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

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Question

(d)

We have,

$x$ and $y$ be two-digit numbers.

Let $x=10 a+b$, where $b$ is units place and $a$ is ten's place.

$\because y=10 b +a$

Vedclass · Accepted Answer

$154$

Vedclass · Answer

(d)

We have,

$x$ and $y$ be two-digit numbers.

Let $x=10 a+b$, where $b$ is units place and $a$ is ten's place.

$\because y=10 b +a$

$x^2-y^2 =m^2$

$\because (10 a+b)^2-(10 b+a)^2 =m^2$

$\Rightarrow(10 a+b+10 b+a)$

$\Rightarrow \quad \begin{aligned}(10 a+b-10 b-a) &=m^2 \ \Rightarrow \quad 11(a+b) \cdot 9(a-b) &=m^2 \end{aligned}$

$\begin{aligned} \Rightarrow & & 11(a+b) \cdot 9(a-b) &=m^2 \ \Rightarrow & & 99\left(a^2-b^2ight) &=m^2 \end{aligned}$

$\Rightarrow \quad 3^2 	imes 11\left(a^2-b^2ight)=m^2$

Now, $3^2 	imes 11\left(a^2-b^2ight)$ is a perfect square.

$\because a^2-b^2=11 \Rightarrow(a+b)(a-b)=11 	imes 1$

$a+b=11$ and $a-b=11$

On solving, we get $a=6, b=5$

$\because$ Number $x=60+5=65$

$y=56$

and $m^2=(65)^2-(56)^2=(65+56)(65-56)$

$\Rightarrow m^2=121 	imes 9 \Rightarrow m=33$

$\because \quad x+y+m=65+56+33=154$

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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