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Problematics | Stolen coins

Dec 30, 2024 11:38 AM IST

Five thieves get unequal shares from a set of coins stolen from the museum. Who gets how many coins?

There is an old puzzle about three thieves who use different methods to calculate their respective shares from a set of mangoes, one after the other, but in the end up with an equal number each. We have not run it in Problematics yet, and even if we do, it can only be the second puzzle, because it is rather low on the difficulty level. In any case, that puzzle is not for today.

Representational image.(Pixabay)
Representational image.(Pixabay)

What we have today is a more intricate puzzle about a different group of thieves, who take their respective shares from a number of objects that is not always divisible by the number of thieves. I hope you will have as much fun solving it as I had creating it.

#Puzzle 123.1

Five thieves break into a museum and lift a certain number of antique coins, which they intend to sell to various collectors. One thief, who has a flight to catch, is the first to take his share, which is an exact one-fifth of the total number of coins. The remaining coins are with the other four, not equally divided as of now, but intended to be fairly shared later.

Plans go awry when the four go to the airport to see off thief #1. After the flight leaves, the cops suddenly arrive and give chase to the other four. Thief #2 is caught, but he is empty-handed, with the other three escaping with all the remaining coins.

The three assemble in their hideout at night. Too tired to divide the loot, they pile the remaining coins on a table. No one knows that thief #2 had no coins with him when he was caught, and as a consequence no one knows that four-fifths of the original loot is still intact. Without counting, they decide to leave the coins on the table and share the spoils in the morning.

Over the night, each thief breaks the agreement one by one. Thief #3 is the first to go to the table. He wants to take one-third secretly now, and another one-third from the remainder in the morning. He finds, however, that the number of coins on the table is not divisible by 3. The problem is solved when he pockets one of the coins. The thief divides the remainder by 3, takes one-third (in addition to the one already in his pocket) and goes back to sleep.

Thief #4 is next. When he comes to the table, he too finds a number that he cannot divide by 3, and he too solves the problem by following the precedent set by thief #3. He pockets one coin, divides the remainder by 3, takes one-third, and goes back to sleep.

As you might have guessed, thief #5 follows suit. He faces the same problem of divisibility and adopts the same SOP, now well established: pocket one coin, divide the remainder by 3, take one-third, go back to sleep.

In the morning, when the thieves reassemble at the table, it is obvious that the pile is smaller than it was at the beginning of the night. But since everyone has cheated, they all keep quiet about it and decide to divide the remaining coins here and now. But guess what? Once again, the coins are not divisible by 3. To solve the problem, they decide to set aside one coin, after which they neatly divide the remaining. The unassigned coin, they decide with full sincerity, will be given to thief #2 when he is released from jail.

What is the minimum number of coins that must have been stolen from the museum, and what is each thief’s final share?

#Puzzle 123.2

A puzzle I came across asks the solver to calculate the angle between the hour and minute hands of a clock when the time is 2:15 (am or pm does not matter, of course). Too tame for Problematics, I thought. Why not enhance it by looking at the angle between the two hands at 1:15, 2:15… all the way to 12:15?

If you calculate the angle at hourly intervals from 1:15 onwards, as mentioned above, does the angle remain the same, or does it increase/decrease in some kind of progression?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 122.1

Kabir Sir,

The answers to the movie anagrams are:

1. DEATH ON THE NILE (1978)

2. ENTER THE DRAGON (1973)

3. THE COLOR PURPLE (1985)

4. A BRIDGE TOO FAR (1977)

5. CHARIOTS OF FIRE (1981)

6.ROBIN AND MARIAN (1976)

7. ROSEMARY'S BABAY (1968)

8. THE PINK PANTHER (1963)

9. CAPTAIN AMERICA (1990)

10. WHERE EAGLES DARE (1968)

— Jaikumar Inder Bhatia & Disha Jaikumar Bhatia, Ulhasnagar (Thane)

#Puzzle 122.1

Dear Mr Kabir,

With a single weighing, the fake coin-stack can be identified as follows.

Let the coin-stacks be numbered from 1 to 10, consisting of 10 coins each. Let's take 1 coin from the first stack, 2 from the second and so on, up to the entire 10 coins of the tenth stack. Then weigh the lot of the collected coins on the given balance against the bona fide weight (given).The excess weight of this collection in number of grams, corresponds to the number of the fake coin-stack. For example, if the collected lot of coins weighs 5g more than it should, then the fake coin-stack must be the fifth stack, from which we had taken 5 coins ( each weighing 1g more than a bona fide coin).

— Shri Ram Aggarwal, Palam (New Delhi)

A few readers have attempted the weighing puzzle by weighing the coins against each other, and therefore required more steps than one. They assumed that we do not have access to weights. Since no such condition was mentioned in the puzzle, their solutions cannot be counted as correct.

Solved both puzzles: Jaikumar Inder Bhatia & Disha Jaikumar Bhatia (Ulhasnagar, Thane), Shri Ram Aggarwal (Delhi) Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), Vickram Crishna & YK Munjal (Delhi), Shri Ram Aggarwal (Delhi), Ajay Ashok (Delhi)

Solved #Puzzle 122.1: Sanjay Gupta (Delhi), Shishir Gupta (Indore), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Manoj Kumar Mech (DGM, Materials/retired, IOCL, Tezpur)

Solved #Puzzle 122.2: Anil Khanna (Ghaziabad), Dr Vivek Jain (Baroda), Yadvendra Somra (Sonipat)

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

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