Question

Sum of the digits of a 2-digit number is 11. The given number is less than the number obtained by interchanging the digits by 9. Find the number.

Answer 1

Hint: We use the variables to find the unit and tens placed digits. We form the number. We find the sum as 11. Then we interchange the digits to get the difference being 9, we form the equation and solve them to get the solution.

Complete step by step solution:
Let the unit and tens place digit of the 2-digit number are x and y respectively.
Sum of the digits of the 2-digit number is 11. So, $x+y=11$.
The number with digits x and y can be represented as $10y+x$.
Now we interchange the digits of the number and get x and y as the tens and unit placed digits respectively.
The new number with digits x and y can be represented as $10x+y$.
The given number is less than the number obtained by interchanging the digits by 9.
Therefore, $\left( 10x+y \right)-\left( 10y+x \right)=9$.
Simplifying we get
$
  \left( 10x+y \right)-\left( 10y+x \right)=9 \\
 \Rightarrow 9x-9y=9 \;
$
Now we divide the both sides of the equation with 9 and get
$
  \dfrac{9x-9y}{9}=\dfrac{9}{9} \\
 \Rightarrow x-y=1 \;
$
We got two equations of $x-y=1$ and $x+y=11$. We add them and get
\[
  \left( x-y \right)+\left( x+y \right)=1+11 \\
 \Rightarrow 2x=12 \\
 \Rightarrow x=6 \;
\]
Putting the value of x in $x+y=11$, we get $y=11-x=11-6=5$.
The number was 56.
So, the correct answer is “56”.

Note: We need to remember that we cannot write the number as simple sun of $xy$ as we have to use their place values. The place values for unit and tens placed digits are 1 and 10 respectively.