Let x and y be two 2-digit numbers such that | vedclass

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(D) Let $x = 10a + b$ and $y = 10b + a$,where $a$ and $b$ are digits $(a, b \in \{1, 2, \dots, 9\})$.Given $x^2 - y^2 = m^2$,we have $(10a + b)^2 - (1

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$154$

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(D) Let $x = 10a + b$ and $y = 10b + a$,where $a$ and $b$ are digits $(a, b \in \{1, 2, \dots, 9\})$.Given $x^2 - y^2 = m^2$,we have $(10a + b)^2 - (10b + a)^2 = m^2$.Using the identity $A^2 - B^2 = (A+B)(A-B)$,we get $(10a + b + 10b + a)(10a + b - 10b - a) = m^2$.$(11a + 11b)(9a - 9b) = m^2$.$99(a^2 - b^2) = m^2$.$9 	imes 11(a^2 - b^2) = m^2$.For this to be a perfect square,$(a^2 - b^2)$ must be of the form $11k^2$. Since $a, b$ are digits,$a^2 - b^2$ can be at most $9^2 - 0^2 = 81$. Thus,$a^2 - b^2 = 11 	imes 1^2 = 11$.$(a-b)(a+b) = 11$. Since $11$ is prime,we must have $a-b = 1$ and $a+b = 11$.Adding these,$2a = 12 \Rightarrow a = 6$. Then $b = 5$.So,$x = 65$ and $y = 56$.$m^2 = 65^2 - 56^2 = (65-56)(65+56) = 9 	imes 121 = 1089 = 33^2$,so $m = 33$.The value of $x + y + m = 65 + 56 + 33 = 154$.

Let $x$ and $y$ be two $2$-digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

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