torch that requires two batteries to light up. But then you realise that you’ve got a total of eight batteries of which four are charged and four are not. You don’t know which. What’s the least number of attempts needed to light up?”)

Least number is 7. Divide 8 into 3 groups two containing 3 each and last one two. This way we are sure one group has two working batteries. Now there are 3 combinations for the first 2 groups (3 batteries each) and 1 combination for the third one with two batteries. Total is 7.  -- V Srinivasa Murthy, vsmurthygm@gmail.com

If we attempt to insert one battery after another, it may take a long while. Let’s divide them into piles so that at least one pile has the possibility of having two charged batteries. This means we can’t divide into 2 piles of 4 each. 3 piles of 3, 3 and 2 torches will be better. The worst case would be for the 2 charged torches to be in the pile of 2 with the remaining divided between the 2 piles of 3. This means we will exhaust 3C2 + 3C2 = 6 attempts before finding a correct pair on the 7th. -- Saishankar Swaminathan, saishankar482@gmail.com

There are three ways two of the three from the first set can be tried AB, BC and AC. If all three attempts fail, at least two of the batteries from the first set are not charged. Repeat with set 2. If again all three attempts fail, two are not charged. That leaves the last pair as the useable, charged ones. So Elkhart might need six unsuccessful attempts or less (if he is lucky with either of the first two sets.) --  J Vaseekhar Manuel, orcontactme@gmail.com

Answer 8 attempts with A, B, C, D, E, F, G, and H as batteries. A + B: dim light (one is charged); C + D: dim light (one is charged); B + C:dim light (one is charged); C + A:either both charged (full bright) or bboth uncharged: no light. Thus, ABCD sorted. Repeat the same for EFGH. (worst luck 8 attempts). Least number 8. -- Narayanan P S, narayananpsn@gmail.com
(The second one was: “Can a pencil shaped (not blunt ended) log of wood be propelled fast enough underwater to heat up and burn?”)

Propelling the log of wood inside water may generate heat due to friction. But the heat is taken away by latent heat of evaporation due to vaporisation of water surrounding the log in addition to convection of heat in water. Since boiling point of water is 100 deg C the temperature under water cannot exceed 100 deg C it cannot burn the wood. -- Raghavendra Rao Hebbani, rao.raghavendrah@gmail.com
All types of wood contain three things -- water, cellulose and minerals. Wood goes through
three stages of burning. (1) Smoking stage up to about 200 degrees C when water and carbon dioxide are driven off. (2) Flame above 200 deg C, when pyrolysis takes place -- volatile gases are driven off, burn and raise the temperature and (3) Carbon burns at 400 deg C, the carbon left behind by this starts to burn causing burn marks. Since the temperature of water flowing under atmospheric conditions cannot exceed 100 deg C, it is not possible to heat up and cause burn marks on the wood by propelling it fast through water. -- K Narayana Murthy k_n_murty@yahoo.com 

But Google This Now
Three chess players agree to play a series of games for a prize which would go to the first player to win two consecutive games. The players draw lots to see which two shall play the first game and from then on the winner plays against the person who just sat out. Assuming the three players to have exactly equal skills, what are their respective chances?
Sharma is a scriptwriter 
and former editor of Science 
Today magazine.
(mukul.mindsport@gmail.com)

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