Problematics | Cat-and-mouse games and betting strategies

A puzzle made from a Russian original, involving cats and mice, and another from betting at a casino, involving wins and losses. Good luck!

Many of the puzzles you have solved here are my own, while there have been others of British or American origin that I have adapted or repackaged. What we need for variety from time to time is a puzzle or two from Russia. I have a decent collection of those, and I hope you will find them as engaging as I do.

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I have adapted the puzzle below from a Russian original, one that Leo Tolstoy is said to have liked very much.

#Puzzle 35.1

#Puzzle 35.1.

The Catwoman rang the bell. “Here are my cats,” she announced as her neighbour let her in.

The two women and the cats trooped inside. “Let’s begin with the kitchen. It has twice as many mice as the bedroom,” the house owner said.

“ ₹5,000 for a day’s work, no matter how many cats or mice are involved,” the Catwoman said.

The day’s work began. The cats set about exterminating the mice in the kitchen. Half the day eventually passed.

“Let us now divide the workforce,” the Catwoman offered, and took half her cats to the bedroom.

The two equal teams now worked separately. At precisely the moment when the workday ended (at sunset), the last of the kitchen mice was taken care of. In the bedroom, however, some mice remained when the team left.

The Catwoman arrived the next morning, this time with only one cat and one reminder: “Same ₹5,000 for a day’s work, no matter how many cats or mice.”

The solo cat finished off the bedroom’s remaining mice precisely at sunset. The bill was settled at ₹10,000.

If each cat killed mice at the same rate, how many cats were in the original team?

#Puzzle 35.2:

When you place a bet at a casino, you either lose your stake or win an amount equal to the stake. A gambler loses as many games as he wins. If at each game he stakes exactly half the money he has at that point, does he win or lose at the end?

Mailbox: Last week’s solvers:

#Puzzle 34.1 The mechanism of picking numbers means that one number from each row but different column gets picked. In the original square, any number in the 1st row is 33 more than the corresponding number in the 2nd row, and 22, 10 and 19 more than corresponding numbers in in the 3rd, 4th and 5th rows. That being the case, the total would always be (total of 1st row) minus (33 + 22+ 10 + 19), or 184 – 84 = 100. — Kanwarjit Singh, Chief Commissioner of Income Tax (retd)

Different solvers have used different approaches, of which I found the above one among the simplest. An even simpler way is by figuring out how the original magic square was created in the first place.

I started by placing 10 numbers outside the grid as shown. In each cell inside the grid, I entered the sum of the numbers in the row and column intersecting at that cell. For example, in row 3, column 4, I entered 13 + 8 = 21.

	11	0	19	8	31
23	34	23	42	31	54
-10	1	-10	9	-2	21
1	12	1	20	9	32
13	24	13	32	21	44
4	15	4	23	12	35

Since the procedure effectively makes you select one number from each column and one number from each row, their total will be the sum of the 10 numbers outside the grid, which is 100.

I came across this curiosity in Mathematical Puzzles and Diversions by Martin Gardner, who uses an example with 59 as the total.

#Puzzle 34.2:  Hi Kabir,  The smallest possible values are a = 10, b = 6, x = 4 and y = 28.   — Akshay Bakhai (Mumbai)

Solved both puzzles: Sunita & Naresh Dhillon (Gurgaon), Dr Sunita Gupta (Delhi), Mayobhav Pathak (Gurgaon), Akshay Bakhai (Mumbai), Amar Lal Miglani (Mohali), Amardeep Singh (Meerut), Shishir Gupta (Indore), Rahul Agarwal (Bay Area, California), Sabornee Jana (Mumbai), Anshul Kumar, Puneet Vashistha

Solved #Puzzle 34.1: Kanwarjit Singh (Delhi)

Solved #Puzzle 34.2: Shrey Gujral (Delhi), Devika Pandey, Ananya Iyer (Mumbai), Toshiba Naidu (Delhi)

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