A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Given :

A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed.

To find :

We have to find the number.

Solution :

Let the number be $10x+y$.

According to the question,

$10x+y=4(x+y)$

$10x+y=4x+4y$

$10x-4x=4y-y$

$6x=3y$

$2x=y$....(i)

It is also given that if 18 is added to the number the digits gets reversed.

Therefore,

$10x+y+18 = 10y+x$

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

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

$9y-x(10-1)=18$

$9y-9x=18$

$9(y-x)=18$

$y-x=\frac{18}{9}$

$y-x=2$

$2x-x=2$   (From (i))

$x=2$

This implies,

$y=2x=2(2)=4$

The original number is $10(2)+4=20+4=24$.

Therefore, the number is 24.