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(C) Let the two integers be $A = 10a+b$ and $G = 10b+a$,where $A$ is the arithmetic mean and $G$ is the geometric mean.Given,$\frac{x+y}{2} = 10a+b$ a

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$130$

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(C) Let the two integers be $A = 10a+b$ and $G = 10b+a$,where $A$ is the arithmetic mean and $G$ is the geometric mean.Given,$\frac{x+y}{2} = 10a+b$ and $\sqrt{xy} = 10b+a$.Thus,$x+y = 2(10a+b)$ and $xy = (10b+a)^2$.We know that $(x-y)^2 = (x+y)^2 - 4xy$.Substituting the values,$(x-y)^2 = 4(10a+b)^2 - 4(10b+a)^2$.$(x-y)^2 = 4[(10a+b)^2 - (10b+a)^2] = 4(10a+b-10b-a)(10a+b+10b+a)$.$(x-y)^2 = 4(9a-9b)(11a+11b) = 4 	imes 9 	imes 11(a-b)(a+b) = 396(a-b)(a+b)$.For $(x-y)^2$ to be a perfect square,$(a-b)(a+b)$ must be of the form $11 	imes k^2$.Since $a$ and $b$ are digits,$a+b \leq 18$ and $a-b Solving $a+b=11$ and $a-b=1$ gives $2a=12 \Rightarrow a=6$ and $b=5$.Thus,$x+y = 2(10(6)+5) = 2(65) = 130$.

The arithmetic mean and the geometric mean of two distinct $2$-digit numbers $x$ and $y$ are two integers,one of which can be obtained by reversing the digits of the other (in base $10$ representation). Then,$x+y$ equals

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