Question

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

• A. 88
• B. 112
• C. 144
• D. 154

Answer 1

Answer: (D)

We have,
$x$ and $y$ be two-digit numbers.
Let $x=10a+b$, where $b$ is units place and $a$ is ten’s place.
$\therefore y=10b+a$
$x^2-y^2=m^2$
$\because (10a+b{)}^2-(10b+a{)}^2=m^2$
$\Rightarrow (10a+b+10b+a)$
$(10a+b-10b-a)=m^2$
$\Rightarrow 11(a+5)\cdot 9(a-b)=m^2$
$\Rightarrow 99(a^2-b^2)=m^2$
$\Rightarrow 3^{2}x11(a^2-b^2)=m^2$
Now, $3^2\times 11(a^2-b^2)$ is a perfect square.
$\because a^2-b^2=11$
$\Rightarrow (a+b)(a-b)=11\times 1$
$a+b=11$ and $a-b=11$
On solving, we get $a=6,b=5$
$\because$ Number $x=60+5=65$
$y=56$
and $m^2=(65{)}^2-(56{)}^2=(65+56)(65-56)$
$\Rightarrow m^2=121\times 9$
$\Rightarrow m^2=33$
$\because x+y+m=65+56+33=154$