Question

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

Answer 1

Hint:
We will formulate equations for values of the number using the formula for the value of a 2 digit number. Then, we will find the value of the number and then the value of the number obtained on reversing the digits. We will equate their sum to be 66 and find the 2-digit number.

Complete step by step solution:
We will assume that the digit at one’s place in the number is \[x\] . As the digits of the number differ by 2, we will assume that the digit at the ten’s place is \[x + 2\] .
The value of a 2 digit number (say \[ab\] ) is the sum of the product of the digit at ten’s place with 10 and the digit at one’s place with 1 is given by \[ab = 10a + b\].
Therefore, Original number \[ = 10\left( {x + 2} \right) + 1 \cdot x\]
We will simplify the above expression:
Original Number \[ = 10x + 20 + x = 11x + 20\]
If we reverse the digits of the number, \[x\] will be at ten’s place and \[x + 2\] will be at one’s place. The value of the number will be:
Reversed number \[ = 10 \cdot x + 1\left( {x + 2} \right)\]
We will simplify the above expression:
Reversed number \[ = 10x + x + 2 = 11x + 2\]
We know that the sum of the original number and the number when its digits are reversed is 66. We will formulate an equation for this and find the value of \[x\]:
Reversed number \[ + \] Original Number \[ = 66\]
Substituting the values in the equation, we get
\[ \Rightarrow 11x + 20 + 11x + 2 = 66\]
Adding the like terms, we get
\[ \Rightarrow 22x + 22 = 66\]
Subtracting the like terms, we get
\[ \Rightarrow 22x = 44\]
Dividing both sides by 22, we get
\[ \Rightarrow x = 2\]
We have calculated the value of \[x\] to be 2. The value of \[x + 2\] will be 4.

$\therefore $ The original number is 42 and the reversed number is 24.
There are 2 such numbers.

Note:
We can also find the solution by creating a linear equation in 2 variables. Here, we will take the digit at one’s place to be \[x\] and the digit at ten’s place to be \[y\] . Then, according to the formula, The value of first number \[ = 10y + x\]
And the value of the second number \[ = 10x + y\]
Now we know that the sum of both numbers is 66. Therefore,
\[\begin{array}{l} \Rightarrow 10x + y + 10y + x = 66\\ \Rightarrow 11x + 11y = 66\end{array}\]
Dividing both sides by 11, we get
\[ \Rightarrow x + y = 6\]………………………\[\left( 1 \right)\]
We also know that the difference between the 2 numbers is 2:
\[ \Rightarrow x - y = 2\]
\[ \Rightarrow x = y + 2\]……………………………..\[\left( 2 \right)\]
Substituting equation (2) in equation (1), we get
\[\begin{array}{l} \Rightarrow y + 2 + y = 6\\ \Rightarrow 2y = 4\end{array}\]
Dividing both sides by 2, we get
\[ \Rightarrow y = 2\]
So, the value of \[x\] will be 4. The required numbers are 42 and 24.