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

Hint: A 2-digit number is always of the form \[10a+b\] where a is ten’s digit and b is one’s digit. Using assumed values of x and y we will find the value of m.

Complete step by step answer:

We are given two 2-digit numbers x and y. Let \[x=10a+b......\left( i \right)\], where a is ten’s digit and b is one’s digit.
Now, we have been given that y is obtained by reversing the digits of x.
Therefore, we get \[y=10b+a......\left( ii \right)\], where b is ten’s digit and a is one’s digit.
Now, we have been given a relation that \[{{x}^{2}}-{{y}^{2}}={{m}^{2}}......\left( iii \right)\].
From (i) and (ii), we get the values of x and y. Putting them in equation (iii), we get,
\[{{\left( 10a+b \right)}^{2}}-{{\left( 10b+a \right)}^{2}}={{m}^{2}}......\left( iv \right)\]
We can use \[{{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \] to expand \[{{\left( 10a+b \right)}^{2}}\] and \[{{\left( 10b+a \right)}^{2}}\]. Therefore, we will get \[{{\left( 10a+b \right)}^{2}}=100{{a}^{2}}+{{b}^{2}}+20ab\] and \[{{\left( 10b+a \right)}^{2}}=100{{b}^{2}}+{{a}^{2}}+20ab\].
So, equation (iv) will become
\[{{m}^{2}}=\left( 100{{a}^{2}}+{{b}^{2}}+20ab \right)-\left( 100{{b}^{2}}+{{a}^{2}}+20ab \right)\]
\[\Rightarrow {{m}^{2}}=100{{a}^{2}}+{{b}^{2}}+20ab-100{{b}^{2}}-{{a}^{2}}-20ab\]
\[\Rightarrow {{m}^{2}}=99{{a}^{2}}-99{{b}^{2}}\]
\[\Rightarrow {{m}^{2}}=99\left( {{a}^{2}}-{{b}^{2}} \right)\]
\[\Rightarrow {{m}^{2}}=9\times 11\left( {{a}^{2}}-{{b}^{2}} \right)......\left( v \right)\]
For m to be a perfect square \[\left( {{a}^{2}}-{{b}^{2}} \right)\] should be at least a multiple of 11 and a perfect square which can also be 1 or greater than that. So, we can write equation (v) as \[{{m}^{2}}=9\times 11\left( a-b \right)\left( a+b \right)\].
Let us consider that \[\left( a+b \right)=11\]. Then, the cases for which \[\left( a+b \right)=11\] are
\[\begin{align}
  & a=1,b=10 \\
 & a=2.b=9 \\
 & a=3,b=8 \\
 & a=4,b=7 \\
 & a=5,b=6 \\
 & a=6,b=5 \\
 & a=7,b=4 \\
 & a=8,b=3 \\
 & a=9,b=2 \\
 & a=10,b=1 \\
\end{align}\]
The only possible case for which \[\left( a+b \right)=11\] and \[\left( a-b \right)\] is a perfect square is \[a=6,b=5\]. Therefore we can write that \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{6}^{2}}-{{5}^{2}} \right)\].
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)=\left( 36-25 \right)\]
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)=\left( 11 \right)......\left( vi \right)\]
Therefore, from equation (v) and (vi), we get
\[{{m}^{2}}=9\times 11\times 11\]
\[\Rightarrow {{m}^{2}}=1089\]
\[\Rightarrow m=\pm 33\]
We have given m is a positive integer, \[m=33\]. Now, by putting the value of a and b, that is, \[a=6,b=5\] in equation (i) and (ii), we get,
\[x=65\] and \[y=56\]
Therefore, \[x+y+m=65+56+33\]
\[\Rightarrow x+y+m=154\]
Hence, option (d) is correct.

Note: If we take y as \[y=10p+q\] the solution will become very long and lengthy. To avoid the complicacy in the solution we have written y in terms of a and b. A mistake that a student can make here is by considering $y=10a-b$ instead of $y=10b+a$.