Question

A two digit number is $3$ more than $4$ times the sum of its digits. If $18$ is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.

Answer 1

Let the digit in one's place be $y$ and digit in ten's place be $x$.

Then the two digit number is $10x+y$.

Case1:

Two digit number is $3$ more than $4$ times the sum of its digits.
$\Rightarrow 10x+y=3+4(x+y)$

$\Rightarrow 10x+y=3+4x+4y$

$\Rightarrow 10x+y-4x-4y=3$

$\Rightarrow 6x-3y=3$

$\Rightarrow 3(2x-y)=3$

$\Rightarrow 2x-y=1$ ---(1)

Case2:

If $18$ is added to this number, the sum is equal to the number obtained by interchanging the digits.

$\Rightarrow 10x+y+18=10y+x$

$\Rightarrow 10x+y-10y-x=-18$

$\Rightarrow 9x-9y=-18$

$\Rightarrow 9(x-y)=-18$

$\Rightarrow x-y=-2$ ----(2)

Lets do (1)-(2)

$\Rightarrow 2x-y-(x-y)=1-(-2)$

$\Rightarrow 2x-y-x+y=1+2$

$\Rightarrow x=3$

Substitute, $x=3$ in $2x-y=1$

$\Rightarrow 2\left(3\right)-y=1$

$\Rightarrow 6-y=1$

$\Rightarrow 6-1=y$

$\Rightarrow 5=y$

Therefore, the number is $10x+y=10\left(3\right)+5=35$

Hence, the required number is $35$.