Mathematical Muses

The talk

A revered professor at the mathematics department at IIT Delhi held a talk in which he would show some of his mathematics tricks. 

He would explain some tricks related to cards, games and the calendar, along with some general talk.

A friend of mine who was active in the mathematical Olympiad circuit as well (for those who don't know, I was deeply involved in mathematical olympiads during my high school days, and they have always fascinated me), texted me one day, about this session. I immediately said yes, and forwarded it to some of my close friends whom I knew that they would probably come.

Though the session didn't have much audience (roughly 14-15 people), and only 3 IIT Delhi students (he had some of his old friends coming as well, along with their families), it was quite good and worth attending.

The friends whom I texted were interested, but either they had classes or assignment deadlines (phew!). Only 2 came.

I have tried to explain some of the tricks that intrigued me, in this writeup.

The numbers 13 and 52 - card mystery

This is a trick based on a deck of cards.

Randomly shuffle a deck of cards. Pick the topmost card and note the number on it. Keep the card aside and from that number, count to 13, and every time you count a number, pick the next card from the remaining deck and place the card on the stack.

For example, if the top card is numbered 4, place it aside, start a stack, and then count 5 to 13, every time placing the top card of the remaining deck on the stack.

After reaching 13, pick a new card from the deck and start a new stack, and repeat the process with it.

After sometime you would realize that the cards are finished but you haven't reached the number 13, and the stack which is being created is incomplete. 

Add all the numbers on the cards in the incomplete stack and for the complete stacks, add all the numbers of the bottom cards on the stack.

The sums are equal.

If there are no incomplete stacks, treat the last stack as the incomplete stack.

Truth Dare Situation

Almost all of us have played this game "21 dares" while at school, in which we have to count till 21 and the person saying 21 loses and has to choose one among "truth", "dare" or "situation".

There is some mirch-masala here - I would teach you how to win (or rather, not to lose) the game. 21 tak to mai pahuncha dunga aapko.

Uske aage kya hoga, mujhe nahi pata, meri zimmedari nahi hai, to koi blame mat karna mujhe. (I can make you reach till 21, but I don't know what would happen after that, and don't take any responsibility so don't blame me). I would teach you how to play, but only till 21.

First I would talk about the specific "21 dares" game, then I would talk about the "n-dares" case (for two people though for the sake of simplicity, but the basic idea applies for a >2-member game as well).

21 dares

Let's assume that me and you are playing the game "21 dares". I would start. I can count "1" or "1-2". You can count 1 or 2 numbers ahead, and so on.

If you reach 20, I will have to speak 21 and you win.

Notice that if I reach 17, you would speak 18 and I would speak 19-20, or you would speak 18-19 and I would speak 20. I would win in any case. So the person who speaks 17, in a way, wins the game in any case.

Revised target: Reach 17. You would win the game.

To reach 17: if I reach 14, you would speak 15 and I would speak 16-17, or you would speak 15-16 and I would speak 17. I would win in any case. So the person who speaks 14, in a way, reaches 17 i.e. the revised target and wins the game.

Further revising the target: Reach 14.

This way, we can keep on revising the target into "mini-targets" like breaking a big problem into smaller problems.

Catch the sequence 2,5,8,11,14,17,20. Any way you start, any player who starts, you can catch the sequence.

n-dares

Take the general case.

Let's assume that we have to count till 'n'. Step size of maximum 'a'.

Now, if I am at n-a-1, you say any number from 0 to a, I can win the game from that situation, irrespective of your choice.

Revised target: n-a-1.

Keep on revising the target, by reducing it by (a+1).

The sequence you need to catch: n-a-1, n-2a-2, n-3a-3,............. (reverse order I am writing).

No, I won't play first!

Consider a special case: What if n is a multiple of (a+1)?

The sequence would be 0, a+1, 2a+2,....................,n-a-1,n

If I start, irrespective of the number k I would increment by, you would increment by a number (a+1-k) and would catch the sequence, and eventually win.

I am assuming that you know how to play the game.

Not a fair game!

Calendar Dates

Enough of truths and dares. Aage badhte hai ab.

I claim that I can calculate the day on any date without a calendar or device.

Follow these three steps. (Illustrating with an example of the day I wrote this - April 13, 2024)

Step 1: Associate some numbers with the months:

• February and March: 0

• All other even-numbered months have the same number with them.

• July (7) -> 11, November (11) -> 7. Easy way to remember: 7-11 (the name of a convenience store chain, found in Singapore, the US and many other countries).

• September (9) -> 5, May (5) -> 9. Easy way to remember: 9-5 job cycle.

• January (1) -> 3.

• For every number, subtract it from 8.

• This number N1 is the number obtained from step 1

In this case, N1 = 4 (April)

Step 2: Take the year. Take the last 2 digits of the year.

• Take the quotient Q and remainder R of the number formed, on dividing by 12.

• Divide the remainder R by 4, and take the quotient Q1.

• Add Q, Q1 and R. Take modulo 7 again.

• This number is the number N2 obtained from step 2

In this case, N2 = 2

Step 3: 

• Add N1 and N2, and today's date

• Add 2 to the sum.

• Again, take modulo 7

• The day of the week is as per the final number.

• The days are numbered from 0 to 6 (Saturday to Friday).

Leap year (exception): If the date falls in a January or a February of a leap year, subtract 1 from the result you get.

Square 2 digit numbers in 10 seconds

Take the number. Suppose it is AB when read from left to right. Tens digit A, ones digit B.

Add B to AB. Multiply by A. Multiply further by 10.

Square B. Add to the number calculated before.

You get the square of the number!

Square 3 digit numbers in 10 seconds

Take the number. Suppose it is ABC when read from left to right. Hundreds digit A, tens digit B, ones digit C.

Multiply the number A00 by (ABC + BC). Add square of (BC) to it.

You get the number!