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$
$x^2 - y^2 = m^2$
$\because (10a + b)^2 - (10b + a)^2 = m^2$
$\Rightarrow (10a + b + 10b + a)$
$(10a + b - 10b - a) = m^2$
$\Rightarrow 11(a + 5) \cdot 9 (a - b) = m^2$
$\Rightarrow 99(a^2 - b^2) = m^2$
$\Rightarrow 3^2 x 11 (a^2 - b^2) = m^2$
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 $m^2 = (65)^2 - (56)^2 = (65+ 56) (65- 56)$
$\Rightarrow m^2 = 121 \times 9$
$\Rightarrow m^2 = 33$
$\because x + y + m = 65 + 56 + 33 = 154$