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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4th mathematics quiz organised by our community

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Mathematical Discussions community has organised 4th mathquiz of it's quiz series. The questions of the quiz can be downloaded by clicking here. So, have a look at the questions. The questions are mostly olympiad-type and require basic mathematics of high school to be solved. The solutions to the questions of the quiz can be downloaded by clicking here. Note that there may be alternative solutions too.

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