The sum of a two-digit number and the number formed by reversing the order of digit is 66.

Solution:

Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.

The two digits of the number are differing by 2 . Thus, we have $x-y=\pm 2$

After interchanging the digits, the number becomes $10 x+y$.

The sum of the numbers obtained by interchanging the digits and the original number is 66 . Thus, we have

$(10 x+y)+(10 y+x)=66$

$\Rightarrow 10 x+y+10 y+x=66$

$\Rightarrow 11 x+11 y=66$

$\Rightarrow 11(x+y)=66$

$\Rightarrow x+y=\frac{66}{11}$

$\Rightarrow x+y=6$

So, we have two systems of simultaneous equations

$x-y=2$

 

$x+y=6$

$x-y=-2$,

 

$x+y=6$

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

$x-y=2$

$x+y=6$

Adding the two equations, we have

$(x-y)+(x+y)=2+6$

$\Rightarrow x-y+x+y=8$

 

$\Rightarrow 2 x=8$

$\Rightarrow x=\frac{8}{2}$

$\Rightarrow x=4$

Substituting the value of x in the first equation, we have

$4-y=2$

$\Rightarrow y=4-2$

 

$\Rightarrow y=2$

Hence, the number is $10 \times 2+4=24$.

(ii) Now, we solve the system

$x-y=-2$,

$x+y=6$

Adding the two equations, we have

$(x-y)+(x+y)=-2+6$

$\Rightarrow x-y+x+y=4$

$\Rightarrow 2 x=4$

$\Rightarrow x=\frac{4}{2}$

$\Rightarrow x=2$

Substituting the value of x in the first equation, we have

$2-y=-2$

$\Rightarrow y=2+2$

$\Rightarrow y=4$

Hence, the number is $10 \times 4+2=42$.

There are two such numbers.