West Bengal RMO 2015 Problem 6 Solution

The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication.  West Bengal RMO 2015 Problem 6 Solution has been written for RMO preparation series. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO.

Solution

Also visit: Math Olympiad program of Cheenta

We can write $latex a=[a]+\{a\},$ where $latex [a]$ denotes the integral part of $latex a.$

Now, we can say that $latex 0<\{a\}<1,$ as $latex a\not\in\mathbb{Z}.$

Let $latex a=[a]+\dfrac{1}{2},$ where $latex [a]$ is odd. Then $latex {a}=\dfrac{1}{2}.$

All such integers, must satisfy the property $latex 2k+1<a<2k+2,$ where $latex k$ is a non-negative integer.

Then $latex a\left(3-{a}\right)=\dfrac{\left(2[a]+1\right)}{2}~\cdot~\left([a]+\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{(2[a]+1)([a]-1)}{2}~.$

Now, $latex [a]=2k+1.$ Means, $latex [a]-1$ is even.

So $latex 2|[a]-1.$

Or, $latex =a(3-{a})=\dfrac{(2[a]+1)([a]-1)}{2}$ is an integer.

Hence, $latex a(3-{a})$ is an integer for all positive reals $latex a$ satisfying $latex -$

$latex (I)~2k+1<a<2k+2,$ for some non-negative integer $latex k.$

$latex (II)~{a}=\dfrac{1}{2}~.$

As $latex k$ takes infinitely many values, number of such positive real numbers $latex a$ is also infinite.

This completes the proof.

Key Idea:  Greatest Integer Function