Some Arithmetic Titbits — as shared by Prof Jayant Narlikar

(Ref: Fun and Fundamentals of Mathematics by Jayant V Narlikar and Mangala Narlikar)

We will look at some interesting games and puzzles involving numbers. They will not entail anything more complicated than the four basic operations of  arithmetic: addition, subtraction, multiplication and division. To set the ball rolling, let us consider a simple three-digit number, any number, whose first and last digits are not the same. Let us take, say,

568

Now, reverse it to get 865

Next, subtract the smaller of the two from the other:

865-568=297

Now, reverse the answer and add to it. Thus,

792+297=1089.

So, you end up with the number 1089. What is so special about it? Nothing, except that, you always get this number as the final answer, no matter how you started! 

Try again with another number, say, with 841. We have the following 4 steps as above:

(1) Reversal: 148

(2) Subtraction: 841-148=693

(3) Reversal: 396

(4) Addition: 693+396=1089.

The answer of the subtraction made at the beginning must, however, be considered as a “three digit number”. If the subtraction gives the answer as 99, it is to be considered as 099. This happens if the difference between the first and the third digits of the original number is 1.

You can use this result as  a guessing game. You can ask your friend to start with any three digit number and perform these operations, in secret, without telling you. Then, you impress him/her by telling him/her the answer!

Race to 50:

Here, is a game that apparently involves addition only, but in reality also requires you to think of a strategy to win. The rules of the game are as follows:

Two players, and B, play it with each alternately adding any number from 1 to 6, to a score, starting from zero. Whosoever adds the number that brings the score to 50 wins.

The game could proceed, for example, as follows:

\begin{array}{cccc} Round \hspace{0.1 in}No & A \hspace{0.1 in }adds & B \hspace{0.1 in}adds & Total \\ 1 & 4 & 0 & 4 \\ 2 & 0 & 5 & 9 \\ 3 & 3 & 0 & 12 \\ 4 & 0 & 6 & 18 \\ 5 & 5 &  & 23 \\ 6 & 0 & 2 & 25 \\ 7 & 6 & 0 & 31 \\ 8 & & 4 & 35 \\ 9 & 1 & & 36 \\ 10 &  & 3 & 39 \\ 11 & 6 & 0 & 45 \\ 12 & 0 & 5 & 50 \end{array}

Thus, B wins.

The above is just an example. Now, you should think of a strategy so that you will necessarily win. Is it possible to ensure victory for both A and B?

More fun later with thanks to Professor Narlikar,

Nalin Pithwa

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