Question

The sum of 2-digit number and the number obtained by reversing the order of the digit is 165. If the digits differ by 3, find the number, when the tens digit is bigger than the unit’s digit.

Answer 1

Hint: We first assume the unit and tens place digits. We reverse the order of the digits. We add these numbers to get 165. Also, it is given that the digits differ by 3. We form the algebraic equations and solve them to find the original number.

Complete step by step answer:
Let the unit and tens place digits of the 2-digit number are $y,x$ respectively. So, the number is $10x+y$. Now if we reverse the order of the digits, the unit and tens place digits of the 2-digit number becomes $x,y$ respectively. The number becomes $10y+x$. The sum of those numbers is 165. This gives
\[\left( 10x+y \right)+\left( 10x+y \right)=165 \\
\Rightarrow 11x+11y=165 \\ \]
Dividing both sides with 11 we get $x+y=\dfrac{165}{11}=15......(i)$.
We are also given that the digits differ by 3 and the tens digit is bigger than the unit’s digit for the original number. So, $x-y=3.......(ii)$
We add the equations to get
$\left( x+y \right)+\left( x-y \right)=15+3 \\
\Rightarrow 2x=18 \\
\Rightarrow x=\dfrac{18}{2}=9 $
Putting the value, we get $y=15-x=15-9=6$.

Hence, the number is 96.

Note:Instead of taking two variables we also could have taken one variable and got the other one using the condition of difference of the digits being 3. So, we take one variable to be $x$ as the tens place digit. We get the unit’s place digit as $x-3$.