Problematics | A winner from long ago

A forgotten memory popped out recently from my scrapbook of newspaper clippings from 1993 and 1994. Then again, something forgotten cannot logically be called a memory: I have no recollection of this puzzle, which I published in January 1994, although I have no doubt that it is my own creation (we all recognise, even if we don’t remember, the signatures we put on so many places).

It is not very easy. I have managed to find a possible solution (there have been times when I cannot solve my own puzzles). I think the solution is unique; do let me know if it isn’t.

#Puzzle 133.1

A tournament of six teams begins with every team playing every other team in a round robin. These 15 league matches are followed by two semifinals, and then the final, a total of 18 matches. Pretty standard format.

In my original puzzle, I had named the six teams Mercury, Mars, Jupiter, Saturn, Neptune and Pluto (which was still considered a planet back in 1994) and there is no reason to change the names now. Here is how the tournament progresses:

1. Every match has a winner and a loser; there are no draws, ties etc.

2. Mercury, Jupiter, Saturn and Pluto are the four semifinalists. Each of these four win the same number of matches in the round robin.

3. Mars finishes bottom of the table, below Neptune. Their consolation: they both defeat the team that will go on to become runner-up.

4. Saturn defeats Jupiter in the league stage.

5. Pluto defeats Mercury in the league and again in the knockouts.

6. Two defeats in the league stage are avenged in the knockout stage.

List the results of all 18 matches.

#Puzzle 133.2

The year is 1977. A customer buys a samosa whose cost is 25 paise. He tries to pay with a 50p coin, but the shopkeeper does not have the required 25p change. The customer hunts his pockets and finds a 1-rupee note. “Give me that,” the shopkeeper says, and gives the customer 75p.

How is this possible?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 132.1

Hi Kabir,

There are three possible combinations. In two of these, it is Mr Chatterjee who never pays a bill on the same day that Ms Das pays the other bill. In the third one it is Ms Chatterjee.

— Professor Anshul Kumar, Delhi

The way I had originally planned the puzzle, the solution was intended to be unique with two different combinations. I had missed the third combination that gives a second possible solution. For fairness, readers who have given any one of these combinations are being counted as correct.

#Puzzle 133.2

Dear Kabir,

Let the room number be FrRr.

r has to be even because 1) Rr is twice Ff and 2) 1.5r is an integer. As twice Ff is Rr, a two-digit number, F has to be equal to or less than 4. As F/1.5 is an integer, F has to be 3. This gives r as 2. So the number is 3fR2.

Now for R2 to be twice of 3f, f can be either 2 or 6. With f as 2, R comes 6 which not acceptable because difference between f and R is 1. So F is 6. This gives R as 7.

The number is therefore 3672.

— Yadvendra Somra, Sonipat

Solved both puzzles: Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Anil Khanna (Ghaziabad), Vinod Mahajan (Delhi), Shri Ram Aggarwal (Delhi), Shishir Gupta (Indore), Aishwarya Rajarathinam (Coimbatore), Sanjay Gupta (Delhi), Dr Sunita Gupta (Delhi), Ajay Ashok (Delhi), Aditya Krishnan (Coimbatore)

Solved #Puzzle 131.1: YK Munjal (Delhi)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.