# 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

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

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

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

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

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

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

So $2|[a]-1.$

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

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

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

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

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

This completes the proof.

Key Idea:  Greatest Integer Function

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