# 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 out the number. How many such numbers are there?

Last updated date: 18th Mar 2023

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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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