Problematics | Digital hunt | Hindustan Times

Problematics | Digital hunt

Updated on: Oct 06, 2025 08:42 AM IST

Five gamblers betting on five digits each, plus a second puzzle in which you again find out missing digits.

Last week’s geometry puzzle, as stated then, is several centuries old. It dates back to the time when Plato recorded Socrates asking a student to double the area of a given square (my tweaked version asked you to halve its area). The underlying principle is the same in both versions: the diagonal of the smaller square is equal to the side of the larger square,

Representational image.(Shutterstock)
Representational image.(Shutterstock)

I came across the puzzle in a recently published study in which researchers posed this puzzle to ChatGPT. The AI initially messed it up, trying an algebraic approach that would have been unknown in Plato’s time. It was stubborn about its answer (something many of us have faced) but ultimately got the right one after several attempts. It turned out that ChatGPT knew of the Socrates-Plato episode, but it had not made the leap to applying that knowledge to the puzzle presented to it.

As a consolation to ChatGPT, I asked it to draw the figure for the answer to my version of the puzzle. YK Munjal had sent a correct but hand-drawn figure, and ChatGPT has touched it up well enough to make it presentable for any geometry textbook.

That will come later, of course. In keeping with the long-established template, Problematics presents new puzzles before going to the solution of the previous week’s ones.

#Puzzle 163.1

Puzzle 163.1
Puzzle 163.1

At a casino game, each gambler has to guess the five digits of a hidden number. The number consists entirely of 1, 2, 3, 9 and 0, but it is not necessary that each digit should appear once. Some digits can repeat themselves, while some may not appear at all. It is also possible that all five digits are different, of course, but no gambler can take that from granted.

The first three digits can be any combination of 1, 2 or 3, including or excluding repeats. The last two digits are any pairing of 9 and 0, containing both digits or only one of them twice: you never know. The croupier who handles the gambling table sets the five digits on a frame, which he hides under the table.

A gambler must place a separate bet on each digit. If all 5 digits match, they obviously win. If no one gets all five correct, the player with four correct digits win. And so on. If two or more players have the same number of correct digits, the win is shared, with each gambler earning exactly twice the money he or she had placed on any matching digit.

The gamblers Al Capone, Bonnie Parker, Clyde Barrow, Jackie Brown and Scarface (Tony Montana, not Capone) bet separately on each digit. The wheel is spun five times, and the winning combination of five digits is found. At the end, it turns out that each of them got a different number of digits correct. That is to say, if any one of them got three digits correct, for example, then none of the others got exactly three correct.

Which gambler got the highest number of digits correct?

#Puzzle 163.2

A, O, R, and T represent different digits. The number TA signifies T in tens place and A in units place, ROT means R in hundreds place, O in tens place and T in units place, and so on. Squaring the number TA gives you TA x TA = ROT.

Can you identify each digit?

MAILBOX: LAST MONTH’S SOLVERS

#Puzzle 162.1

Puzzle 162.1
Puzzle 162.1

Dear Kabir,

If the side of the original square is x, its area is x², and each diagonal is √2x. The new square should have an area of x²/2, so its side must be x/√2. The diagonal of the new square will be √2(x/√2) = x. In other words, the side of the original square is equal to the diagonal of the new square. Also, half the diagonal of the original square = √2x/2 = x/√2 = side of the new square.

Let the original square be ABCD. Using the edge of the protractor, draw diagonals AC and BD intersecting at O. The square is divided into 4 equal triangular parts, each 1/4 of the area of the square. Using the protractor on any one triangle, say, AOB, draw perpendiculars from point A and point B, intersecting at say point E. This forms a new triangle AEB, with an area equal to triangle AOB. The new square AEBO so formed, comprising triangles AOB and triangle AEB, is half the area of the original square.

— YK Munjal, New Delhi

#Puzzle 162.2

Hello Kabir,

Money spent on the eggs = 30 (Reena) + 18 (Neera) = 48. So, the value of each sister’s omelette share = 48/3 = 16. This implies that the eldest sister’s contribution of 8 glasses of coke is equivalent to 16, which means each glass is equivalent to 2.

Reena spent 30 and ate 16 worth of omelletes, a net contribution of 14. Neera spent 18 and ate 16 worth of omelette, a net contribution of 2. So the division of coke should have been 14:2 or 7:1, i.e. 7 glasses for Reena and 1 glass for Neera. (Since Neera has already taken 2 glasses, Reena will have taken a loss even if she asks for all 4 remaining glasses).

— Dr Sunita Gupta, New Delhi

Solved both puzzles: Dr Sunita Gupta (Delhi), Vinod Mahajan (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat)

Solved Puzzle #162.1: Y K Munjal (Delhi), Anil Khanna (Ghaziabad), Ajay Ashok (Delhi), Shri Ram Aggarwal (Delhi)

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

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