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

Question:

The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there?

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 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 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.