Happy Numbers make Happy Programmers ! :-)

Here is one question which one of my students, Vedant Sahai asked me. It appeared in his computer subject exam of his recent ICSE X exam (Mumbai):

write a program to accept a number from the user, and check if the number is a happy number or not; and the program has to display a message accordingly:

A Happy Number is defined as follows: take a positive number and replace the number by the sum of the squares of its digits. Repeat the process until the number equals 1 (one). If the number ends with 1, then it is called a happy number.

For example: 31

Solution : 31 replaced by 3^{2}+1^{2}=10 and 10 replaced by 1^{2}+0^{2}=1.

So, are you really happy? 🙂 🙂 🙂

Cheers,

Nalin Pithwa.

Yet another special number !

The eminent British mathematician had once remarked: Every integer was a friend to Srinivasa Ramanujan.

Well, we are mere mortals, yet we can cultivate some “friendships with some numbers”. Let’s try:

Question:

Squaring 12 gives 144. By reversing the digits of 144, we notice that 441 is also a perfect square. Using C, C++, or python, write a program to find all those integers m, such that 1 \leq m \leq N, verifying this property.

PS: in order to write some simpler version of the algorithm, start playing with small, particular values of N.

Reference: 1001 Problems in Classical Number Theory, Indian Edition, AMS (American Mathematical Society), Jean-Marie De Konick and Armel Mercier.

Amazon India link:

https://www.amazon.in/1001-Problems-Classical-Number-Theory/dp/0821868888/ref=sr_1_1?s=books&ie=UTF8&qid=1509189427&sr=1-1&keywords=1001+problems+in+classical+number+theory

Cheers,

Nalin Pithwa.

 

A Special Number

Problem:

Show that for each positive integer n equal to twice a triangular number, the corresponding expression \sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}} represents an integer.

Solution:

Let n be such an integer, then there exists a positive integer m such that n=(m-1)m=m^{2}-m. We then have n+m=m^{2} so that we have successively

\sqrt{n+m}=m; \sqrt{n + \sqrt{n+m}}=m; \sqrt{n+\sqrt{n+\sqrt{n+m}}}=m and so on. It follows that

\sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}}=m, as required.

Comment: you have to be a bit aware of properties of triangular numbers.

Reference:

1001 Problems in Classical Number Theory by Jean-Marie De Koninck and Armel Mercier, AMS (American Mathematical Society), Indian Edition:

Amazon India link:

https://www.amazon.in/1001-Problems-Classical-Number-Theory/dp/0821868888/ref=sr_1_1?s=books&ie=UTF8&qid=1508634309&sr=1-1&keywords=1001+problems+in+classical+number+theory

Cheers,

Nalin Pithwa.

Math concept(s) : simple yet subtle

Most math concepts are intuitive, simple, yet subtle. A similar opinion is expressed by Prof. Michael Spivak in his magnum opus, Differential Geometry (preface). It also reminds me — a famous quote of the ever-quotable Albert Einstein: “everything should be as simple as possible, and not simpler.”

I have an illustrative example of this opinion(s) here:

Consider the principle of mathematical induction:

Most students use the first version of it quite mechanically. But, is it really so? You can think about the following simple intuitive argument which when formalized becomes the principle of mathematical induction:

Theorem: First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

  1. The integer 1 belongs to S.
  2. Whenever the integer k is in S, the next integer k+1 must also be in S.

Then, S is the set of all positive integers.

The proof of condition 1 is called basis step for the induction. The proof of 2 is called the induction step. The assumptions made in carrying out the induction step are known as induction hypotheses. The induction situation has been likened to an infinite row of dominoes all standing on edge and arranged in such a way that when one falls it knocks down the next in line. If either no domino is pushed over (that is, there is no basis for the induction), or if the spacing is too large (that is, the induction step fails), then the complete line will not fall. 

So, also remember that the validity of the induction step does not necessarily depend on the truth of the statement that one is endeavouring to prove.

More later,

Nalin Pithwa.