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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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Verified Answer
The correct answer is:
154
$X \rightarrow a b$ or $x=10 a+b$
$y \rightarrow b a$ or $y=10 b+a$
Now $x^{2}-y^{2}=(10 a+b)^{2}-(10 b+a)^{2}$
$$
\begin{array}{l}
=99\left(a^{2}-b^{2}\right) \\
=3^{2} \times 11(a+b)(a-b) \quad ......\text {(1)}
\end{array}
$$
According of $Q$
$(a+b)(a-b)=11$ and $a-b=1$
$\Rightarrow a+b=11$ and $a-b=1$
$\quad \Rightarrow a=6, b=5$
Hence
$x=65$
$y=56$
and $\quad m=33 \quad \Rightarrow x+y+m=154$
$y \rightarrow b a$ or $y=10 b+a$
Now $x^{2}-y^{2}=(10 a+b)^{2}-(10 b+a)^{2}$
$$
\begin{array}{l}
=99\left(a^{2}-b^{2}\right) \\
=3^{2} \times 11(a+b)(a-b) \quad ......\text {(1)}
\end{array}
$$
According of $Q$
$(a+b)(a-b)=11$ and $a-b=1$
$\Rightarrow a+b=11$ and $a-b=1$
$\quad \Rightarrow a=6, b=5$
Hence
$x=65$
$y=56$
and $\quad m=33 \quad \Rightarrow x+y+m=154$
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