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.

 

Fundamental theorem of algebra: RMO training

It is quite well-known that any positive integer can be factored into a product of primes in a unique way, up to an order. (And, that 1 is neither prime nor composite) —- we all know this from our high school practice of “tree-method” of prime factorization, and related stuff like Sieve of Eratosthenes. But, it is so obvious, and so why it call it a theorem, that too “fundamental” and yet it seems it does not require a proof. It was none other than the prince of mathematicians of yore, Carl Friedrich Gauss, who had written a proof to it. It DOES require a proof — there are some counter example(s). Below is one, which I culled for my students:

Question:

Let E= \{a+b\sqrt{-5}: a, b \in Z\}

(a) Show that the sum and product of elements of E are in E.

(b) Define the norm of an element z \in E by ||z||=||a+b\sqrt{-5}||=a^{2}+5b^{2}. We say that an element p \in E is prime if it is impossible to write p=n_{1}n_{2} with n_{1}, n_{2} \in E, and ||n_{1}||>1, ||n_{2}||>1; we say that it is composite if it is not prime. Show that in E, 3 is a prime number and 29 is a composite number.

(c) Show that the factorization of 9 in E is not unique.

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.

Another cute proof: square root of 2 is irrational.

Reference: Elementary Number Theory, David M. Burton, Sixth Edition, Tata McGraw-Hill.

(We are all aware of the proof we learn in high school that \sqrt{2} is irrational. (due Pythagoras)). But, there is an interesting variation of that proof.

Let \sqrt{2}=\frac{a}{b} with gcd(a,b)=1, there must exist integers r and s such that ar+bs=1. As a result, \sqrt{2}=\sqrt{2}(ar+bs)=(\sqrt{2}a)r+(\sqrt{2}b)s=2br+2bs. This representation leads us to conclude that \sqrt{2} is an integer, an obvious impossibility. QED.

RMO 2017 Warm-up: Two counting conundrums

Problem 1:

There are n points in a circle, all joined with line segments. Assume that no three (or more) segments intersect in the same point. How many regions inside the circle are formed in this way?

Problem 2:

Do there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a re-ordering of their digits?

Solutions will be put up in a couple of days.

Nalin Pithwa.