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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SIMPLE QUESTION FROM RMO 1995

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Question: If a,b,c and d are positive integers such that abcd=1, then show that (1+a)(1+b)(1+c)(1+d)≥16

Detailed solution: To solve the question, you just need basic knowledge of AM-GM inequality.

Using the AM-GM inequality for the terms of LHS inside brackets, we have

(1+a)/2≥√(1×a), which gives (1+a)≥2√a    …….(1)

(1+b)/2≥√(1×b), which gives (1+b)≥2√b    …….(2)

(1+c)/2≥√(1×c), which gives (1+c)≥2√c   …….(3)

(1+d)/2≥√(1×d), which gives (1+d)≥2√d    …….(4)

Multiplying equations (1),(2),(3) and(4), we have

(1+a)(1+b)(1+c)(1+d)≥16√(abcd)

or,(1+a)(1+b)(1+c)(1+d)≥16 [Since abcd=1]

Hence proved

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