Simple Problem

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  Question: Find all two-digit positive integers such that the sum of the number and the number obtained by reversing the order of digits is a perfect square. Detailed solution: Let $xy$ be any such two-digit number (Note: Here $xy$ is just the form of a two-digit number like 25,43,etc., where $x$ and $y$ are digits). Therefore, $xy$ can be written as $10x+y$ (in expanded form). Number obtained by reversing the order of digits of $xy$ is $yx=10y+x$. According to question, The sum of $xy$ and the number obtained by reversing the order of digits of $xy$ i.e., $yx$ is a perfect square, which means $(10x+y)+(10y+x)=$Perfect square or, $11x+11y=$Perfect square or, $11(x+y)=$Perfect square .....$(1)$ But, maximum value of $x$ and $y$ are 9 each, so maximum value of $x+y$ is $9+9=18$. Observe that, to satisfy equation $(1), x+y=11k²$, for some positive integer $k$. For $k=1,x+y=11$ and for $k>1$ i.e., for $k≥2, x+y≥44$, which is not possible. Therefore, we get that $x+y=11$ is the only...
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Solutions to IYMC Qualification Round 2020

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The International Youth Math Challenge (IYMC) is one of the biggest online math competition for students from all around the world. To know more about IYMC, you can visit this website https://iymc.info/en/. The questions of the qualification round of IYMC, 2020 can be downloaded by clicking here and the solutions can be downloaded by clicking here.

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PUTNAM MATHEMATICS COMPETITION 2018

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