## Category Archives: Basic Set Theory and Logic

### The infinite hotel paradox : due Jeff Dekofsky

This hardcore stuff about “infinity” , quite nicely, explained was pointed out to me by my ISC XII student, Mr. Utkarsh Malhotra! 🙂

### Logicalympics — 100 meters!!!

Just as you go to the gym daily and increase your physical stamina, so also, you should go to the “mental gym” of solving hard math or logical puzzles daily to increase your mental stamina. You should start with a laser-like focus (or, concentrate like Shiva’s third eye, as is famous in Hindu mythology/scriptures!!) for 15-30 min daily and sustain that pace for a month at least. Give yourself a chance. Start with the following:

The logicalympics take place every year in a very quiet setting so that the competitors can concentrate on their events — not so much the events themselves, but the results. At the logicalympics every event ends in a tie so that no one goes home disappointed 🙂 There were five entries in the room, so they held five races in order that each competitor could win, and so that each competitor could also take his/her turn in 2nd, 3rd, 4th, and 5th place. The final results showed that each competitor had duly taken taken their turn in finishing in each of the five positions. Given the following information, what were the results of each of the five races?

The five competitors were A, B, C, D and E. C didn’t win the fourth race. In the first race A finished before C who in turn finished after B. A finished in a better position in the fourth race than in the second race. E didn’t win the second race. E finished two places  behind C in the first race. D lost the fourth race. A finished ahead of B in the fourth race, but B finished before A and C in the third race. A had already finished before C in the second race who in turn finished after B again. B was not first in the first race and D was not last. D finished in a better position in the second race than in the first race and finished before B. A wasn’t second in the second race and also finished before B.

So, is your brain racing now to finish this puzzle?

Cheers,

Nalin Pithwa.

PS: Many of the puzzles on my blog(s) are from famous literature/books/sources, but I would not like to reveal them as I feel that students gain the most when they really try these questions on their own rather than quickly give up and ask for help or look up solutions. Students have finally to stand on their own feet! (I do not claim creating these questions or puzzles; I am only a math tutor and sometimes, a tutor on the web.) I feel that even a “wrong” attempt is a “partial” attempt; if u can see where your own reasoning has failed, that is also partial success!

### Pick’s theorem to pick your brains!!

Pick’s theorem:

Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as $(B/2)+I-1$ square units, where B is number of lattice points on the boundary, I is number of lattice points in the interior of the polygon. Prove this theorem!

Do you like this challenge?

Nalin Pithwa.

### Genius of Srinivasa Ramanujan

1. In December 1914, Ramanujan was asked by his friend P.C. Mahalanobis to solve a puzzle that appeared in Strand magazine as “Puzzles at a Village Inn”. The puzzle stated that n houses on one side of the street are numbered sequentially starting from 1. The sum of the house numbers on the left of a particular house having the number m, equals that of the houses on the right of this particular house. It is given that n lies between 50 and 500 and one has to determine the values of m and n. Ramanujan immediately rattled out a continued fraction generating all possible values of m without having any restriction on the values of n. List the first five values of m and n.
2. Ramanujan had posed the following problem in a journal: $\sqrt{1+2\sqrt{1+3\sqrt{\ldots}}}=x$, find x. Without receiving an answer from the readers, after three months he gave answer as 3. This he could say because he had an earlier general result stating $1+x=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}}$ is true for all x. Prove this result, then $x=2$ will give the answer to Ramanujan’s problem.

Try try until you succeed!!

Nalin Pithwa.

### Men, monkey and coconuts problem

Five men and a monkey collected a heap of coconuts in a desert island. The men went to sleep in the night. One of them got up in the middle of the night and divided the heap into five equal parts and found an extra coconut, which he gave to the monkey. He hid his share and made one heap combining the rest of the coconuts. Then, the second person got up and divided this new heap into five equal parts and again found an extra coconut, which he gave to the monkey. He also hid his share and made a new heap of the remaining coconuts. This process continued until the last person did exactly the same thing as the others. What must be the minimum number of coconuts in the original heap if

(a) the next morning when they divided the last remaining heap into five equal parts, there was no coconut left for the monkey.

(b) the next morning they divided the last remaining heap into five equal parts, they again found an extra one, which they gave to the monkey.

Cheers,

Nalin Pithwa.

PS: One good way to increase concentration, motivation, intellectual stamina for solving such demanding puzzles or math problems for the IITJEE or RMO/INMO or even Mensa challenges is to give yourself a “reward” after you actually solve that problem. So, you may go ahead and have chilled coconut juice after you solve this puzzle. Anyway, these are the hot summer days in India.

### Alice in Wonderland and probability stuff !!

Lewis Carroll of Alice’s Adventures in the Wonderland was a mathematician in Cambridge. He had posed the following problem in one of his books:

A box contains a handkerchief, which is rather white or black. You put a white handkerchief in this box and mix up the contents. Then you draw a handkerchief, which turns out to  be white. Now, if you draw the remaining handkerchief, what is the probability of this one being white?

How likely is that you would enjoy such questions ?! 🙂

Nalin Pithwa.

### Gruffs

The reporter for the local Gazette was at the national dog show yesterday. Unfortunately, he also had to report on quite a few other events on the same day. He made some notes when he was at the dog show which are shown below, but the editor of the Gazette has asked for a list of the 26 finalists and the positions in which they finished overall. Using the reporter’s notes below, see if you can construct the list for the editor, as the reporter didn’t have time to make a note of the overall positions.

The Afghan Hound finished before the Alsatian, the Poodle and the Beagle. The Beagle finished before the Bull Mastiff and the Labrador. The Labrador finished before the Poodle and the Pug. The Pug finished before the Alsatian, the Dobermann Pinscher, the St. Bernard, the Sheepdog and the Griffon. The Griffon finished before the Sheepdog. The Sheepdog finished after the Poodle. The Poodle finished before the St. Bernard, the Collie, the Pug, the Griffon, the Alsatian and the Dobermann Pinscher. The Dobermann Pinscher finished after the Alsatian. The Alsatian finished after the Chow. The Chow finished before the Afghan Hound, the Bulldog, the Chihuahua, the Poodle, the Beagle and the Dachshund. The Dachshund finished before the Spaniel. The Spaniel finished before the Foxhound.

The Foxhound finished before the Labrador. The Labrador finished after the Whippet. The Whippet finished before the Great Dane. The Great Dane finished after the Bull Terrier and before the Chow. The Chow finished before the Bull Mastiff. The Bull Mastiff finished before the Foxhound. The Foxhound finished after the Yorkshire Terrier. The Yorkshire Terrier finished before the Greyhound. The Greyhound finished before the Dachshund. The Dachshund finished after the Kind Charles Spaniel. The King Charles Spaniel finished before the Bull Mastiff, the Greyhound, the Pug, the Chihuahua and the Afghan Hound. The Afghan Hound finished after the Dalmatian. The Dalmatian finished before the Pug, the Labrador, the Poodle and the Collie.

The Collie finished before the Pug. The Pug finished after the Retriever. The Retriever finished before the Bull Terrier, the Chow, the Yorkshire Terrier and the Whippet. The Whippet finished before the Chow, the Spaniel, the Dalmatian and the Yorkshire Terrier. The Yorkshire Terrier finished after the Bulldog. The Bulldog finished before the Chihuahua and the Dalmatian. The Dalmatian finished after the Great Dane. The Great Dane finished before the Yorkshire Terrier. The Yorkshire Terrier finished before the Collie, the King Charles Spaniel, the Spaniel and the Dalmatian. The Dalmatian finished after the Spaniel. The Griffon finished before the Alsatian. The Alsatian finished after the Sheepdog. The Griffon finished after the St. Bernard. The Greyhound finished before the Chihuahua. The Chihuahua finished before the Dachshund. The Bull Terrier finished before the Whippet and the Yorkshire Terrier. The Afghan Hound finished before the Pug and the Foxhound.

So, if u try this, it might ‘hound’ u !!! Good puzzles ain’t easy! But, such puzzles help u develop the habit and power of sustained thinking on a problem for long hours.

Nalin Pithwa.

### I have a sweet tooth !!!

1. Tracey will not get the chocolate-covered mints unless Neil has the plain mints.
2. Alan will not get the toffee unless Robert has the mint-flavoured toffee.
3. Neil will not get the plain mints unless Alan has the mint-flavoured toffee.
4. Robert will not get the toffee unless Tracey gets the plain mints.
5. Robert will not get the chocolate-covered mints unless Tracey gets the toffee.
6. Tracey will not get the toffee unless Neil gets the chocolate.
7. Robert will not get the mint-flavoured toffee unless James gets the plain mints.
8. Alan will not get the plain mints unless Tracey gets the toffee.
9. Tracey will not get the plain mints unless Alan gets the chocolate-covered mints.
10. Alan will not get the mint-flavoured toffee unless Tracey gets the chocolate-covered mints.
11. James will not get the plain mints unless Tracey gets the toffee.
12. Neil will not get the chocolate unless Alan gets the chocoate-covered mints.
13. Roberts will not get the plain mints unless James gets the toffee.
14. James will not get the toffee unless Tracey gets the plain mints.
15. Neil will not get the toffee unless Tracey gets the toffee.
16. Tracey will not get the plain mints unless Roberts gets the toffee.
17. Alan will not get the chocolate-covered mints unless James gets the plain mints.

Who will get what ?

Oh, to be precise, I do not have a sweet tooth. I have sweet-teeth! Ich habe eine stucke schokolade sehr gehn!

Auf wiedersehen.

Nalin Pithwa.

### People’s Pets

Consider the following:

1. Five men each have different first names and different surnames, have five different pets and live at five different addresses. All five pets have a different name.
2. Tom’s surname is Williams and the fish is not  called Spike, Benson or Rodney.
3. Harry has a pet cat and the budgie is called Percy.
4. George’s surname is not Hudson or Smith.
5. Mr. Thompson owns the dog and the owner of the rabbit lives in Pine Avenue.
6. The cat is not called Benson and one of the five men has a pet called Fred.
7. Mr. Anderson does not live in Cedar Road.
8. Mr. Hudson lives in Willow Street.
9. Mr. Anderson owns a pet called Percy and John lives in Cedar Road.
10. Bill’s pet is called Rodney and is not  the dog; the owner of the fish lives in Maple Grove.

So, who lives in Chestnut Crescent and what is the name of their pet?

-Nalin Pithwa.