Answer

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Hint: Consider the number as $xy$ . Try to understand the place values of the number. Apply the conditions carefully and try to make an equation.

Complete step-by-step answer:

Let us assume that the number is $xy$ .

In one place we have $y$ . In tens place we have $x$ . That means:

$xy=\left( 10\times x \right)+\left( 1\times y \right)=10x+y$

Now if we reverse the number, the number becomes $yx$ .

Now in ones place we have $x$ and in tens place we have $y$ . That means:

$yx=\left( 10\times y \right)+\left( 1\times x \right)=10y+x$

It is given in the question that if we add these two numbers we will get 66. So this is our first condition.

$\left( 10x+y \right)+\left( 10y+x \right)=66$

Now, add the terms with same variables:

$\begin{align}

& \Rightarrow \left( 10x+x \right)+\left( 10y+y \right)=66 \\

& \Rightarrow 11x+11y=66 \\

\end{align}$

Take 11 common from the left hand side,

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

Divide both sides by 11,

$\begin{align}

& \Rightarrow \dfrac{11\left( x+y \right)}{11}=\dfrac{66}{11} \\

& \Rightarrow x+y=6........(1) \\

\end{align}$

Now the second condition is the two digits differ by 2.

So either $x$ is greater than $y$ or $y$ is greater than $x$ . The difference between them is 2.

Let us first assume that $x$ is greater than $y$ . So we will get:

$\begin{align}

& x-y=2 \\

& \Rightarrow x=2+y........(2) \\

\end{align}$

Put the value of $x$ in equation (1)

$\begin{align}

& (y+2)+y=6 \\

& \Rightarrow (y+y)+2=6 \\

& \Rightarrow 2y+2=6 \\

& \Rightarrow 2y=6-2 \\

& \Rightarrow 2y=4 \\

& \Rightarrow y=\dfrac{4}{2} \\

& \Rightarrow y=2 \\

\end{align}$

Now put the value of $y$ in (2)

$x=2+2=4$

Hence, $x=4,y=2$

Therefore the number is 42.

If $y$ is greater than $x$ ,

$\begin{align}

& y-x=2 \\

& \Rightarrow y=2+x...........(3) \\

\end{align}$

Now put the value of $y$ in equation (1). We will get,

$\begin{align}

& x+(x+2)=6 \\

& \Rightarrow 2x+2=6 \\

& \Rightarrow 2x=6-2 \\

& \Rightarrow 2x=4 \\

& \Rightarrow x=\dfrac{4}{2} \\

& \Rightarrow x=2 \\

\end{align}$

Put the value of $x$ in equation (3) to get the value of $y$ .

$\begin{align}

& y=2+2 \\

& \Rightarrow y=4 \\

\end{align}$

Hence, $x=2,y=4$

Therefore the number is 24.

We can get two such numbers. One is 24 and another one is 42.

Note: We can assume the number either as $xy$ or as $yx$ . If we take the number as $yx$ , then $x$ will be our ones position and $y$ will be our tens position. We have to apply the conditions accordingly. In both the cases we will get the same answer.

Complete step-by-step answer:

Let us assume that the number is $xy$ .

In one place we have $y$ . In tens place we have $x$ . That means:

$xy=\left( 10\times x \right)+\left( 1\times y \right)=10x+y$

Now if we reverse the number, the number becomes $yx$ .

Now in ones place we have $x$ and in tens place we have $y$ . That means:

$yx=\left( 10\times y \right)+\left( 1\times x \right)=10y+x$

It is given in the question that if we add these two numbers we will get 66. So this is our first condition.

$\left( 10x+y \right)+\left( 10y+x \right)=66$

Now, add the terms with same variables:

$\begin{align}

& \Rightarrow \left( 10x+x \right)+\left( 10y+y \right)=66 \\

& \Rightarrow 11x+11y=66 \\

\end{align}$

Take 11 common from the left hand side,

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

Divide both sides by 11,

$\begin{align}

& \Rightarrow \dfrac{11\left( x+y \right)}{11}=\dfrac{66}{11} \\

& \Rightarrow x+y=6........(1) \\

\end{align}$

Now the second condition is the two digits differ by 2.

So either $x$ is greater than $y$ or $y$ is greater than $x$ . The difference between them is 2.

Let us first assume that $x$ is greater than $y$ . So we will get:

$\begin{align}

& x-y=2 \\

& \Rightarrow x=2+y........(2) \\

\end{align}$

Put the value of $x$ in equation (1)

$\begin{align}

& (y+2)+y=6 \\

& \Rightarrow (y+y)+2=6 \\

& \Rightarrow 2y+2=6 \\

& \Rightarrow 2y=6-2 \\

& \Rightarrow 2y=4 \\

& \Rightarrow y=\dfrac{4}{2} \\

& \Rightarrow y=2 \\

\end{align}$

Now put the value of $y$ in (2)

$x=2+2=4$

Hence, $x=4,y=2$

Therefore the number is 42.

If $y$ is greater than $x$ ,

$\begin{align}

& y-x=2 \\

& \Rightarrow y=2+x...........(3) \\

\end{align}$

Now put the value of $y$ in equation (1). We will get,

$\begin{align}

& x+(x+2)=6 \\

& \Rightarrow 2x+2=6 \\

& \Rightarrow 2x=6-2 \\

& \Rightarrow 2x=4 \\

& \Rightarrow x=\dfrac{4}{2} \\

& \Rightarrow x=2 \\

\end{align}$

Put the value of $x$ in equation (3) to get the value of $y$ .

$\begin{align}

& y=2+2 \\

& \Rightarrow y=4 \\

\end{align}$

Hence, $x=2,y=4$

Therefore the number is 24.

We can get two such numbers. One is 24 and another one is 42.

Note: We can assume the number either as $xy$ or as $yx$ . If we take the number as $yx$ , then $x$ will be our ones position and $y$ will be our tens position. We have to apply the conditions accordingly. In both the cases we will get the same answer.

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