RMO 2015 Problem 6 Solution

Solution 6

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 \(2k+1<a<2k+2,\) 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 \(-\)

\((I)~2k+1<a<2k+2,\) 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.

One Reply to “RMO 2015 Problem 6 Solution”

Leave a Reply

Your email address will not be published. Required fields are marked *