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Problematics | Cubic calendar

Jan 20, 2025 02:29 PM IST

Two cubes, a little larger than ludo dice, carry some digits that you can see and some that you cannot. What are the ones you cannot see?

A couple of readers have noted that last week’s puzzles were too simple. While simple and tough are subjective, I must repeat something I have said before and will say later: the objective of Problematics is to give puzzle solvers a fun experience, and this is something that depends more on the way the puzzle is constructed than on the level of difficulty. That said, I have always sought a medium range on the Problematics scale of hardness, with puzzles I think are neither too easy nor too difficult, but the perspective of readers, of course, can differ on that too.

Welcome to Problematics! (Shutterstock)
Welcome to Problematics! (Shutterstock)

This week, we have a puzzle that may come across to most readers as easy. I got it from the writings of Martin Gardner, who attributes it to other sources. Something about the solution that Gardner provides, however, has been bothering me. I cannot disclose that now, but I will say this much: something claimed by Gardner as necessary to the solution does not appear necessary to me.

Below, I have added some extra information that was not in the original puzzle, so that the solution will match what Gardner described. After receiving your solutions, I shall let you know what has been bothering me; and if necessary we can take it up for discussion in next week’s column.

#Puzzle 126.1

Puzzle 126.1
Puzzle 126.1

If you excuse my drawing skills (or lack of them) with the simple app I have used, you will possibly guess that the image represents a calendar. The date shown, of course, is the deadline by which you must submit the solution to this puzzle. The calendar is in the form of two cubes, whose front digits together give you the date on any given day of a given month.

To ensure that the puzzle is clearly understood, let us oversimplify every aspect of the calendar.

1. Each of the 6 faces in either cube contains one digit, leaving the calendar setter with a total of 12 digits at their disposal.

2. Some digits are unique to one cube, while some others appear on both cubes. For example, 0 appears on both cubes.

3. The digits are such that any date from 01 to 31 can be shown by the two cubes. The dates from the 1st to 9th of any month, too, are represented by two digits: 01, 02, 03… etc.

4. It does not matter which cube is placed to the right and which one is placed to the left.

In the figure, you can see 5 of the 12 digits. What are the remaining 4 digits on the left cube and the remaining 3 digits on the right cube?

VENUS

– EARTH

= MARS

This puzzle is not mine, but it appears in more than one publication. The answer may be out there on the internet, so I recommend that you don’t search for it and give up the fun of solving it yourself.

Each letter in the above subtraction stands for a specific digit. One letter represents only one digit however many times it appears. One digit is represented only by its own unique letter.

Convert the planetary subtraction into “digital” form.

MAILBOX: LAST WEEK’S REPLIES

#Puzzle 125.1

Hi Kabir,

For a total of 896 books which are finally distributed equally among 7 shelves in the manner described in the puzzle, the initial number of books (Nₘ) on shelf #m can be determined from the following equation:

Nₘ = 7 x 2⁷⁻ᵐ + 1; or

N₁ = 7 x 2⁶ + 1 = 7 x 64 + 1 = 449

N₂ = 7 x 2⁵ + 1 = 7 x 32 + 1 = 225

N₃ = 7 x 2⁴ + 1 = 7 x 16 + 1 = 113

N₄ = 7 x 2³ + 1 = 7 x 8 + 1 = 57

N₅ = 7 x 2² + 1 = 7 x 4 + 1 = 29

N₆ = 7 x 2¹ + 1 = 7 x 2 + 1 = 15

N₇ = 7 x 2¹ + 1 = 7 x 1 + 1 = 8

The equation can be derived as follows. First, let us take a general case of a grand total of N books. Now

let us see what happens, in general, on shelf #m during the 7 redistribution steps.

In each of the first (m – 1) steps, Nₘ doubles. Therefore, before step #m, the number of books on shelf #m is 2ᵐ⁻¹Nₘ.

In step #m, shelf #m gives away books whose total is the sum of the number of books on the other 6 shelves. This total is equal to the grand total (= N) minus the number existing on this shelf (= 2ᵐ⁻¹Nₘ) before this step. After these are removed, the number of books remaining on shelf #m is

2ᵐ⁻¹Nₘ – (N – 2ᵐ⁻¹Nₘ) = 2ᵐNₘ – N

In each of the remaining (7 – m) steps, the number of books on this shelf will double. This leads to a final count of

2⁷⁻ᵐ(2ᵐNₘ – N)

= 2⁷Nₘ – 2⁷⁻ᵐN.

This is the general expression for the final number of books on each of 7 shelves. In the present case, this number is given to be 128 or 2⁷, for all m. Also N = 896 = 7 x 128 = 7x2⁷. So:

2⁷Nₘ – 2⁷⁻ᵐ(7x2⁷) = 2⁷; or

Nₘ – 7x2⁷⁻ᵐ = 1; therefore,

Nₘ = 7x2⁷⁻ᵐ + 1

This is the equation we used above to determine the initial number of books on each shelf.

— Professor Anshul Kumar, Delhi

#Puzzle 125.2

Dear Kabir Sir,

Since a non-leap year starts on any one of 7 days (Sunday to Saturday), there are 7 unique calendars for non-leap years. Similarly, a leap year can also start on any one of 7 days, giving 7 unique calendars for leap years. Therefore, the total number of unique calendars is 14.

— Aishwarya Rajarathinam, Coimbatore

Solved both puzzles: Professor Anshul Kumar (Delhi), Aishwarya Rajarathinam (Coimbatore), Dr Sunita Gupta (Delhi), Avanti Kashikar (Mumbai), Shishir Gupta (Indore), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Sanjay Gupta (Delhi), Ajay Ashok (Delhi)

Solved #Puzzle 125.2: Sabornee Jana (Mumbai)

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

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