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

The correct option is D $154$

Let $x=10a+b$ and $y=10b+a$

$x^2-y^2=m^2$

$\therefore (10a+b{)}^2-(10b+a{)}^2=m^2$

$\therefore (10a+b+10b+a)(10a+b-10b-a)=m^2$

$\therefore 11(a+b)9(a-b)=m^2$

$\therefore 99(a+b)(a-b)=m^2$

Since a and b are digits from $1$ to $9,a+b=11$ and $a-b=1$ is the only possibility for m to be an integer.

$\therefore a=6$ and $b=5$

$\Rightarrow x=65,y=56,m=33$

$\therefore x+y+m=65+56+33=154$