Question

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

Answer 1

Step 1: Let’s assume the digit at unit’s place as $x$ and ten’s place as $y$

Thus from the question, the number needed to be found is $10y+x$.

From the question it’s told as, the two digits of the number are differing by $2$.

Thus, we can write

$x–y=\pm 2\dots \dots \dots \dots ..\left(i\right)$

Now after reversing the order of the digits, the number becomes $10x+y$.

Again from the question it’s given that, the sum of the numbers obtained by reversing the digits and the original number is $66$.

Thus, this can be written as;

$\begin{array}{rcl}& \Rightarrow & (10x+y)+(10y+x)=66\end{array}$

$\begin{array}{rcl}& \Rightarrow & 10x+y+10y+x=66\end{array}$

$\begin{array}{rcl}& \Rightarrow & 11x+11y=66\end{array}$

$\begin{array}{rcl}& \Rightarrow & 11(x+y)=66\end{array}$

$\begin{array}{rcl}& \Rightarrow & x+y=\frac{66}{11}\end{array}$

$\begin{array}{rcl}& \Rightarrow & x+y=6\dots \dots \dots \dots ..\left(ii\right)\end{array}$

Now, we have two sets of systems of simultaneous equations

$x–y=2$ and $x+y=6$

$x–y=-2$ and $x+y=6$

Step 2: Let’s first solve the first set of system of equations

$$\begin{gathered} x–y=2\dots \dots \dots \dots .\left(iii\right) \\ x+y=6\dots \dots \dots \dots ..\left(iv\right) \end{gathered}$$

On adding the equations (iii) and (iv), we get;

$\begin{array}{rcl}& \Rightarrow & (x–y)+(x+y)=2+6\end{array}$

$\begin{array}{rcl}& \Rightarrow & x–y+x+y=8\end{array}$

$\begin{array}{rcl}& \Rightarrow & 2x=8\end{array}$

$\begin{array}{rcl}& \Rightarrow & x=\frac{8}{2}\end{array}$

$\begin{array}{rcl}& \Rightarrow & x=4\end{array}$

Putting the value of x in equation (iii), we get

$\begin{array}{rcl}4-y& =& 2\\ y& =& 4-2\\ y& =& 2\end{array}$

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

Step 3: Let’s first solve the first set of system of equations

$$\begin{gathered} x–y=-2\dots \dots \dots \dots .\left(v\right) \\ x+y=6\dots \dots \dots \dots ..\left(vi\right) \end{gathered}$$

On adding the equations (v) and (vi), we get

$\Rightarrow (x–y)+(x+y)=-2+6$

$\Rightarrow x–y+x+y=4$

$\Rightarrow 2x=4$

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

$\Rightarrow x=2$

Step 4: Putting the value of $x$ in equation $5$

We get,

$\Rightarrow 2–y=-2$

$\Rightarrow y=2+2$

$\Rightarrow y=4$

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

Therefore, there are two such possible numbers i.e, $24$ and $42$.