Let $x & y$ 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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112

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144

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154

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Solution

The correct option is D
154

$x \to a b$ or $x = 10 a + b$

$y \to b a$ or $y = 10 b + a$

Now $x^{2} - y^{2} = (10 a + b)^{2} - (10 b + a)^{2}$

$= 99 (a^{2} - b^{2})$

$= 3^{2} \times 11 (a + b) (a - b)$ ------ (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: $x = 65; y = 56 & m = 33$

$\Rightarrow x + y + m = 154$

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Let x & y be two 2-digit numbers such that y is obtained by reversing the digits of x. Suppose they also satisfy x2–y2=m2 for some positive integer m. The value of x+y+m is.

The correct option is D 154 x→ab or x=10a+b y→ba or y=10b+a Now x2−y2=(10a+b)2−(10b+a)2 =99(a2−b2) =32×11(a+b)(a−b) ------ (1) According of Q (a+b)(a−b)=11 and a−b=1 ⇒a+b=11 and a−b=1 ⇒a=6,b=5 Hence: x=65; y=56 & m=33 ⇒x+y+m=154

Let $x & y$ 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.