Question

Let x and y be two 2- digit number such that y is obtain 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

$X\to ab$ or $x=10a+b$
$y\to ba$ or $y=10b+a$
Now $x^2-y^2=(10a+b{)}^2-(10b+a{)}^2$
=$99(a^2-b^2)$
=$3^2\times 11(a+b)(a-b)$..........(1)
According of Q
$(a+b)(a-b)=11$ and $a-b=1$
$\Rightarrow a+b=11$ and $a-b=1$
$\Rightarrow a=6,b=5$
Hence
x=65
y=56 and m=33
$\Rightarrow x+y+m=154$