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Problematics | A clockwork montage

Nov 11, 2024 10:23 AM IST

This week, two puzzles involving clocks, both unique and associated with illustrious personalities in mathematics.

With our mobile phones always at hand, we are depending less and less on other timekeeping devices. The wristwatch, once a necessity of life, has become an affordable fashion accessory to some and an expensive luxury item to others. Thankfully, the wall clock has not yet gone out of fashion, remaining easier to consult than bringing our phones out of our pockets or handbags.

Representational Image.
Representational Image.

Clocks have long been a feature of puzzles, some of which have appeared in Problematics over the last couple of years. But there are plenty more of them. Both of this week’s puzzles involve clocks, and come from illustrious sources. The first one, in particular, is of a unique kind.

#Puzzle 116.1

The late Martin Gardner, the American writer of popular mathematics who introduced readers to puzzles from around the world, also created some diversions himself. The following trick is one of Gardner’s own. It begins, like so many mathematical tricks, by asking someone to think of a number. In this case, Gardner suggests, let the number be less than 50 in order to save time.

Show a clock, or a drawing of one, to a friend. Also give her a die, and ask her to think of any number less than 50. While you turn your back, your friend rolls the die. Your back still turned, she starts tapping silently on the numbers depicting the hours on the clock, starting at the number that appears on the die and going clockwise, round and round as many times as necessary. That is to say, if the number rolled on the die is 4, she counts IV on the clockface as 1, then V as 2, VI as 3, and so on. She stops the count when she reaches the number she has thought of, and writes down the number from the clockface on which the count has ended.

The friend then goes back to the number on which she had started, i.e. the number that appeared on the die. She starts counting again from there, but this time going anticlockwise. If 4 was the number that appeared on the die, IV on the clockface is again counted as 1, but now it is followed by III as 2, II as 1, and so on. Again, she stops when she reaches the number that she has thought of. She writes down this number on the clockface too, then adds it to the previous number she wrote down.

She announces the total. You immediately announce the number that appeared on the die.

In his explanation for how the trick is performed, Gardner gives two slightly different rules. If the total announced by your friend is below a certain limit, you perform a certain operation; if it is above that limit, you perform a slightly different operation.

How is the trick done?

#Puzzle 116.2

In the early days of Problematics, we had a puzzle in which you determined the number of times when the two hands of a clock coincide. Here is a variation, sourced from the writings of the Russian mathematician Yakov Perelman who, in turn, attributes it to a conversation between Albert Einstein and his biographer, A Moszkowski.

Take the time 12:00, when both hands are at the same place. If you interchanged the hands, it would still be a valid time. On the other hand, take 6:00, when the hour hand is at 6 and the minute hand at 12. If you interchanged the hands now, they would be in a position that never appears in a normal clock.

There are, however, certain times that remain valid even if the hands are interchanged, although they need not necessarily be the same times indicated by the hands before the interchange.

To keep it simple, let’s not go into a 24-hour cycle. In a span of 12 hours, how many such times exist?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 115.1

Dear Mr Kabir

Since each person buys as many puzzles as the number of rupees they pay for one puzzle, the amount each person spends for x puzzles is x². Let one husband buy y puzzles while his wife buys z puzzles, spending y² and z² respectively. It is given that

y² – z² = 63 => (y + z) (y – z) = 63

The left hand side can be (21 x 3), (63 x 1) or (9 x 7). Corresponding to these possibilities, the values of (y, z) are respectively (12, 9), (32, 31) and (8, 1).

Since Mr Faridpur bought 23 puzzles more than Ms Beed, it means they bought 32 and 9 respectively. Since Mr Erode bought 11 puzzles more than Ms Agra, it means they bought 12 and 1 respectively. This gives us the couples and the number of puzzles bought:

Mr Faridpur and Ms Churu (32, 31)

Mr Erode and Ms Beed (12, 9)

Mr Dewas and Ms Agra (8, 1)

— Shri Ram Aggarwal, Delhi

 

#Puzzle 115.2

Hi Kabir,

Greetings. We need 3 moves to get all glasses right side up. Let D be the position where the glasses are face down and U be the position when they are face up.

Initial Position: DDDDDDD

1st move: UUUDDDD

2nd move: DUUUUDD

3rd move: UUUUUUU

— Sanjay S, Coimbatore

Solved both puzzles: Shri Ram Aggarwal (Delhi), Sanjay S (Coimbatore), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Anil Khanna (Ghaziabad), Sanjay Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Shishir Gupta (Indore), Professor Anshul Kumar (Delhi), YK Munjal (Delhi)

Solved #Puzzle 115.1: Dr Vivek Jain (Baroda), Jaikumar Bhatia & Inder Bhatia (Ulhasnagar, Thane)

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

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