(Yes, there was a small goof-up last week when some copy got accidentally repeated but then
haven’t you also been separated at birth from your evil twin so many times before?)
Have you also ever been eight or nine years old and forcibly taken by your parents to a large gathering of people who have nothing better to do on a Sunday but to collect a bunch of cards with various numbers printed on them in various boxes. Up in front stands a person with a round rotating cage next to him in which are a whole lot of small balls with on particular number printed on them. He picks one at random, yells it out, and if you have the number on your card, you strike it out. The first person who strikes all of them out wins. Wow! And you go home waiting for Monday.
Luckily, there’s a complex method of manufacturing tickets for the game of Housie (also known as Lotto or Tambola). Every set of six tickets contain all the 90 numbers from 1 through 90. Every ticket contains 15 numbers, every row five and every column at least one. Exactly how many different Housie tickets are possible?
(Gimme dat ol’ time religion problem was: “A population starts with a single amoeba. For this and for the next generation, there is a 3/4 probability that an individual amoeba will split to create two amoebae and a 1/4 probability that it will die without producing an offspring. What’s the probability the family tree of original amoeba will go on forever?”)
For the amoeba family to survive, its offspring lineage must survive. There are three possibilities here:: both dying, one of them dying or both of them splitting. The family tree continues in two of the three cases. The same argument goes for each generation. So the probability is 2/3. Saifuddin S F Khomosi, Dubai.
Let the probability of an amoeba splitting into two be p and the probability of everlasting amoeba family tree be P. So, the probability of both second-generation amoebae perishing becomes (1 - P)^2. (And so on and so forth until . . . -- MS) With p = 3/4, we get P = 2/3.
Abhay Prakash, firstname.lastname@example.org
(The second one was: “Solve: TWO + THREE + SEVEN = TWELVE, given that each alphabet stands for a distinct digit 0 thru 9 both inclusive.”)
There are 10 alphabets in the equation and the values of the alphabets work out to be as follows- T = 1, E = 2, L = 3, N = 4, V = 5, O = 6, R = 7, S = 8, H = 9, W = 0. In the problem TWO + THREE + SEVEN = TWELVE, if you replace the alphabets with the corresponding numbers you get 106 + 19722 + 82524 = 102352 and this solves the problem. You can also interchange the values of O and N as O = 4 and N = 6, and still get the problem solved. -- Gnanaseharan Ponnaiah, email@example.com
Obviously T = 1, S = 8, H = 9,W = 0. V has to be odd, sum of O + N =10. All letters represent distinct digits, so a little trial and error gives V= 5, E = 2, R = 7, L = 3, O = 4 OR 6, N = 6 OR 4. -- V K Bargah, firstname.lastname@example.org
(The third problem was about completing two vaguely science worded ‘fizycks’ limericks.”)
The answers to the fill-in-the-blanks questions given under “Fyzicks Limericks and all that jazz” is given below: (1) I know they are there “just the same”(three words); (2) At the end of which “grinned Archimedes” (two words). -- Narayana Murty Karri, email@example.com
I used a very student friendly, light-hearted text book called Physics For You by Keith Johnson. My answers, hence, are not original but what I had enjoyed with a lot of students. Thank you for the trip down memory lane. -- J Vaseekhar Manuel, firstname.lastname@example.org
But Google This Now
1. Five hungry friends decided to share a large pizza with each receiving just one slice. Now each of them has one odd habit. A person eats a slice that is exactly 1/N of the entire pizza. One of these five guys ate a slice whose size was 1/19th of the whole. What were the sizes of the slices of his four friends if the entire pizza was eaten?
2. Below are 13 five-lettered nouns, each of which has had two of its letters removed. In total there are 26 letters from A to Z. The remaining letters in each word are in the correct order. They are: clh,
dir, ine, eba, ual, toe, yat, coh, rea, okr, aor, ree, sam.